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#### Page No 156:

#### Question 1:

Can the centre of mass of a body be at a point outside the body?

#### Answer:

Yes, the centre of mass can be at a point outside the body.

For example, the centre of mass of a ring lies at its centre, which is not a part of the ring.

#### Page No 156:

#### Question 2:

If all the particles of a system lie in *X-Y* plane, is it necessary that the centre of mass be in *X-Y *plane?

#### Answer:

Yes, if all the particles of a system lie in the *X–Y* plane, then it's necessary that its centre of mass lies in the *X–Y* plane.

${z}_{\mathit{cm}}=\frac{{m}_{1}{z}_{1}+{m}_{2}{z}_{2}+...{m}_{n}{z}_{n}}{{\displaystyle \sum _{}}m}$

As all the particles lie in the *X–Y* plane, their *z-*coordinates are zero.

Therefore, for the whole system, *z*_{cm} = 0; i.e., its centre of mass lies in the *X–Y* plane.

#### Page No 156:

#### Question 3:

If all the particle of a system lie in a cube, is it necessary that the centre of mass be in the cube?

#### Answer:

Yes. As a cube is a 3-dimensional body, all the particles of a system lying in a cube lie in the *x,y* and *z* plane.

Let the *i*^{th} element of mass ∆*m*_{i}_{ }is located at the point (*x*_{i},*y*_{i},*z*_{i}).

The co-ordinates of the centre of mass are given as:

$X=\frac{1}{M}{\sum}_{i=1}^{i=n}\left(\u2206{m}_{i}\right){x}_{i}\phantom{\rule{0ex}{0ex}}Y=\frac{1}{M}{\sum}_{i=1}^{i=n}\left(\u2206{m}_{i}\right){y}_{i}\phantom{\rule{0ex}{0ex}}Z=\frac{1}{M}{\sum}_{i=1}^{i=n}\left(\u2206{m}_{i}\right){z}_{i}$

*X, Y* and *Z* lie inside the cube because it is a weighted mean.

#### Page No 156:

#### Question 4:

The centre of mass is defined as $\overrightarrow{R}=\frac{1}{M}\sum _{i}{m}_{i}\overrightarrow{{r}_{i}}$. Suppose we define "centre of charge" as ${\overrightarrow{R}}_{c}=\frac{1}{Q}\sum _{i}{q}_{i}\overrightarrow{{r}_{i}}$ where *q*_{i} represents the *i*th charge placed at ${\overrightarrow{r}}_{i}$ and *Q* is the total charge of the system.

(a) Can the centre of charge of a two-charge system be outside the line segment joining the charges?

(b) If all the charges of a system are in *X-Y* plane, is it necessary that the centre of charge be in *X-Y* plane?

(c) If all the charges of a system lie in a cube, is it necessary that the centre of charge be in the cube?

#### Answer:

(a) Yes

Consider a charge distributed in *X-Y *plane.

${X}_{cm}=\frac{-6q\times 0+q\times 5a}{-6q+q}=-a$

(b) Yes. Because the *z-*coordinates of all the charges are zero, the centre of charge lies in X-Y plane.

(c) No, it is not necessary that the centre of charge lies in the cube because charge can be either negative or positive.

#### Page No 156:

#### Question 5:

The weight *Mg* of an extended body is generally shown in a diagram to act through the centre of mass. Does it mean that the earth does not attract other particles?

#### Answer:

In order to simplify the situation, we consider that the weight *M*g of an extended body acts through its centre of mass.

Although the earth attracts all the particles, the net effect can be assumed to be at the centre of mass.

#### Page No 156:

#### Question 6:

A bob suspended from the ceiling of a car which is accelerating on a horizontal road. The bob stays at rest with respect to the car with the string making an angle θ with the vertical. The linear momentum of the bob as seen from the road is increasing with time. Is it a violation of conservation of linear momentum? If not, where is the external force changes the linear momentum?

#### Answer:

There is no violation of conservation of momentum because in the earth's frame the component of tension is acting in the horizontal direction.

#### Page No 156:

#### Question 7:

You are waiting for a train on a railway platform. Your three-year-old niece is standing on your iron trunk containing the luggage. Why does the trunk not recoil as she jumps off on the platform?

#### Answer:

The trunk does not recoil as the girl jumps off on the platform because the force exerted by the girl is less than the limiting friction between the platform and the iron trunk.

#### Page No 156:

#### Question 8:

In a head-on collision between two particles, is it necessary that the particles will acquire a common velocity at least for one instant?

#### Answer:

Yes.

For example, consider particle-1 at a velocity of 4 ms^{-1} and particle-2 at a velocity of 2 ms^{-1} undergo a head-on collision.

The velocity of particle-1 decreases but particle-2 increases. Therefore, at an instant, their velocities will be equal.

#### Page No 156:

#### Question 9:

A collision experiment is done on a horizontal table kept in an elevator. Do you expect a change in the result if the elevator is accelerated up or down because of the noninertial character of the frame?

#### Answer:

No. As the collision experiment is being done on a horizontal table in the elevator that is accelerating up or down in vertical direction, no extra force is experienced in the horizontal direction. Hence, the objects in the horizontal direction remain unaffected.

#### Page No 156:

#### Question 10:

Two bodies make an elastic head-on collision on a smooth horizontal table kept in a car. Do you expect a change in the result if the car is accelerated in a horizontal road because of the non inertial character of the frame? Does the equation "Velocity of separation = Velocity of approach" remain valid in an accelerating car? Does the equation "final momentum = initial momentum" remain valid in the accelerating car?

#### Answer:

No, due to the non-inertial character of the frame and the presence of a pseudo force, both the equations, i.e., Velocity of separation = Velocity of approach and Final momentum = Initial momentum, do not remain valid in the accelerating car.

#### Page No 156:

#### Question 11:

If the total mechanical energy of a particle is zero, is its linear momentum necessarily zero? Is it necessarily nonzero?

#### Answer:

No. As the potential energy can have a negative value, the total energy of the system may sum up to zero.

For example:

Two masses *A* and *B *having masses 2 kg and 4 kg respectively move with a velocity of 4 ms^{-1} in opposite directions.

Kinetic energy of system (*A* and *B*)

$=\frac{1}{2}\times 2\times {4}^{2}+\frac{1}{2}\times 4\times {4}^{2}\phantom{\rule{0ex}{0ex}}=48\mathrm{J}$

If the gravitational potential energy of the system is −48 J, the total energy of the system will be zero. However, the linear momentum will be non-zero.

#### Page No 156:

#### Question 12:

If the linear momentum of a particle is known, can you find its kinetic energy? If the kinetic energy of a particle is know can you find its linear momentum?

#### Answer:

Yes, the kinetic energy of the particle can be determined if the value of linear momentum is known.

The kinetic energy is calculated using the formula:

$\mathrm{K}.\mathrm{E}=\frac{1}{2}m{v}^{2}=\frac{{p}^{2}}{2m}\phantom{\rule{0ex}{0ex}}w\mathrm{here},p\mathit{}\mathrm{is}\mathrm{the}\mathrm{linear}\mathrm{momemtum}\mathrm{having}\mathrm{value}mv.$

But linear momentum cannot be determined even if the kinetic energy is known because linear momentum is a vector quantity, whereas kinetic energy is a scalar quantity. Thus, the direction of the linear momentum remains unknown, however its magnitude can be calculated.

#### Page No 156:

#### Question 13:

What can be said about the centre of mass of a uniform hemisphere without making any calculation? Will its distance from the centre be more than *r*/2 or less than *r*/2?

#### Answer:

The distance of centre of mass of a uniform hemisphere from its centre will be less than *r*/2 because the portion of the hemisphere lying below *r*/2 from the diameter is heavier than the portion lying above *r*/2.

#### Page No 156:

#### Question 14:

You are holding a cage containing a bird. Do you have to make less effort if the bird flies from its position in the cage and manages to stay in the middle without touching the walls of the cage? Does it makes a difference whether the cage is completely closed or it has rods to let air pass?

#### Answer:

More effort is needed when the cage is closed, while less effort is required when the cage has rods to let the air pass. When a bird flies from its position, it pushes the air downwards. Thus, when the bird is in a cage, the net downward force will be equal to the weight of the cage plus the downward force due to air (the weight of the bird).

However, if the cage has rods to let air pass, the downward force exerted by air become less. Therefore, less effort will be required to hold the cage.

#### Page No 157:

#### Question 15:

A fat person is standing on a light plank floating on a calm lake. The person walks from one end to the other on the plank. His friend sitting on the shore watches him and finds that the person hardly moves any distance because the plank moves backward about the same distance as the person moves on the plank. Explain.

#### Answer:

According to the question, the weight of plank is very less as compared to the fat person. Therefore, the centre of mass of the whole system effectively lies on the person. As the net external force on the system is zero, the centre of mass of the system does not move.

#### Page No 157:

#### Question 16:

A high-jumper successfully clears the bar. Is it possible that his centre of mass crossed the bar from below it? Try it with appropriate figures.

#### Answer:

From the figure, it can be seen that when a high-jumper successfully clears the bar, it is possible that her centre of mass crosses the bar from below it because the legs as well as the arms of the high-jumper are below the bar.

Hence, the point shown in the figure can be her centre of mass.

#### Page No 157:

#### Question 17:

Which of the two persons shown in figure is more likely to fall down? Which external force is responsible for his falling down?

#### Answer:

The person shown on the right hand side of the figure is more likely to fall down because in the given cart frame the pseudo force will be in backward direction.

#### Page No 157:

#### Question 18:

Suppose we define a quantity 'Linear momentum' as linear momentum = mass × speed.

The linear momentum of a system of particles is the sum of linear momenta of the individual particles. Can we state principle of conservation of linear momentum as "linear momentum of a system remains constant if no external force acts on it"?

#### Answer:

It is not necessary that the linear momentum of a system remains constant even if no external force acts on it because during collision, the sum of magnitudes of momenta does not remain constant.

#### Page No 157:

#### Question 19:

Use the definition of linear momentum from the previous question. Can we state the principle of conservation of linear momentum for a single particle?

#### Answer:

Yes, if the external force applied on the particle is zero, its speed does not change and hence, the momentum remains constant.

#### Page No 157:

#### Question 20:

To accelerate a car we ignite petrol in the engine of the car. Since only an external force can accelerate the centre of mass, is it proper to say that "the force generated by the engine accelerates the car"?

#### Answer:

Yes, it's proper to say that the force generated by the engine accelerates the car. When petrol burns inside the engine, the piston moves, which in turn rotates the wheel. As the wheel rotates, the frictional forces from the road moves the car.

#### Page No 157:

#### Question 21:

A ball is moved on a horizontal table with some velocity. The ball stops after moving some distance. Which external force is responsible for the change in the momentum of the ball?

#### Answer:

The frictional force acting between the surface of the table and the ball is responsible for the change in momentum of the ball. As the force opposes the motion of the ball, it stops after moving some distance.

#### Page No 157:

#### Question 22:

Consider the situation of the previous problem. Take "the table plus the ball" as the system. friction between the table and the ball is then an internal force. As the ball slows down, the momentum of the system decreases. Which external force is responsible for this change in the momentum?

#### Answer:

Considering the table plus the ball as a system, it can be said that the frictional force is responsible for the change in the momentum. As the force acts between the surface of the table and ground, it opposes the motion of the table plus the ball. Hence, the ball slows down and the momentum of the system decreases.

#### Page No 157:

#### Question 23:

When a nucleus at rest emits a beta particle, it is found that the velocities of the recoiling nucleus and the beta particle are not along the same straight line. How can this be possible in view of the principle of conservation of momentum?

#### Answer:

In view of the principle of conservation of momentum, the given situation is possible because as a beta particle is ejected, another particle called an antineutrino is also ejected.

#### Page No 157:

#### Question 24:

A van is standing on a frictionless portion of a horizontal road. To start the engine, the vehicle must be set in motion in the forward direction. How can be persons sitting inside the van do it without coming out and pushing from behind?

#### Answer:

According to the question, the van is standing on a frictionless surface. When throwing something in backward direction, the persons sitting inside the van sets the van in motion in the forward direction according to the principle of conservation of linear momentum.

#### Page No 157:

#### Question 25:

In one-dimensional elastic collision of equal masses, the velocities are interchanged. Can velocities in a one-dimensional collision be interchanged if the masses are not equal?

#### Answer:

No. If the masses are different, the velocities in a one-dimensional collision cannot be interchanged because that would be violation of the principle of conservation of momentum.

#### Page No 157:

#### Question 1:

Consider the following the equations:

(A) $\overrightarrow{R}=\frac{1}{M}\sum _{i}{m}_{i}\overrightarrow{{r}_{i}}$

and

(B) ${\overrightarrow{a}}_{CM}=\frac{\overrightarrow{F}}{M}$

In a noninertial frame

(a) both are correct

(b) both are wrong

(c) *A* is correct but *B* is wrong

(d) *B* is correct but *A* is wrong.

#### Answer:

(c) *A* is correct but *B* is wrong

In a non-inertial frame, the position of centre of mass of the particle does not change but an additional pseudo force acts on it.

#### Page No 157:

#### Question 2:

Consider the following two statements:

(A) Linear momentum of the system remains constant.

(B) Centre of mass of the system remains at rest.

(a) *A* implies *B* and *B* implies *A.*

(b) *A* does not imply *B* and *B* does not imply *A*.

(c) *A* implies *B* but *B* does not imply *A.*

(d) *B* implies* A* but *A* does not imply *B*.

#### Answer:

(d) *B* implies* A* but *A* does not imply *B*.

The centre of mass of a system is given by,

$\overrightarrow{R}=\frac{1}{M}\sum _{}{m}_{i}{\overrightarrow{r}}_{i}$

On differentiating the above equation with respect to time, we get:

$\frac{\mathrm{d}\overrightarrow{R}}{\mathrm{d}t}=\frac{1}{M}\sum _{}{m}_{i}\frac{d{\overrightarrow{r}}_{i}}{dt}$

As the centre of mass of the system remains at rest, we have:

$\frac{1}{M}\sum _{}{m}_{i}\frac{d{\overrightarrow{r}}_{i}}{dt}=0\phantom{\rule{0ex}{0ex}}\sum _{}{m}_{i}{\overrightarrow{v}}_{i}=0$

This implies that the linear momentum of the system remains constant.

#### Page No 157:

#### Question 3:

Consider the following two statements:

(A) Linear momentum of a system of particles is zero.

(B) Kinetic energy of a system of particles is zero.

(a) *A* implies *B* and *B* implies *A*.

(b) *A* does not imply *B* and *B* does not imply *A*.

(c) *A* implies *B* but *B* does not imply *A*.

(d) *B* implies *A* but *A* does not imply *B*.

#### Answer:

(d) *B* implies *A* but *A* does not imply *B*.

If the linear momentum of a system is zero,

$\Rightarrow {m}_{1}{\overrightarrow{v}}_{1}+{m}_{2}{\overrightarrow{v}}_{2}+...$=0

Thus, for a system of comprising two particles of same masses,

${\overrightarrow{v}}_{1}=-{\overrightarrow{v}}_{2}$ ...(1)

The kinetic energy of the system is given by,

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}m{\overrightarrow{v}}_{1}^{2}+\frac{1}{2}m{\overrightarrow{v}}_{2}^{2}$

Using equation (1) to solve above equation, we can say:

$\mathrm{K}.\mathrm{E}.\ne 0$

*i.e.* *A* does not imply *B.*

Now,

If the kinetic energy of the system is zero,

$\Rightarrow \frac{1}{2}m{\overrightarrow{v}}_{1}^{2}+\frac{1}{2}m{\overrightarrow{v}}_{2}^{2}=0$

${v}_{1}=\pm {v}_{2}$

On calculating the linear momentum of the system, we get:

$\overrightarrow{P}=m{\overrightarrow{v}}_{1}+m{\overrightarrow{v}}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{taking}{v}_{1}=-{v}_{2},\mathrm{we}\mathrm{can}\mathrm{write}:\phantom{\rule{0ex}{0ex}}\overrightarrow{P}=0$

Hence, we can say, *B* implies *A* but *A* does not imply *B*.

#### Page No 157:

#### Question 4:

Consider the following two statements:

(A) The linear momentum of a particle is independent of the frame of reference.

(B) The kinetic energy of a particle is independent of the frame of reference.

(a) Both *A* and B are true.

(b) *A* is true but *B* is false.

(c) *A* is false but *B* is true.

(d) both *A* and *B* are false.

#### Answer:

(d) both *A* and *B* are false.

As the velocity of the particle depends on the frame of reference, the linear momentum as well as the kinetic energy is dependent on the frame of reference.

#### Page No 157:

#### Question 5:

All the particles of a body are situated at a distance *R* from the origin. The distance of the centre of mass of the body from the origin is

(a) = *R*

(b) ≤ *R*

(c) > *R*

(d) ≥ *R*

#### Answer:

(b) ≤ *R*

Distance of the centre of mass from the origin is given by,

$R\text{'}=\frac{1}{M}{\sum}_{i=1}^{n}{m}_{i}{\overrightarrow{r}}_{i}$

Let half of the particles lie on +Y-axis and the rest of the particles lie on +X-axis.

${Y}_{\mathrm{CM}}=\frac{{m}_{3}R{\displaystyle \overrightarrow{j}}+{m}_{4}R{\displaystyle \overrightarrow{j}}}{{m}_{1}+{m}_{2}+{m}_{3}+{m}_{4}+...}\phantom{\rule{0ex}{0ex}}=\frac{R{\displaystyle \overrightarrow{j}}}{2}$

Similarly,* ${X}_{\mathrm{CM}}=\frac{R\overrightarrow{i}}{2}$*

Therefore, the coordinates of centre of mass are $\left(\frac{R}{2},\frac{R}{2}\right)$.

Distance of the centre of mass from the origin$=\frac{R}{\sqrt{2}}$

For the given situation, $R\text{'}<R$.

In general, *R'* ≤ *R*.

#### Page No 157:

#### Question 6:

A circular plate of diameter *d* is kept in contact with a square plate of edge *d* as show in figure. The density of the material and the thickness are same everywhere. The centre of mass of the composite system will be

(a) inside the circular plate

(b) inside the square plate

(c) at the point of contact

(d) outside the system.

#### Answer:

(b) inside the square plate

Let *m*_{1} be the mass of circular plate and *m*_{2} be the mass of square plate.

The thickness of both the plates is *t*.

$\mathrm{mass}=\mathrm{densit}y\times \mathrm{volume}\phantom{\rule{0ex}{0ex}}{m}_{1}=\rho \pi {\left(\frac{d}{2}\right)}^{2}t\phantom{\rule{0ex}{0ex}}{m}_{2}=\rho {d}^{2}t$

Centre of mass of the circular plate lies at its centre.

Let the centre of circular plate be the origin.

${\overrightarrow{r}}_{1}=0$

Centre of mass of the square plate lies at its centre.

${\overrightarrow{r}}_{2}=2d\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}R=\frac{{m}_{1}{{\displaystyle \overrightarrow{r}}}_{1}+{m}_{2}{{\displaystyle \overrightarrow{r}}}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{{m}_{1}\times 0+\rho {\mathit{d}}^{2}t\times 2d}{\rho \pi {\left({\displaystyle \frac{d}{2}}\right)}^{2}t+\rho {d}^{2}t}\phantom{\rule{0ex}{0ex}}=\frac{2d}{{\displaystyle \frac{\pi}{4}}+1}=1.12d\phantom{\rule{0ex}{0ex}}\Rightarrow Rd\phantom{\rule{0ex}{0ex}}$

$\therefore $ Centre of mass of the system lies in the square plate.

#### Page No 158:

#### Question 7:

Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration *a*. The centre of mass has an acceleration

(a) zero

(b) $\frac{1}{2}\overrightarrow{a}$

(c) $\overrightarrow{a}$

(d) $2\overrightarrow{a}$

#### Answer:

(b) $\frac{1}{2}\overrightarrow{a}$

Acceleration of centre of mass of a two-particle system is given as,

${\overrightarrow{a}}_{cm}=\frac{{m}_{1}{{\displaystyle \overrightarrow{a}}}_{1}+{m}_{2}{\displaystyle \overrightarrow{{a}_{2}}}}{{m}_{1}+{m}_{2}}$ ...(1)

According to the question,

${m}_{1}={m}_{2}=m\phantom{\rule{0ex}{0ex}}{a}_{1}=0\phantom{\rule{0ex}{0ex}}{a}_{2}=a$

Substituting these values in equation (1), we get:

${\overrightarrow{a}}_{cm}=\frac{m\times 0+m{\displaystyle \overrightarrow{a}}}{2m}=\frac{1}{2}\overrightarrow{a}$

#### Page No 158:

#### Question 8:

Internal forces can change

(a) the linear momentum but not the kinetic energy

(b) the kinetic energy but not the linear momentum

(c) linear momentum as well as kinetic energy

(d) neither the linear momentum nor the kinetic energy

#### Answer:

(b) the kinetic energy but not the linear momentum

Internal forces can not change the position of centre of mass of a system. Therefore, linear momentum of the system is constant, whereas kinetic energy of the system is not.

#### Page No 158:

#### Question 9:

A bullet hits a block kept at rest on a smooth horizontal surface and gets embedded into it. Which of the following does not change?

(a) linear momentum of the block

(b) kinetic energy of the block

(c) gravitational potential energy of the block

(d) temperature of the block.

#### Answer:

(c) gravitational potential energy of the block

When the block kept at rest is hit by a bullet, the block acquires certain velocity by the conservation of linear momentum.

Therefore, the linear momentum and the kinetic energy of the block change.

As some of the kinetic energy carried by the bullet transforms into heat energy, its temperature also changes.

However, the gravitational potential energy of the block does not change, as the height of the block does not change in this process.

#### Page No 158:

#### Question 10:

A uniform sphere is placed on a smooth horizontal surface and a horizontal force *F* is applied on it at a distance *h* above the surface. The acceleration of the centre

(a) is maximum when *h* = 0

(b) is maximum when *h* = *R*

(c) is maximum when *h* = 2*R*

(d) is independent of *h*.

#### Answer:

(d) is independent of *h*.

As the uniform sphere is placed on a smooth surface, the sphere only slips.

The acceleration of the centre of the sphere is given by,

$\frac{\mathrm{Applied}\mathrm{force}}{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{sphere}}$ ,which is independent of height *h.*

#### Page No 158:

#### Question 11:

A body falling vertically downwards under gravity breaks in two parts of unequal masses. The centre of mass of the two parts taken together shifts horizontally towards

(a) heavier piece

(b) lighter piece

(c) does not shift horizontally

(d) depends on the vertical velocity at the time of breaking.

#### Answer:

(c) does not shift horizontally

As the body falls vertically downwards, no external force acts in the horizontal direction.

Hence, the centre of mass does not shift horizontally.

#### Page No 158:

#### Question 12:

A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The velocity of the centre of mass

(a) of the box remains constant

(b) of the box plus the ball system remains constant

(c) of the ball remains constant

(d) of the ball relative to the box remains constant.

#### Answer:

(b) of the box plus the ball system remains constant

Consider the box and the ball a system. As no external force acts on this system, the velocity of the centre of mass of the system remains constant.

#### Page No 158:

#### Question 13:

A body at rest breaks into two pieces of equal masses. The parts will move

(a) in same direction

(b) along different lines

(c) in opposite directions with equal speeds

(d) in opposite directions with unequal speeds.

#### Answer:

(c) in opposite directions with equal speeds

Using the principle of conservation of linear momentum, we can write:

Initial momenta = Final momenta

$\Rightarrow M\times 0=\frac{M}{2}{v}_{1}+\frac{M}{2}{v}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}=-{v}_{2}\phantom{\rule{0ex}{0ex}}$

where *M *is the initial mass of the body, at rest.

The final mass of the two pieces moving with equal speeds in opposite direction is equal to $\frac{M}{2}$.

#### Page No 158:

#### Question 14:

A heavy ring of mass *m* is clamped on the periphery of al light circular disc. A small particle having equal mass is clamped at the centre of the disc. The system is rotated in such a way that the centre moves in a circle of radius *r* with a uniform speed *v*. We conclude that an external force

(a) $\frac{m{v}^{2}}{r}$ must be acting on the central particle

(b) $\frac{2m{v}^{2}}{r}$ must be acting on the central particle

(c) $\frac{2m{v}^{2}}{r}$ must be acting on the system

(d) $\frac{2m{v}^{2}}{r}$ must be acting on the ring.

#### Answer:

(c) $\frac{2m{v}^{2}}{r}$ must be acting on the system

Total mass of the system = 2*m*

To move the centre of the system in a circle of radius *r* with a uniform speed *v, *the external force required is $\frac{2m{v}^{2}}{R}$.

#### Page No 158:

#### Question 15:

The quantities remaining constant in a collisions are

(a) momentum, kinetic energy and temperature

(b) momentum and kinetic energy but not temperature

(c) momentum and temperature but not kinetic energy

(d) momentum, but neither kinetic energy nor temperature.

#### Answer:

(d) momentum, but neither kinetic energy nor temperature

Linear momentum of a system remains constant in a collision. However, the kinetic energy and temperature of the system may vary, as their values depend on the type of collision.

#### Page No 158:

#### Question 16:

A nucleus moving with a velocity $\overrightarrow{v}$ emits an α-particle. Let the velocities of the α-particle and the remaining nucleus be *v*_{1} and *v*_{2} and their masses be *m*_{1} and *m*_{2}.

(a) $\overrightarrow{v},{\overrightarrow{v}}_{1}\mathrm{and}{\overrightarrow{v}}_{2}$ must be parallel to each other.

(b) None of the two of $\overrightarrow{v},{\overrightarrow{v}}_{1}\mathrm{and}{\overrightarrow{v}}_{2}$ should be parallel to each other.

(c) $\overrightarrow{{v}_{1}}+\overrightarrow{{v}_{2}}$ must be parallel to $\overrightarrow{v}$

(d) ${m}_{1}\overrightarrow{{v}_{1}}+{m}_{2}\overrightarrow{{v}_{2}}$ must be parallel to $\overrightarrow{v}$

#### Answer:

(d) ${m}_{1}\overrightarrow{{v}_{1}}+{m}_{2}\overrightarrow{{v}_{2}}$ must be parallel to $\overrightarrow{v}$

By the law of conservation of linear momentum, we can write:

$\mathrm{Initial}\mathrm{momentum}=\mathrm{Final}\mathrm{momentum}\phantom{\rule{0ex}{0ex}}\Rightarrow m\overrightarrow{v}={m}_{1}{\overrightarrow{v}}_{1}+{m}_{2}{\overrightarrow{v}}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow ({m}_{1}{\overrightarrow{v}}_{1}+{m}_{2}{\overrightarrow{v}}_{2})\mathrm{must}\mathrm{be}\mathrm{parallel}\mathrm{to}\overrightarrow{v}$

#### Page No 158:

#### Question 17:

A shell is fired from a cannon with a velocity *V* at an angle θ with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediately after the explosion is

(a) 3*V* cos θ

(b) 2*V* cos θ

(c) $\frac{3}{2}$ *V* cos θ

(d) *V* cos θ

#### Answer:

(a) 3*V* cos θ

The linear momentum is conserved in horizontal direction.

$\therefore $ Initial momentum = Final momentum

$\Rightarrow mv\mathrm{cos}\theta =-\frac{m}{2}v\mathrm{cos}\theta +\frac{m}{2}v\text{'}\phantom{\rule{0ex}{0ex}}\Rightarrow v\text{'}=mv\mathrm{cos}\theta $

#### Page No 158:

#### Question 18:

In an elastic collision

(a) the initial kinetic energy is equal to the final kinetic energy

(b) the final kinetic energy is less than the initial kinetic energy

(c) the kinetic energy remains constant

(d) the kinetic energy first increases then decreases.

#### Answer:

(a) the initial kinetic energy is equal to the final kinetic energy

As no energy is lost into heat in an elastic collision, the initial kinetic energy is equal to the final kinetic energy.

#### Page No 158:

#### Question 19:

In an inelastic collision

(a) the initial kinetic energy is equal to the final kinetic energy

(b) the final kinetic energy is less than the initial kinetic energy

(c) the kinetic energy remains constant

(d) the kinetic energy first increases then decreases.

#### Answer:

(b) the final kinetic energy is less than the initial kinetic energy

As some energy is loss into heat in an inelastic collision, the final kinetic energy is less than the initial kinetic energy.

#### Page No 158:

#### Question 1:

The centre of mass of a system of particles is at the origin. It follows that

(a) the number of particles to the right of the origin is equal to the number of particles to the left

(b) the total mass of the particles to the right of the origin is same as the total mass to the left of the origin

(c) the number of particles on *X*-axis should be equal to the number of particles on *Y*-axis

(d) if there is a particle on the positive *X*-axis, there must be at least one particle on the negative *X*-axis

#### Answer:

None.

The centre of mass of a system of particles depends on the product of individual masses and their distances from the origin.

Therefore, we may say about the given statements:

(a) Distance of particles from origin is not known.

(b) Masses are same but the distance of particles from the origin is not given.

(c) Distance of particles from origin is not given.

(d) It is not necessary that least one particle lies on the negative *X*-axis. The particles can be above the negative X-axis on X-Y plane.

#### Page No 159:

#### Question 1:

Three particles of masses 1.0 kg, 2.0 kg and 3.0 kg are placed at the corners *A*, *B* and *C* respectively of an equilateral triangle *ABC* of edge 1 m. Locate the centre of mass of the system.

#### Answer:

Taking BC as the X-axis and point B as the origin, the positions of masses *m*_{1}_{ }= 1 kg, *m*_{2} = 2 kg and *m*_{3} = 3 kg are

$\left(0,0\right),\left(1,0\right)\mathrm{and}\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ respectively.

The

${X}_{\mathrm{cm}},{Y}_{\mathrm{cm}}=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}+{m}_{3}{x}_{3}}{{m}_{1}+{m}_{2}+{m}_{3}},\frac{{m}_{1}{y}_{1}+{m}_{2}{y}_{2}+{m}_{3}{y}_{3}}{{m}_{1}+{m}_{2}+{m}_{3}}$

$=\frac{1\times 0+2\times 1+3\times {\displaystyle \frac{1}{2}}}{1+2+3},\frac{1\times 0+2\times 0+3\times {\displaystyle \frac{\sqrt{3}}{2}}}{1+2+3}\phantom{\rule{0ex}{0ex}}=\frac{7}{12}\mathrm{m},\frac{\sqrt{3}}{4}\mathrm{m}$

#### Page No 159:

#### Question 2:

The structure of a water molecule is shown in figure. Find the distance of the centre of mass of the molecule from the centre of the oxygen atom.

Figure

#### Answer:

Let OX be the x-axis, OY be the Y-axis and O be the origin.

$\mathrm{Mass}\mathrm{of}\mathrm{O}\mathrm{atom},{m}_{1}=16\mathrm{unit}$

Let the position of oxygen atom be origin.

$\Rightarrow {x}_{1}={y}_{1}=0\phantom{\rule{0ex}{0ex}}\mathrm{Mass}\mathrm{of}\mathrm{H}\mathrm{atom},{m}_{2}=1\mathrm{unit}\phantom{\rule{0ex}{0ex}}{x}_{2}=-0.96\times {10}^{-10}\mathrm{sin}52\xb0\mathrm{m}\phantom{\rule{0ex}{0ex}}{y}_{2}=-0.96\times {10}^{-10}\mathrm{cos}52\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},{m}_{3}=1\mathrm{unit}\phantom{\rule{0ex}{0ex}}{x}_{3}=0.96\times {10}^{-10}\mathrm{sin}52\xb0\mathrm{m}\phantom{\rule{0ex}{0ex}}{y}_{3}=-0.96\times {10}^{-10}\mathrm{cos}52\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{The}X\mathrm{coordinate}\mathrm{of}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{mass}\mathrm{is}\mathrm{given}\mathrm{by}:\phantom{\rule{0ex}{0ex}}{x}_{\mathrm{cm}}=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}+{m}_{3}{x}_{3}}{{m}_{1}+{m}_{2}+{m}_{3}}\phantom{\rule{0ex}{0ex}}=\frac{16\times 0+1\times \left(-0.96\times {10}^{-10}\mathrm{sin}52\xb0\right)+1\times 0.96\times {{10}^{-}}^{10}\mathrm{sin}52\xb0}{1+1+16}=0\phantom{\rule{0ex}{0ex}}\mathrm{The}Y\mathrm{coordinate}\mathrm{of}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{mass}\mathrm{is}\mathrm{given}\mathrm{by}:\phantom{\rule{0ex}{0ex}}{y}_{\mathrm{cm}}=\frac{{m}_{1}{y}_{1}+{m}_{2}{y}_{2}+{m}_{3}{y}_{3}}{{m}_{1}+{m}_{2}+{m}_{3}}\phantom{\rule{0ex}{0ex}}=\frac{16\times 0+2\times 0.96\times {10}^{-10}\mathrm{cos}52\xb0}{1+1+16}\phantom{\rule{0ex}{0ex}}=\frac{2\times 0.96\times {10}^{-10}\mathrm{cos}52\xb0}{18}\phantom{\rule{0ex}{0ex}}=6.4\times {10}^{-12}\mathrm{m}$

Hence, the distance of centre of mass of the molecule from the centre of the oxygen atom is $(0,6.4\times {10}^{-12}\mathrm{m})$.

#### Page No 159:

#### Question 3:

Seven homogeneous bricks, each of length *L*, are arranged as shown in figure. Each brick is displaced with respect to the one in contact by *L*/10. Find the x-coordinate fo the centre of mass relative to the origin shown.

Figure

#### Answer:

Let OX be the X-axis and point O (0, 0) be the origin of the system.

The mass of each brick is *M*.

The length of each brick is *L.*

Each brick is displaced with respect to another in contact by a distance $\frac{\mathrm{L}}{10}$.

∴ The X-coordinate of the centre of mass is given as,

${\mathrm{X}}_{\mathrm{cm}}={\frac{1}{7m{\displaystyle \frac{\mathrm{L}}{10}}}}^{\left\{\frac{m\mathrm{L}}{2}+m\left(\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{10}\right)+..........+m\left(\frac{\mathrm{L}}{2}\right)\right\}}\phantom{\rule{0ex}{0ex}}{\mathrm{X}}_{\mathrm{cm}}=\frac{{\displaystyle \frac{\mathrm{L}}{2}}+{\displaystyle \frac{\mathrm{L}}{2}}+{\displaystyle \frac{\mathrm{L}}{10}}+{\displaystyle \frac{\mathrm{L}}{5}}+{\displaystyle \frac{\mathrm{L}}{2}}+{\displaystyle \frac{3\mathrm{L}}{10}}+{\displaystyle \frac{\mathrm{L}}{2}}+{\displaystyle \frac{\mathrm{L}}{5}}+{\displaystyle \frac{\mathrm{L}}{2}}+{\displaystyle \frac{\mathrm{L}}{10}}+{\displaystyle \frac{\mathrm{L}}{2}}}{7}\phantom{\rule{0ex}{0ex}}=\frac{7{\displaystyle \frac{\mathrm{L}}{2}}+5{\displaystyle \frac{\mathrm{L}}{10}}+2{\displaystyle \frac{\mathrm{L}}{5}}}{7}=\frac{22\mathrm{L}}{35}$

The X-coordinate of the centre of mass relative to the origin is $\frac{22\mathrm{L}}{35}$.

#### Page No 159:

#### Question 4:

A uniform disc of radius *R* is put over another uniform disc of radius 2*R* of the same thickness and density. The peripheries of the two discs touch each other. Locate the centre of mass of the system.

#### Answer:

Let the centre O (0, 0) of the bigger disc be the origin.

Radius of bigger disc = 2*R*

Radius of smaller disc = *R*

Now,

*m*_{1} = μR × T × ρ

*m*2 = μ (2R)^{2} × T × ρ

where T is the thickness of the two discs, and

ρ is the density of the two discs.

Position of the centre of mass is calculated as:

${x}_{1}=\mathrm{R},{y}_{1}=0\phantom{\rule{0ex}{0ex}}{x}_{2}=0,{y}_{2}=0$

$=\left(\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}},\frac{{m}_{1}{y}_{1}+{m}_{2}{y}_{2}}{{m}_{1}+{m}_{2}}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{\mathrm{\pi R}}^{2}\mathrm{T\rho R}+0}{{\mathrm{\mu R}}^{2}\mathrm{T\rho R}+\mathrm{\mu}(2\mathrm{R}{)}^{2}\mathrm{T\rho}},\frac{0}{{m}_{1}+{m}_{2}}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{\mathrm{\pi R}}^{2}\mathrm{T\rho R}}{5{\mathrm{\mu R}}^{2}\mathrm{T\rho}},0\right)=\left(\frac{\mathrm{R}}{5},0\right)\phantom{\rule{0ex}{0ex}}$

$\Rightarrow $The centre of mass lies at distance $\frac{\mathrm{R}}{5}$ from the centre of bigger disc, towards the centre of smaller disc.

Hence, the centre of mass of the system is $\left(\frac{\mathrm{R}}{5},0\right)$.

#### Page No 159:

#### Question 2:

A body has its centre of mass at the origin. the *x*-coordinates of the particles

(a) may be all positive

(b) may be all negative

(c) may be all non-negative

(d) may be positive for some cases and negative in other cases.

#### Answer:

(c) may be all non-negative

(d) may be positive for some cases and negative for other cases

According to the question, the centre of mass is at origin.

$\therefore $ $X=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}+{m}_{3}{x}_{3}+...}{{m}_{1}+{m}_{2}+{m}_{3}+.......}=0$

$\Rightarrow {m}_{1}{x}_{1}+{m}_{2}{x}_{2}+{m}_{3}{x}_{3}+...=0$

From the above equation, it can be concluded that all the x-coordinates may be non-negative.

In other words, they may be positive for some cases and negative for others.

#### Page No 159:

#### Question 3:

In which of the following cases the centre of mass of a rod is certainly not at its centre?

(a) the density continuously increases from left to right

(b) the density continuously decreases from left to right

(c) the density decreases from left to right upto the centre and then increases

(d) the density increases from left to right upto the centre and then decreases.

#### Answer:

(a) the density continuously increases from left to right

(b) the density continuously decreases from left to right

As the density continuously increases/decreases from left to right, there will be difference in the masses of rod that lie on either sides of the centre of mass. Thus, the centre of mass of a rod in such a case will certainly not be at its centre.

#### Page No 159:

#### Question 4:

If the external force acting on a system have zero resultant, the centre of mass

(a) must not move

(b) must not accelerate

(c) may move

(d) may accelerate.

#### Answer:

(b) must not accelerate

(c) may move

If the external force acting on a system has zero resultant,

then, acceleration of centre of mass$=\frac{{{\displaystyle \overrightarrow{F}}}_{\mathrm{net}}}{M}=0$.

However, it may move uniformly with constant velocity.

#### Page No 159:

#### Question 5:

A nonzero external force acts on a system of particles. The velocity and the acceleration of the centre of mass are found to be *v*_{0} and *a*_{0} at instant *t*. It is possible that

(a) *v*_{0} = 0, *a*_{0} = 0

(b) *v*_{0} = 0, *a*_{0} ≠ 0

(c) *v*_{0} ≠ 0, *a*_{0} = 0

(d) *v*_{0} ≠ 0, *a*_{0} ≠ 0

#### Answer:

(b) *v*_{0} = 0, *a*_{0} ≠ 0

(d) *v*_{0} ≠ 0, *a*_{0} ≠ 0

If a non-zero external force acts on a system of particles, it causes the centre of mass of the system to accelerate with acceleration *a*_{0} at any instant *t.* In such a case, the velocity of centre of mass of the system of particles is either *v*_{0} or zero.

#### Page No 159:

#### Question 6:

Two balls are thrown simultaneously in air. The acceleration of the centre of mass of the two balls while in air

(a) depends on the direction of the motion of the balls

(b) depends on the masses of the two balls

(c) depends on the speeds of the two balls

(d) is equal to *g*.

#### Answer:

(d) is equal to *g*

The acceleration of the centre of mass of two balls having masses *m*_{1} and *m*_{2} is given by

${a}_{\mathrm{cm}}=\frac{{m}_{1}{{\displaystyle \overrightarrow{a}}}_{1}+{m}_{2}{{\displaystyle \overrightarrow{a}}}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{{m}_{1}g+{m}_{2}g}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}=g$

#### Page No 159:

#### Question 7:

A block moving in air breaks in two parts and the parts separate

(a) the total momentum must be conserved

(b) the total kinetic energy must be conserved

(c) the total momentum must change

(d) the total kinetic energy must change

#### Answer:

(a) the total momentum must be conserved

(d) the total kinetic energy must change

As no external force acts on the block, the linear momentum is conserved.

Some energy is used to break the block, thus the total kinetic energy must change.

#### Page No 159:

#### Question 8:

In an elastic collision

(a) the kinetic energy remains constant

(b) the linear momentum remains constant

(c) the final kinetic energy is equal to the initial kinetic energy

(d) the final linear momentum is equal to the initial linear momentum.

#### Answer:

(b) the linear momentum remains constant

(c) the final kinetic energy is equal to the initial kinetic energy

(d) the final linear momentum is equal to the initial linear momentum.

During an elastic collision, all of the above statements are valid.

#### Page No 159:

#### Question 9:

A ball hits a floor and rebounds after an inelastic collision. In this case

(a) the momentum of the ball just after the collision is same as that just before the collision

(b) the mechanical energy of the ball remains the same during the collision

(c) the total momentum of the ball and the earth is conserved

(d) the total energy of the ball and the earth remains the same

#### Answer:

(c) the total momentum of the ball and the earth is conserved

(d) the total energy of the ball and the earth remains the same

As the ball rebounds after hitting the floor, its velocity changes.

*i.e. *Velocity of ball before collision ≠ Velocity of ball after collision

Therefore, the momentum of the ball just after the collision is not same as that just before the collision.

The mechanical energy of the ball also changes during the collision.

However, the total momentum of the system (earth plus ball) and the total energy of the system remain conserved.

#### Page No 159:

#### Question 10:

A body moving towards a finite body at rest collides with it. It is possible that

(a) both the bodies come to rest

(b) both the bodies move after collision

(c) the moving body comes to rest and the stationary body starts moving

(d) the stationary body remains stationary, the moving body changes its velocity.

#### Answer:

(b) both the bodies move after collision

(c) the moving body comes to rest and the stationary body starts moving

By using the law of conservation of linear momentum we can write:

Initial momentum of the two bodies = Final momentum of the two bodies

$\Rightarrow {p}_{i}={p}_{f}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{1}u={m}_{1}{v}_{1}+{m}_{2}{v}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}={v}_{2}\ne 0\phantom{\rule{0ex}{0ex}}\mathrm{However},\mathrm{it}\mathrm{may}\mathrm{be}\mathrm{possible}\mathrm{that}{\mathrm{v}}_{1}\mathrm{becomes}0\mathrm{and}{\mathrm{v}}_{2}\mathrm{becomes}\frac{{\mathrm{m}}_{1}\mathrm{u}}{{\mathrm{m}}_{2}}.$

#### Page No 159:

#### Question 11:

In a head-on elastic collision of two bodies of equal masses

(a) the velocities are interchanged

(b) the speeds are interchanged

(c) the momenta are interchanged

(d) the faster body slows down and the slower body speeds up.

#### Answer:

All of the above.

(a) the velocities are interchanged

(b) the speeds are interchanged

(c) the momenta are interchanged

(d) the faster body slows down and the slower body speeds up

#### Page No 160:

#### Question 5:

A disc of radius *R* is cut out from a larger disc of radius 2R in such a way that the edge of the hole touches the edge of the disc. Locate the centre of mass of the residual disc.

#### Answer:

Let O be the origin of the system (smaller disc plus bigger disc).

The density of the rods is ρ.

The thickness of rods is *T*.

Let *m*_{1}* *be the mass and *R* be the radius of the smaller disc, and

*m*_{2}* *be the mass and 2*R* be the radius of the bigger disc.

According to the question, the smaller disc is cut out from the bigger disc.

From the figure given below, we can write:

${m}_{1}=\pi {R}^{2}T\rho \mathit{}\phantom{\rule{0ex}{0ex}}\mathrm{Position}\mathrm{of}\mathrm{centre}\mathrm{of}\mathrm{mass}\mathrm{of}\mathrm{smaller}\mathrm{disc},\phantom{\rule{0ex}{0ex}}{x}_{1}=R,{y}_{1}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{m}_{2}=\pi \mathit{(}2R{\mathit{)}}^{\mathit{2}}T\rho \phantom{\rule{0ex}{0ex}}\mathrm{Position}\mathrm{of}\mathrm{centre}\mathrm{of}\mathrm{mass}\mathrm{of}\mathrm{bigger}\mathrm{disc},\phantom{\rule{0ex}{0ex}}{x}_{2}=0,{y}_{2}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{position}\mathrm{of}\mathrm{center}\mathrm{of}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{system},\phantom{\rule{0ex}{0ex}}{X}_{\mathrm{cm}}=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}},{Y}_{\mathrm{cm}}=\frac{{m}_{1}{y}_{1}+{m}_{2}{y}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Since}\mathrm{the}\mathrm{smaller}\mathrm{disc}\mathrm{is}\mathrm{removed}\mathrm{from}\mathrm{the}\mathrm{bigger}\mathrm{disc},\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{disc}\mathrm{is}\mathrm{taken}\mathrm{as}\mathrm{negative}(-{m}_{\mathit{1}}).\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{X}_{\mathrm{cm}}=\frac{-\pi {R}^{2}T\rho R+0{m}_{2}}{-\pi {R}^{2}T\rho +\pi (2R{)}^{2}T\rho},{Y}_{\mathrm{cm}}=\frac{0+0}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}{X}_{\mathrm{cm}}=\frac{-\pi {R}^{2}T\rho R}{3\pi {R}^{2}T\rho},{Y}_{\mathrm{cm}}=0\phantom{\rule{0ex}{0ex}}{X}_{\mathrm{cm}}=\frac{-R}{3},{Y}_{\mathrm{cm}}=0$

Hence, the centre of mass of the system lies at distance $\frac{R}{3}$ from the centre of bigger disc, away from centre of the hole.

#### Page No 160:

#### Question 6:

A square plate of edge *d* and a circular disc of diameter *d* are placed touching each other at the midpoint of an edge of the plate as shown in figure. Locate the centre of mass of the combination, assuming same mass per unit area for the two plates.

figure

#### Answer:

Let *m* be the mass per unit area of the square plate and the circular disc.

$\Rightarrow $Mass of the square plate, *M*_{1} = *d*^{2}*m*

Mass of the circular disc, *M*_{2} = $\frac{\mathrm{\pi}{d}^{\mathit{2}}}{4}m$

Let the centre of the circular disc be the origin of the system.

$\Rightarrow $ *x*_{1} = *d*, *y*_{1} = 0

*x*_{2} = 0, *y*_{2} = 0

$\Rightarrow $ Position of the centre of mass of circular disc and square plate:

$=\left(\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}},\frac{{m}_{1}{y}_{1}+{m}_{2}{y}_{2}}{{m}_{1}+{m}_{2}}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{d}^{2}md+\mathrm{\pi}\left({d}^{2}/4\right)m\times 0}{{d}^{2}m+\mathrm{\pi}\left({d}^{\mathit{2}}\mathit{/}\mathit{4}\right)m},\frac{0+0}{{m}_{1}+{m}_{2}}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{d}^{3}m}{{d}^{2}m(1+\mathrm{\pi}/4)},0\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{4d}{\mathrm{\pi}+4},0\right)$

Hence, the new centre of mass of the system (circular disc plus square plate) lies at distance $\frac{4d}{(\mathrm{\pi}+4)}$ from the centre of circular disc, towards right.

#### Page No 160:

#### Question 7:

Calculate the velocity of the centre of mass of the system of particles shown in figure.

Figure

#### Answer:

From the figure, the velocities of different masses can be written as:

$\mathrm{For}{m}_{1}=1.0\mathrm{kg},\phantom{\rule{0ex}{0ex}}\mathrm{Velocity},{\overrightarrow{v}}_{1}=\left(-1.5\mathrm{cos}37\xb0\hat{i}-1.5\mathrm{sin}37\xb0\hat{j}\right)=-1.2\hat{i}-0.9\hat{j}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{For}{m}_{2}=1.2\mathrm{kg},\phantom{\rule{0ex}{0ex}}\mathrm{Velocity},{\overrightarrow{v}}_{2}=0.4\overrightarrow{j}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{For}{m}_{\mathit{3}}=1.5\mathrm{kg},\phantom{\rule{0ex}{0ex}}\mathrm{Velocity},{\overrightarrow{v}}_{3}=-1.0\mathrm{cos}37\xb0\hat{i}+1.0\mathrm{sin}37\xb0\hat{j}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{For}{m}_{4}=0.50\mathrm{kg},\phantom{\rule{0ex}{0ex}}\mathrm{Velocity},{\overrightarrow{v}}_{4}=3.0\hat{i}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{For}{m}_{5}=1.0\mathrm{kg},\phantom{\rule{0ex}{0ex}}\mathrm{Velocity},{\overrightarrow{v}}_{5}=2.0\mathrm{cos}37\xb0\hat{i}-2.0\mathrm{sin}37\xb0\hat{j}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}(\mathrm{cos}37\xb0=\frac{4}{5}\mathrm{and}\mathrm{sin}37\xb0=\frac{3}{5})\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{V}}_{\mathrm{cm}}=\frac{{m}_{1}{\overrightarrow{v}}_{1}+{m}_{2}{\overrightarrow{v}}_{2}+{m}_{3}{\overrightarrow{v}}_{3}+{m}_{4}{\overrightarrow{v}}_{4}+{m}_{5}{\overrightarrow{v}}_{5}}{{m}_{1}+{m}_{2}+{m}_{3}+{m}_{4}+{m}_{5}}\phantom{\rule{0ex}{0ex}}={\frac{1}{1.0+1.2+1.5+1.0+0.50}}^{\left[1.0\left(-1.5\times \frac{4}{5}\overrightarrow{i}-1.5\times \frac{3}{5}\overrightarrow{j}\right)+...-2.0\times \frac{3}{5}\overrightarrow{j}\right]}\phantom{\rule{0ex}{0ex}}$

On solving the above equation, we get:

V_{cm} is 0.20 m/s , at 45° below the direction, towards right.

#### Page No 160:

#### Question 8:

Two blocks of masses 10 kg and 20 kg are placed on the *X*-axis. The first mass is moved on the axis by a distance of 2 cm. By what distance should the second mass be moved to keep the position of the centre of mass unchanged?

#### Answer:

Let the two masses *m*_{1} and *m*_{2} be placed on the X-axis.

It is given that:

*m*_{1} = 10 kg

*m*_{2} = 20 kg

The first mass is displaced by a distance of 2 cm.

$\therefore {\mathrm{X}}_{\mathrm{cm}}=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{X}}_{\mathrm{cm}}=\frac{10\times 2+20{\mathrm{x}}_{2}}{30}$

As the position of the centre of mass remains unchanged,

X_{cm} = 0

$\Rightarrow 0=\frac{20+20{\mathrm{x}}_{2}}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow 20+20{\mathrm{x}}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 20=-20{\mathrm{x}}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{x}}_{2}=-1$

Therefore, to keep the position of centre of mass unchanged, the block of mass 20 kg should be moved by a distance of 1 cm, towards left.

#### Page No 160:

#### Question 9:

Two blocks of masses 10 kg and 30 kg are placed along a vertical line. The first block is raised through a height of 7 cm. By what distance should the second mass be moved to raise the centre of mass by 1 cm?

#### Answer:

Let the two masses *m*_{1} and *m*_{2} are kept along a vertical line.

It is given that:

*m*_{1} = 10 kg

*m*_{2} = 30 kg

First block is raised through a height of 7 cm.

Thus, to raise the centre of mass by 1 cm:

$Y\mathrm{cm}=\frac{{m}_{1}{y}_{1}+{m}_{2}{y}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{10\times 7+30\times {y}_{2}}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow 70+30{y}_{2}=40\phantom{\rule{0ex}{0ex}}\Rightarrow 30{y}_{2}=-30\phantom{\rule{0ex}{0ex}}\Rightarrow {y}_{2}=-1$

Therefore, the 30 kg block (*m*_{2}_{)} should be moved by 1 cm downwards in order to raise the centre of mass by 1 cm.

#### Page No 160:

#### Question 10:

Consider a gravity-free hall in which a tray of mass *M*, carrying a cubical block of ice of mass *m* and edge *L*, is at rest in the middle. If the ice melts, by what distance does the centre of mass of "the tray plus the ice" system descend?

#### Answer:

As there is no gravity or other external forces acting on the system, the melting ice tends to acquire a spherical shape. Therefore, the centre of mass of the system does not move.

#### Page No 160:

#### Question 11:

Find the centre of mass of a uniform plate having semicircular inner and outer boundaries of radii *R*_{1} and *R*_{2}.

Figure

#### Answer:

Let the mass of the plate be* M*.

Consider a small semicircular portion of mass *dm* and radius *r*, as shown in fig.

$dm=\frac{M\pi rdr}{{\displaystyle \frac{\pi \left({R}_{2}^{2}-{R}_{1}^{2}\right)}{2}}}=\frac{M}{{\displaystyle \frac{\left({R}_{2}^{2}-{R}_{1}^{2}\right)}{2}}}rdr$

The centre of mass is given as:

${y}_{cm}=\frac{\int ydm}{M}\phantom{\rule{0ex}{0ex}}$

${y}_{cm}={\int}_{{R}_{1}}^{{R}_{2}}\left(\frac{2r}{\pi}\right).\frac{M}{{\displaystyle \frac{\left({R}_{2}^{2}-{R}_{1}^{2}\right)}{2}}}\times \frac{rdr}{M}\phantom{\rule{0ex}{0ex}}=\frac{2}{\pi {\displaystyle \frac{\left({R}_{2}^{2}-{R}_{1}^{2}\right)}{2}}}{\int}_{{R}_{1}}^{{R}_{2}}{r}^{2}dr\phantom{\rule{0ex}{0ex}}=\frac{2}{\pi {\displaystyle \frac{\left({R}_{2}^{2}-{R}_{1}^{2}\right)}{2}}}\left[\frac{\left({R}_{2}^{3}-{R}_{1}^{3}\right)}{3}\right]\phantom{\rule{0ex}{0ex}}=\frac{4\left({R}_{1}^{2}+{R}_{1}{R}_{2}+{R}_{2}^{2}\right)}{3\pi \left({R}_{1}+{R}_{2}\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 160:

#### Question 12:

Mr. Verma (50 kg) and Mr. Mathur (60 kg) are sitting at the two extremes of a 4 m long boat (40 kg) standing still in water. To discuss a mechanics problem, they come to the middle of the boat. Neglecting friction with water, how far does the boat move on the water during the process?

#### Answer:

It is given that:

Mass of Mr. Verma, *m*_{1} = 50 kg

Mass of Mr. Mathur, *m*_{2} = 60 kg

Mass of boat, *m*_{3} = 40 kg

Let A be the origin of the system (boat plus two men).

Initially, Mr. Verma and Mr. Mathur were at two extremes of the boat.

∴ Distance of the centre of mass:

${X}_{\mathrm{cm}}=\frac{{m}_{1}\times {x}_{1}+{m}_{2}\times {x}_{2}+{m}_{3}\times {x}_{3}}{{m}_{1}+{m}_{2}+{m}_{3}}\phantom{\rule{0ex}{0ex}}=\frac{60\times 0+50\times 4+40\times 2}{60+50+40}\phantom{\rule{0ex}{0ex}}=\frac{280}{150}=1.87\mathrm{m}\mathrm{from}\mathrm{A}$

As no external force is experienced in longitudinal direction, the centre of mass would not shift.

Initially, the centre of mass lies at a distance of 2 m from A.

When the men move towards the middle of the boat, the centre of mass shifts and lies at 1.87 m from A.

Therefore, the shift in centre of mass = 2 − 1.87 = 0.13 m, towards right

Hence, the boat moves 13 cm or 0.13 m towards right.

#### Page No 160:

#### Question 13:

A car of mass *M* is at rest on a frictionless horizontal surface and a pendulum bob of mass *m* hangs from the roof of the cart. The string breaks, the bob falls on the floor, makes serval collisions on the floor and finally lands up in a small slot made in the floor. The horizontal distance between the string and the slot is *L*. Find the displacement of the cart during this process.

Figure

#### Answer:

The mass of the bob is *m*.

The mass of the cart is M.

Considering the bob falls at point A.

Initial distance of centre of mass of the system from P is given as

$x=\frac{m\times \mathrm{L}+\mathrm{M}\times 0}{\mathrm{M}+m}=\frac{m}{\mathrm{M}+m}\mathrm{L}$

When the bob falls in the slot, the distance of centre of mass of the system from P becomes zero.

$\therefore \mathrm{Shift}\mathrm{in}\mathrm{centre}\mathrm{of}\mathrm{mass}=0-\frac{m\mathrm{L}}{\mathrm{M}+m}\phantom{\rule{0ex}{0ex}}=-\frac{m\mathrm{L}}{\mathrm{M}+m}\mathrm{towards}\mathrm{left}\phantom{\rule{0ex}{0ex}}=\frac{m\mathrm{L}}{\mathrm{M}+m}\mathrm{towards}\mathrm{right}\phantom{\rule{0ex}{0ex}}$

Therefore, the cart moves a distance of $\frac{m\mathrm{L}}{\mathrm{M}+m}$ towards right.

#### Page No 160:

#### Question 14:

The balloon, the light rope and the monkey shown in figure are at rest in the air. If the monkey reaches the top of the rope, by what distance does the balloon descend? Mass of the balloon = *M*, mass of the monkey = *m* and the length of the rope ascended by the monkey = *L*.

Figure

#### Answer:

Given:

The mass of monkey is *m*.

The mass of balloon is* M.*

Initially, the monkey, balloon and the rope are at rest.

Let the centre of mass is at a point P.

When the monkey descends through a distance *L*,

The centre of mass shifts.

${l}_{0}=\frac{m\times L+M\times 0}{M+m}\phantom{\rule{0ex}{0ex}}=\frac{mL}{M+m}\mathrm{from}\mathrm{P}$

Therefore, the balloon descends through a distance $\frac{mL}{M+m}$.

#### Page No 160:

#### Question 15:

Find the ratio of the linear momenta of two particles of masses 1.0 kg and 4.0 kg if their kinetic energies are equal.

#### Answer:

Let the masses of the two particles be *m*_{1} and *m*_{2}.

Given:

*m*_{1} = 1 kg

*m*_{2} = 4 kg

Now,

Kinetic energy of the first particle = Kinetic energy of the second particle

$\left(\frac{1}{2}\right){m}_{1}{v}_{1}^{2}=\left(\frac{1}{2}\right){m}_{2}{v}_{2}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{m}_{1}}{{m}_{2}}=\frac{{v}_{2}^{2}}{{v}_{1}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{v}_{2}}{{v}_{1}}=\sqrt{\frac{{m}_{1}}{{m}_{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{v}_{1}}{{v}_{2}}=\sqrt{\frac{{m}_{2}}{{m}_{1}}}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{ratio}\mathrm{of}\mathrm{linear}\mathrm{momenta}\left(mv\mathit{\right)}\mathit{}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{particles},\phantom{\rule{0ex}{0ex}}\frac{{P}_{1}}{{P}_{2}}=\frac{{m}_{1}{v}_{1}}{{m}_{2}{v}_{2}}=\frac{{m}_{1}}{{m}_{2}}\sqrt{\frac{{m}_{2}}{{m}_{1}}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{{m}_{1}}{{m}_{2}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {P}_{1}:{P}_{2}=1:2$

Therefore, the ratio of linear momenta is 1:2.

#### Page No 160:

#### Question 16:

A uranium-238 nucleus, initially at rest, emits an alpha particle with a speed of 1.4 × 10^{7} m/s. Calculate the recoil speed of the residual nucleus thorium-234. Assume that the mass of a nucleus is proportional to the mass number.

#### Answer:

According to the question, uranium 238 nucleus emits an alpha-particle with a speed of 1.4 × 10^{7} m/s.

Let the speed of the residual nucleus thorium 234 be *v*_{2}.

By the law of conservation of linear momentum, we have:

${m}_{1}{v}_{1}={m}_{2}{v}_{2}$

Here, *m*_{1} and *v*_{1} are the mass and velocity of the alpha-particle repectively, and *m*_{2} is the mass of the residual nucleus.

$\Rightarrow 4\times 1.4\times {10}^{7}=234\times {v}_{2}\phantom{\rule{0ex}{0ex}}{v}_{2}=\frac{4\times 1.4\times {10}^{7}}{234}=2.4\times {10}^{5}\mathrm{m}/\mathrm{s}$

Therefore, the speed of the residual nucleus is $2.4\times {10}^{5}\mathrm{m}/\mathrm{s}$.

#### Page No 160:

#### Question 17:

A man of mass 50 kg starts moving on the earth and acquires a speed 1.8 m/s. With what speed does the earth recoil? Mass of earth = 6 × 10^{24} kg.

#### Answer:

By the law of conservation of linear momentum, we have:

${m}_{1}{v}_{1}={m}_{2}{v}_{2}$

Here, *m*_{1} and *v*_{1} are the mass and velocity of the man respectively, and *m*_{2}_{ }and *v*_{2} are the mass and velocity of the Earth respectively.

$\Rightarrow 50\times 1.8=6\times {10}^{24}\times {v}_{2}\phantom{\rule{0ex}{0ex}}\therefore {v}_{2}=\frac{50\times 1.8}{6\times {10}^{24}}=15\times {10}^{-24}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{2}=1.5\times {10}^{-23}\mathrm{m}/\mathrm{s}$

Hence, the earth recoils with a speed of 1.5 × 10^{−23} m/s.

#### Page No 160:

#### Question 18:

A neutron initially at rest, decays into a proton, an electron and an antineutrino. The ejected electron has a momentum of 1.4 × 10^{−26} kg-m/s and the antineutrino 6.4 × 10^{−27} kg-m/s. Find the recoil speed of the proton (a) if the electron and the antineutrino are ejected along the same direction and (b) if they are ejected along perpendicular directions. Mass of the proton = 1.67 × 10^{−27} kg.

#### Answer:

It is given that:

Mass of proton, *m*_{p} = 1.67 × 10^{−27} kg

Momentum of electron = 1.4 × 10^{−26} kg m/s

Momentum of antineutrino = 6.4 × 10^{−27} kg m/s

Let the recoil speed of the proton be *V*_{p}.

(a) When the electron and the antineutrino are ejected in the same direction,

Applying the law of conservation of momentum, we get:

${m}_{p}{V}_{p}={p}_{\mathrm{electron}}+{p}_{\mathrm{antineutrino}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}1.67\times {10}^{-27}\times {V}_{\mathrm{P}}=1.4\times {10}^{-26}+6.4\times {10}^{-27}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{\mathrm{P}}=\left(\frac{20.4}{1.67}\right)=12.2\mathrm{m}/\mathrm{s},\mathrm{in}\mathrm{the}\mathrm{opposite}\mathrm{direction}$

(b) When the electron and the antineutrino are ejected perpendicular to each other,

Total momentum of electron and antineutrino is given as,

$p=\sqrt{{p}_{\mathrm{electron}}+{p}_{\mathrm{antineutrino}}}=\sqrt{(14{)}^{2}+(6.4{)}^{2}}\times {10}^{-27}=15.4\times {10}^{-27}\mathrm{kg}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{p}=\frac{15.4\times {10}^{-27}\mathrm{kgm}/\mathrm{s}}{1.67\times {10}^{-27}\mathrm{kg}}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{p}=9.2\mathrm{m}/\mathrm{s}$

#### Page No 161:

#### Question 19:

A man of mass *M* having a bag of mass *m* slips from the roof of a tall building of height *H* and starts falling vertically. When at a height *h* from the ground, the notices that the ground below him is pretty hard, but there is a pond at a horizontal distance *x* from the line of fall. In order to save himself he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water. If the man just succeeds to avoid the hard ground, where will the bag land?

Figure

#### Answer:

Mass of man = M

Initial velocity of the man = 0

Mass of bag = *m*

Let the man throws the bag towards left with a velocity *v* and himself moves towards right with a velocity *V*.

Using the law of conservation of momentum,

$mv=MV\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{MV}{m}...\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{the}\mathrm{total}\mathrm{time}\mathrm{he}\mathrm{takes}\mathrm{to}\mathrm{reach}\mathrm{ground}\mathrm{be}{t}_{1}.\phantom{\rule{0ex}{0ex}}\Rightarrow {t}_{1}=\sqrt{\frac{2H}{g}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{the}\mathrm{total}\mathrm{time}\mathrm{he}\mathrm{takes}\mathrm{to}\mathrm{reach}\mathrm{the}\mathrm{height}\mathit{}h\mathrm{be}{t}_{2}.\phantom{\rule{0ex}{0ex}}\Rightarrow {t}_{\mathit{2}}=\sqrt{\frac{2(H-h)}{g}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{The}\mathrm{time}\mathrm{of}\mathrm{flying}\mathrm{in}\mathrm{covering}\mathrm{the}\mathrm{remaining}\mathrm{height}h\mathrm{is},\phantom{\rule{0ex}{0ex}}t={t}_{1}-{t}_{2}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}t=\sqrt{\frac{2H}{g}}-\sqrt{\frac{2(H-h)}{g}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2}{g}}\left(\sqrt{H}-\sqrt{H-h}\right)$

During this time, the man covers a horizontal distance *x *and lands in the water.

$\Rightarrow x=V\times t\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{x}{t}\phantom{\rule{0ex}{0ex}}\therefore v=\frac{M}{m}\frac{x}{t}\left[\mathrm{using}\mathrm{equation}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{M}{m}\frac{x}{\sqrt{{\displaystyle \frac{2}{g}}}\left(\sqrt{H}-\sqrt{H-h}\right)}$

Thus, the minimum horizontal velocity imparted to the bag, such that the man lands in the water is $\frac{M}{m}\frac{x}{\sqrt{{\displaystyle \frac{2}{g}}}\left(\sqrt{H}-\sqrt{H-h}\right)}$.

Let the bag lands at a distance *x' *towards left from actual line of fall.

As there is no external force in horizontal direction, the *x*-coordinate of the centre of mass will remain same.

$\Rightarrow 0=\frac{M\times \left(x\right)+m\times x\text{'}}{M+m}\phantom{\rule{0ex}{0ex}}\Rightarrow x\text{'}=-\frac{M}{m}x$

Therefore, the bag will land at a distance $\frac{M}{m}x$.

#### Page No 161:

#### Question 20:

A ball of mass 50 g moving at a speed of 2.0 m/s strikes a plane surface at an angle of incidence 45°. The ball is reflected by the plane at equal angle of reflection with the same speed. Calculate (a) the magnitude of the change in momentum of the ball (b) the change in the magnitude of the momentum of the ball.

#### Answer:

It is given that:

Mass of the ball = 50 g = $0.05\mathrm{Kg}$

Speed of the ball, *v* = 2.0 m/s

Incident angle = 45˚

$v=2\mathrm{cos}45\xb0\hat{i}-2\mathrm{sin}45\xb0\hat{j}\phantom{\rule{0ex}{0ex}}v\text{'}=-2\mathrm{cos}45\xb0\hat{i}-2\mathrm{sin}45\xb0\hat{j}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{a}\right)\mathrm{Change}\mathrm{in}\mathrm{momentum}=\mathrm{m}\overrightarrow{\mathrm{v}}-\mathrm{m}\overrightarrow{\mathrm{v}}\text{'}\phantom{\rule{0ex}{0ex}}=0.05\left(2\mathrm{cos}45\xb0\hat{i}-2\mathrm{sin}45\xb0\hat{j}\right)-0.05\left(-2\mathrm{cos}45\xb0\hat{i}-2\mathrm{sin}45\xb0\hat{j}\right)\phantom{\rule{0ex}{0ex}}=0.1\mathrm{cos}45\xb0\hat{i}-0.1\mathrm{sin}45\xb0\hat{j}+0.1\mathrm{cos}45\xb0\hat{i}+0.1\mathrm{sin}45\xb0\hat{j}\phantom{\rule{0ex}{0ex}}=0.2\mathrm{cos}45\xb0\hat{i}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Magnitude}=\frac{0.2}{\sqrt{2}}=0.14\mathrm{Kg}\mathrm{m}/\mathrm{s}$

(b) The change in magnitude of the momentum of the ball,

$\left|{\overrightarrow{P}}_{2}\right|-\left|{\overrightarrow{P}}_{1}\right|=2\times 0.5-2\times 0.5=0$

*i.e.* There is no change in magnitude of the momentum of the ball.

#### Page No 161:

#### Question 21:

Light in certain cases may be considered as a stream of particles called photons. Each photon has a linear momentum *h*/λ where *h* is the Planck's constant and λ is the wavelength of the light. A beam of light of wavelength λ is incident on a plane mirror at an angle of incidence θ. Calculate the change in the linear momentum of a photon as the beam is reflected by the mirror.

#### Answer:

It is given that:

Wavelength of light = λ

Momentum of each photon = *h*/λ

Angle of incidence = *θ*

${\overrightarrow{\mathrm{P}}}_{Incidence}=\left(\frac{h}{\lambda}\right)\mathrm{cos}\mathrm{\theta}\hat{i}-\left(\frac{\mathrm{h}}{\mathrm{\lambda}}\right)\mathrm{sin}\mathrm{\theta}\hat{j}\phantom{\rule{0ex}{0ex}}{\overrightarrow{\mathrm{P}}}_{\mathrm{Reflected}}=-\left(\frac{h}{\mathrm{\lambda}}\right)\mathrm{cos}\mathrm{\theta}\hat{i}-\left(\frac{h}{\mathrm{\lambda}}\right)\mathrm{sin}\mathrm{\theta}\hat{j}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{change}\mathrm{in}\mathrm{momentum}\mathrm{will}\mathrm{only}\mathrm{be}\mathrm{in}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{x}-\mathrm{axi}si.e.,\phantom{\rule{0ex}{0ex}}\left|\mathrm{\Delta P}\right|=\left(\frac{h}{\mathrm{\lambda}}\right)\mathrm{cos}\mathrm{\theta}-\left(-\frac{h}{\mathrm{\lambda}}\mathrm{cos}\mathrm{\theta}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{2h}{\mathrm{\lambda}}\right)\mathrm{cos}\mathrm{\theta}$

#### Page No 161:

#### Question 22:

A block at rest explodes into three equal parts. Two parts start moving along *X* and *Y* axes respectively with equal speeds of 10 m/s. Find the initial velocity of the third part.

Figure

#### Answer:

As the block is exploded only because of its internal energy, the net external force on the system is zero.

Thus, the centre of mass of does not change.

Let the body was at the origin of the co-ordinate system during explosion.

Resultant velocity of two bodies of equal mass moving at a speed of 10 m/s in + *x-*axis and + *y*-axis direction, is given as:

$v=\sqrt{{10}^{2}+{10}^{2}+2.10.10\mathrm{cos}90\xb0}=10\sqrt{2}\mathrm{m}/\mathrm{s},45\xb0\mathrm{w}.\mathrm{r}.\mathrm{t}x-\mathrm{axis}$

If the centre of mass is at rest, the third part having equal mass as that of the other two masses will move in the opposite direction (*i.e.* $135\xb0$ w.r.t. +*x*-axis) at the same velocity of $10\sqrt{2}\mathrm{m}/\mathrm{s}.$.

#### Page No 161:

#### Question 23:

Two fat astronauts each of mass 120 kg are travelling in a closed spaceship moving at a speed of 15 km/s in the outer space far removed from all other material objects. The total mass of the spaceship and its contents including the astronauts is 660 kg. If the astronauts do slimming exercise and thereby reduce their masses to 90 kg each, with what velocity will the spaceship move?

#### Answer:

According to the question, the spaceship is removed from all other material objects and kept totally isolated from the surroundings. Thus, the mass loss by astronauts couldn't slip away from the spaceship. Therefore, the total mass of the spaceship remains unchanged and so does its velocity.

Hence, the spaceship moves with the speed of 15 km/s.

#### Page No 161:

#### Question 24:

During a heavy rain, hailstones of average size 1.0 cm in diameter fall with an average speed of 20 m/s. Suppose 2000 hailstones strike every square meter of a 10 m × 10 m roof perpendicularly in one second and assume that the hailstones do not rebound. Calculate the average force exerted by the falling hailstones on the roof. Density of a hailstone is 900 kg/m^{3}.

#### Answer:

It is given that:

Diameter of hailstone = 1 cm = 0.01 m

⇒ Radius of hailstone, *r* = 0.005 m

Average speed of hailstone = 20 m/s

Density of hailstone = 900 kg/m^{3} = 0.9 g/cm^{3}

$\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{hailstones}\mathrm{is}\mathrm{given}\mathrm{as},\phantom{\rule{0ex}{0ex}}V=\frac{4}{3}\mathrm{\pi}{r}^{3}\mathit{}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}V\mathit{=}\mathit{}\frac{4}{3}\mathrm{\pi}(0.005{)}^{3}=5.235\times {10}^{-7}{\mathrm{m}}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{Mass}=v\mathrm{olume}\times \mathrm{density}=5.235\times {10}^{-7}\times 900\phantom{\rule{0ex}{0ex}}=4.711\times {10}^{-4}\mathrm{kg}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Mass}\mathrm{of}2000\mathrm{hailstone}=2000\times 4.711\times {10}^{-4}=0.9422\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{momentum}=0.9422\times 20\approx 19\mathrm{N}/{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}$

$\therefore $ The total force exerted on the roof = $19\times 100=1900\mathrm{N}$

#### Page No 161:

#### Question 25:

A ball of mass *m* is dropped onto a floor from a certain height. The collision is perfectly elastic and the ball rebounds to the same height and again falls. Find the average force exerted by the ball on the floor during a long time interval.

#### Answer:

It is given that the mass of the ball is *m*.

Let the ball be dropped from a height *h*.

The speed of ball before the collision is *v _{1}.*

$\therefore {v}_{1}=\sqrt{2gh}\phantom{\rule{0ex}{0ex}}$

The speed of ball after the collision is

*v*

_{2}.

${v}_{2}=-\sqrt{2gh}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{velocity}=\mathrm{acceleration}\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{2\sqrt{2gh}}{t}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Force},F=\frac{m\times 2\sqrt{2gh}}{t}\dots \left(1\right)\phantom{\rule{0ex}{0ex}}$

Using Newton's laws of motion, we can write:

$v=\sqrt{2gh},s=h,u=0\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{2gh}=gt$

$\Rightarrow t=\sqrt{\frac{2h}{g}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Total}\mathrm{time}=2\sqrt{\frac{2h}{g}}$

Substituting this value of time

*t*in equation (1), we get:

*F*=

*mg*

#### Page No 161:

#### Question 26:

A railroad car of mass *M* is at rest on frictionless rails when a man of mass *m* starts moving on the car towards the engine. If the car recoils with a speed *v* backward on the rails, with what velocity is the man approaching the engine?

#### Answer:

Given:

The mass of the railroad car is *M*.

The mass of the man is *m*.

The car recoils with a speed *v*, backwards on the rails.

Let the man of mass *m* approaches towards the engine with a velocity *v'* w.r.t the engine.

∴ The velocity of man w.r.t earth is *v'* − *v*, towards right.

${V}_{\mathrm{centre}\mathrm{of}\mathrm{mass}}=0(\mathrm{Initially}\mathrm{at}\mathrm{rest})\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore 0=-Mv+m(v\text{'}-v)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow Mv=m(v\text{'}-v)\phantom{\rule{0ex}{0ex}}\Rightarrow mv\text{'}=Mv+mv\phantom{\rule{0ex}{0ex}}\Rightarrow v\text{'}=\left(\frac{M+m}{m}\right)v\phantom{\rule{0ex}{0ex}}\Rightarrow v\text{'}=\left(1+\frac{M}{m}\right)v$

#### Page No 161:

#### Question 27:

A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is 50 *m* where *m* is the mass of one shell. If the velocity of the shell with respect to the gun (in its state before firing) is 200 m/s, what is the recoil speed of the car after the second shot? Neglect friction.

#### Answer:

It is given that:

Mass of the car, the gun, the shells and the operator = 50*m*

Mass of one shell = *m*

Muzzle velocity of the shells, *v* = 200 m/s

Let the speed of car be *v.*

On applying the law of conservation of linear momentum, we get:

$49mV+mv=0\phantom{\rule{0ex}{0ex}}\Rightarrow 49m\times V+m\times 200=0\phantom{\rule{0ex}{0ex}}\Rightarrow V=-\frac{200}{49}\mathrm{m}/\mathrm{s}(\text{'}-\text{'}\mathrm{sign}\mathrm{indicates}\mathrm{direction}\mathrm{towards}\mathrm{left})$

Thus, when another shell is fired, the velocity of the car with respect to the platform is,

$\mathrm{V}=\frac{200}{49}\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{left}$

When one more shell is fired, the velocity of the car with respect to the platform is,

${\mathrm{V}}_{1}=\frac{200}{48}\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{left}$

∴ Velocity of the car w.r.t the earth $=200\left(\frac{1}{49}+\frac{1}{48}\right)\mathrm{m}/\mathrm{s}$

#### Page No 161:

#### Question 28:

Two persons each of mass *m* are standing at the two extremes of a railroad car of mass *M* resting on a smooth track. The person on left jumps to the left with a horizontal speed *u* with respect to the state of the car before the jump. Thereafter, the other person jumps to the right, again with the same horizontal speed *u* with respect to the state of the car before his jump. Find the velocity of the car after both the persons have jumped off.

Figure

#### Answer:

It is given that:

Mass of each persons = *m*

Mass of railroad car = M

Let** **the velocity of the railroad w.r.t. earth, when the man on the left jumps off be *V*.

By the law of conservation of momentum:

$0=-mu+(M+m)V\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow V=\left(\frac{mu}{M+m}\right),\mathrm{towards}\mathrm{right}$

When** **the man on the right jumps, his velocity w.r.t.** **the car is *u*.

$0=mu-MV\text{'}\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}=\frac{mu}{M}$

(V is the change in velocity of the platform when the platform itself is taken as reference, assuming the car to be at rest.)

∴ Net velocity towards left, (*i.e.* the velocity of the car)

$V\text{'}-V=\frac{mu}{M}-\frac{mu}{(M+m)}\phantom{\rule{0ex}{0ex}}=\frac{mMu+{m}^{2}u-Mmu}{M(M+m)}\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}-V=\frac{{m}^{2}u}{M(M+m)}$

#### Page No 161:

#### Question 29:

Figure shows a small block of mass *m* which is started with a speed *v* on the horizontal part of the bigger block of mass *M* placed on a horizontal floor. The curved part of the surface shown in semicircular. All the surfaces are frictionless. Find the speed of the bigger block when the smaller block reaches the point *A* of the surface.

Figure

#### Answer:

Given:

The mass of the small block is *m.*

Initial speed of this block is *v*.

The mass of the bigger block is *M. *

Initial speed of this block is zero.

At point A, as the small block transfers its momentum to the bigger block, both the blocks move with a common velocity *V* (say) in the same direction as *v.*

Using the law of conservation of linear momentum, we can write:

Initial momentum = final momentum

$mv+M\times O=(m+M)V\phantom{\rule{0ex}{0ex}}$

$\Rightarrow V=\frac{mv}{m+M}$

Therefore, the speed of the bigger block when the smaller block reaches point *A* of the surface is$\frac{mv}{m+M}$.

#### Page No 162:

#### Question 30:

In a typical Indian *Bugghi* (a luxury cart drawn by horses), a wooden plate is fixed on the rear on which one person can sit. A bugghi of mass 200 kg is moving at a speed of 10 km/h. As it overtakes a school boy walking at a speed of 4 km/h, the boy sits on the wooden plate. If the mass of the boy is 25 kg, what will be the plate. If the mass of the boy is 25 kg, what will be the new velocity of the bugghi?

#### Answer:

It is given that:

Mass of the bugghi, *m*_{b}_{ }= 200 kg

Velocity of the bugghi, V_{b}_{ }= 10 km/h

Mass of the boy, m_{boy} = 25 kg

Velocity of the boy, V_{Boy} = 4 km/h

Consider the boy and the bugghi as a system.

The total momentum before the process of sitting remains same after the process of sitting.

Using the law of conservation of momentum, we can write:

${\mathrm{m}}_{b}{\mathrm{V}}_{b}+{m}_{\mathrm{boy}}\mathit{}{\mathrm{V}}_{\mathrm{boy}}=({m}_{b}+{m}_{\mathrm{boy}})\mathrm{V}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 200\times 10+25\times 4=(200+25)\times \mathrm{V}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{V}=\frac{2100}{225}=\frac{28}{3}\mathrm{Km}/\mathrm{h}$

#### Page No 162:

#### Question 31:

A ball of mass 0.50 kg moving at a speed of 5.0 m/s collides with another ball of mass 1.0 kg. After the collision the balls stick together and remain motionless. What was the velocity of the 1.0 kg block before the collision?

#### Answer:

It is given that:

Speed of the ball, *v*_{1} = 5.0 m/s

Mass of the ball, *m*_{1} = 0.5 kg

Mass of another ball, *m*_{2} = 1 kg

Let the velocity of this ball be *v*_{2} m/s.

On applying the law of conservation of momentum, we get:

${m}_{1}{v}_{1}+{m}_{2}{v}_{2}=0\phantom{\rule{0ex}{0ex}}0.5\times 5+1\times {v}_{2}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{2}=-2.5\mathrm{m}/\mathrm{s}$

Hence, the velocity of second ball is 2.5 m/s, opposite to the direction of motion of the first ball.

#### Page No 162:

#### Question 32:

A 60 kg man skating with a speed of 10 m/s collides with a 40 kg skater at rest and they cling to each other. Find the loss of kinetic energy during the collision.

#### Answer:

It is given that:

Mass of the skater who is skating, *m*_{1} = 60 kg

Initial speed of this man, *v*_{1} = 10 m/s

Mass of the skater at rest, *m*_{2} = 40 kg

Initial speedof this man, *v*_{2} = 0

Let the velocity of both men after collision be *v.*

Using the law of conservation of momentum, we can write:

${m}_{1}{v}_{1}+{m}_{2}{v}_{2}=({m}_{1}+{m}_{2})v\phantom{\rule{0ex}{0ex}}\Rightarrow 60\times 10+0=100\times v\phantom{\rule{0ex}{0ex}}\Rightarrow v=6\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{loss}\mathrm{in}\mathrm{kinetic}\mathrm{energy}\mathrm{during}\mathrm{collision}\mathrm{is}\mathrm{given}\mathrm{as},\phantom{\rule{0ex}{0ex}}\u2206KE=\frac{1}{2}{m}_{1}{v}_{1}^{2}-\frac{1}{2}({m}_{1}+{m}_{2}){v}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206KE=\frac{1}{2}\times 60\times (10{)}^{2}-\frac{1}{2}\times 100\times 36\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206KE=1200\mathrm{J}$

#### Page No 162:

#### Question 33:

Consider a head-on collision between two particles of masses *m*_{1} and *m*_{2}. The initial speeds of the particles are *u*_{1} and u_{2} in the same direction. the collision starts at *t* = 0 and the particles interact for a time interval ∆*t*. During the collision, the speed of the first particle varies as $v\left(t\right)={u}_{1}+\frac{t}{\u2206t}({v}_{1}-{u}_{1})$

Find the speed of the second particle as a function of time during the collision.

#### Answer:

It is given that:

Speed of the first particle during collision, $v\left(t\right)={u}_{1}+\frac{t}{\u2206t}({v}_{1}-{u}_{1})$

Let *v' *be the speed of the second particle, during collision.

On applying the law of conservation of linear momentum on both particles, we get:

*m*_{1}*u*_{1} + *m*_{2}*u*_{2} = *m*_{1}*v*(*t*) + *m*_{2}*v*'

$\Rightarrow {m}_{1}{u}_{1}+{m}_{2}{u}_{2}={m}_{1}{u}_{1}+{m}_{1}\times \left(\frac{t}{\u2206t}\right)({v}_{1}-{u}_{1})+{m}_{2}v\text{'}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{the}\mathrm{sides}\mathrm{by}{m}_{2},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}{u}_{2}=\frac{{m}_{1}}{{m}_{2}}\left(\frac{t}{\u2206t}\right)({v}_{1}-{u}_{1})+v\text{'}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow v\text{'}={u}_{2}-\frac{{m}_{1}}{{m}_{2}}\left(\frac{t}{\u2206t}\right)({v}_{1}-{u}_{1})$

The speed of the second particle during collision can be written as a function of time and is given by the expression, ${u}_{2}-\frac{{m}_{1}}{{m}_{2}}\left(\frac{t}{\u2206t}\right)({v}_{1}-{u}_{1})$.

#### Page No 162:

#### Question 34:

A bullet of mass *m* moving at a speed *v* hits a ball of mass *M* kept at rest. A small part having mass *m* breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed *v*_{1} in the direction of the bullet. Find the velocity of the bullet after the collision.

#### Answer:

Given:

The mass of bullet moving with speed *v* is *m*.

The mass of the ball is *M* and it is at rest.

*m'* is the fractional mass of the ball that sticks with the bullet.

The remaining mass of the ball moves with the velocity *v*_{1}.

Let *v*_{2} be the final velocity of the bullet plus fractional mass system.

On applying the law of conservation of momentum, we get:

*mv* + 0 = (*m'* + *m*)*v*_{2} + (*M* − *m*') *v*_{1}

$\Rightarrow {v}_{2}=\frac{mv-(M-m\text{'}){v}_{1}}{m+m\text{'}}$

Therefore, the velocity of the bullet after the collision is $\frac{mv-(M-m\text{'}){v}_{1}}{m+m\text{'}}$.

#### Page No 162:

#### Question 35:

A ball of mass *m* moving at a speed *v* makes a head-on collision with an identical ball at rest. The kinetic energy of the balls after the collision is three fourths of the original. Find the coefficient of restitution.

#### Answer:

Given:

The mass of the both balls is *m.*

Initial speed of first ball* = v *

Initial speed of second ball* = *0

Let the final velocities of balls be *v*_{1} and *v*_{2} respectively.

$e=\frac{\mathrm{velocity}\mathrm{of}\mathrm{separation}}{\mathrm{velocity}\mathrm{of}\mathrm{approach}}\phantom{\rule{0ex}{0ex}}\Rightarrow e=\frac{{v}_{1}-{v}_{2}}{v}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}-{v}_{2}=ev...\left(1\right)$

On applying the law of conservation of linear momentum, we get:

$m({v}_{1}+{v}_{2})=mv\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}+{v}_{2}=v...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{According}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{condition},\phantom{\rule{0ex}{0ex}}\mathrm{Final}\mathrm{K}.\mathrm{E}.=\frac{3}{4}\mathrm{Initial}\mathrm{K}.\mathrm{E}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}m{v}_{1}^{2}+\frac{1}{2}m{v}_{2}^{2}=\frac{3}{4}\times \frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathit{}{v}_{\mathit{1}}^{\mathit{2}}+{\mathrm{v}}_{2}^{2}=\frac{3}{4}{v}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{({v}_{1}+{v}_{2}{)}^{2}+({v}_{1}-{v}_{2}{)}^{2}}{2}=\frac{3}{4}{v}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\left(1+{e}^{2}\right){v}^{2}}{2}=\frac{3}{4}{v}^{2}\left[\mathrm{using}\mathrm{the}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 1+{e}^{2}=\left(\frac{3}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {e}^{2}\mathit{=}\mathit{}\frac{\mathit{1}}{\mathit{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow e=\frac{1}{\sqrt{2}}$

Hence, the coefficient of restitution is found to be $\frac{1}{\sqrt{2}}$.

#### Page No 162:

#### Question 36:

A block of mass 2.0 kg moving 2.0 m/s collides head on with another block of equal mass kept at rest. (a) Find the maximum possible loss in kinetic energy due to the collision. (b) If he actual loss in kinetic energy is half of this maximum, find the coefficient of restitution.

#### Answer:

It is given that:

Mass of first block, m_{1}_{ }= 2 kg

Initial speed,*v*_{1} = 2.0 m/s

Mass of second block, *m*_{2} = 2 kg

Initial speed of this block = 0

For maximum possible loss in kinetic energy, we assume that the collision is elastic and both the blocks move with same final velocity *v* (say).

On applying the law of conservation of linear momentum, we get:

*m*_{1}*v*_{1}* + **m*_{2}* ×0 = (**m*_{1}*+**m*_{2})*v*

2 × 2 = (2 + 2)*v*

⇒ *v* = 1 m/s

Loss in K.E. in elastic collision is give by,

$=\frac{1}{2}{m}_{1}{v}_{1}^{2}-\frac{1}{2}\left({m}_{1}+{m}_{2}\right)v\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}\right)\times 2\times {2}^{2}-\left(\frac{1}{2}\right)(2+2)\times (1{)}^{2}\phantom{\rule{0ex}{0ex}}=4-2=2\mathrm{J}$

$\left(\mathrm{b}\right)\mathrm{The}\mathrm{actual}\mathrm{loss}\mathrm{in}\mathrm{K}.\mathrm{E}.=\frac{\mathrm{Maximum}\mathrm{loss}\mathrm{in}\mathrm{K}.\mathrm{E}.}{2}=1\mathrm{J}$

Let the final velocities of the blocks be *v*_{1} and *v*_{2} respectively.

The coefficient of restitution is *e.*

$\therefore $ The loss in K.E. is given by,

$\left(\frac{1}{2}\right)\times 2\times (2{)}^{2}-\left(\frac{1}{2}\right)2\times {v}_{1}^{2}\mathit{-}\mathit{}\left(\frac{1}{2}\right)\mathit{}\times 2{v}_{2}^{2}=1\phantom{\rule{0ex}{0ex}}\Rightarrow 4-\left({v}_{1}^{2}+{v}_{2}^{2}\right)=1\phantom{\rule{0ex}{0ex}}\Rightarrow 4-\frac{\left(1+{e}^{2}\right)\times 4}{2}=1\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(1+{e}^{2}\right)=3\phantom{\rule{0ex}{0ex}}\Rightarrow 1+{e}^{2}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {e}^{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{e}=\frac{1}{\sqrt{2}}$

#### Page No 162:

#### Question 37:

A particle of mass 100 g moving at an initial speed *u* collides with another particle of same mass kept initially at rest. If the total kinetic energy becomes 0.2 J after the collision, what could be the minimum and the maximum value of *u*.

#### Answer:

It is given that:

Mass of particles = 100 g

Initial speed of the first particle = *u*

Final K.E. of the system after collision = 0.2J

Initial K.E. of the system, before collision = $\frac{1}{2}m{u}^{2}+0\phantom{\rule{0ex}{0ex}}$

*i.e*. Initial K.E. = $\frac{1}{2}\times 0.1\times {u}^{2}=0.05{u}^{2}$

Let *v*_{1} and *v*_{2} be the final velocities of the first and second block respectively.

By law of conservation of momentum, we know:

$m{v}_{1}+m{v}_{2}=mu$

$\Rightarrow {v}_{1}+{v}_{2}=u...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}({v}_{1}-{v}_{2})+e({u}_{1}-{u}_{2})=0\phantom{\rule{0ex}{0ex}}\Rightarrow eu={v}_{2}-{v}_{1}...\left(2\right)[\mathrm{Putting}{u}_{2}=0,{u}_{1}=u]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\mathrm{the}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}2{v}_{2}=(1+e)\mathrm{u}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{v}}_{2}=\left(\frac{u}{2}\right)(1+e)\phantom{\rule{0ex}{0ex}}\therefore {\mathrm{v}}_{1}=u-\frac{\mathrm{u}}{2}(1+e)\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{1}=\frac{\mathrm{u}}{2}(1-e)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{According}\mathrm{to}\mathrm{given}\mathrm{condition},\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}_{1}^{2}+\frac{1}{2}m{v}_{2}^{2}=0.2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}^{2}+{v}_{2}^{2}=4\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{u}^{2}}{2}\left(1+{e}^{2}\right)=4\phantom{\rule{0ex}{0ex}}\Rightarrow {u}^{2}=\frac{8}{1+{e}^{2}}$

For maximum value of *u*, denominator should be minimum in the above equation.

*i.e.* *e* = 0

⇒ *u*^{2} = 8

$\Rightarrow u=2\sqrt{2}\mathrm{m}/\mathrm{s}$

For minimum value of *u*, denominator should have maximum value.

*i.e*. e = 1

⇒ *u*^{2}^{ }= 4

⇒ *u* = 2 m/s

#### Page No 162:

#### Question 38:

Two friends *A* and *B* (each weighing 40 kg) are sitting on a frictionless platform some distance *d* apart. *A* rolls a ball of mass 4 kg on the platform towards *B* which *B* catches. Then *B* rolls the ball towards *A* and *A* catches it. The ball keeps on moving back and forth between *A* and *B*. The ball has a fixed speed of 5 m/s on the platform. (a) Find the speed of *A* after he catches the ball for the first time. (c) Find the speeds of *A* and *B* after the all has made 5 round trips and is held by *A*. (d) How many times can *A* roll the ball? (e) Where is the centre of mass of the system "*A* + *B* + ball" at the end of the *n*th trip?

#### Answer:

It is given that:

Weight of A = Weight of B = 40 kg

Velocity of ball = 5 m/s

(*a*) Case-1: Total momentum of the man A and ball remains constant.

∴ 0 = 4 × 5 − 40 × *v*

⇒ *v* = 0.5 m/s, towards left

(*b*) Case-2: When B catches the ball, the momentum between B and the ball remains constant.

⇒ 4 × 5 = 44 *v*

⇒ $v=\left(\frac{20}{44}\right)\mathrm{m}/\mathrm{s}$

Case-3: When B throws the ball,

On applying the law of conservation of linear momentum, we get:

$\Rightarrow 44\times \left(\frac{20}{44}\right)=-4\times 5+40\times v\phantom{\rule{0ex}{0ex}}\Rightarrow v=1\mathrm{m}/\mathrm{s},(\mathrm{towards}\mathrm{right})$

Case-4**:** When A catches the ball,

Applying the law of conservation of liner momentum, we get:

$-4\times 5+(-0.5)\times 40=44v\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{40}{44}=\frac{10}{11}\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{left}$

(*c*) Case-5: When A throws the ball,

Applying the law of conservation of linear momentum, we get:

$44\times \left(\frac{10}{11}\right)=4\times 5+40\times v\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{60}{40}=\frac{3}{2}\mathrm{m}/\mathrm{s}(\mathrm{towards}\mathrm{left})$

Case-6: When B receives the ball,

Applying the law of conservation of linear momentum, we get:

$40\times 1+4\times 5=44\times v\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{60}{44}\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{right}$

Case-7: When B throws the ball,

On applying the law of conservation of linear momentum, we get:

$\Rightarrow v=44\times \left(\frac{60}{44}\right)\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{right}$

Case-8: When A catches the ball,

On applying the law of conservation of linear momentum, we get:

$-4\times 5+40\left(\frac{3}{2}\right)=-44v\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{V}=-\frac{40}{44}=\left({\displaystyle \frac{10}{11}}\right)\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{left}$

Similarly, after 5 round trips,

The velocity of A will be $\left(\frac{50}{11}\right)$ m/s and the velocity of B will be 5 m/s.

(*d*) As after 6 round trips, the velocity of A becomes $\frac{60}{11}$ i.e. > 5 m/s, it cannot catch the ball. Thus, A can only roll the ball six times.

(*e*) Let the ball and the body A be at origin, in the initial position.

$\therefore {\mathrm{X}}_{\mathrm{c}}=\frac{40\times 0+4\times 0+40\times d}{40+40+4}\phantom{\rule{0ex}{0ex}}=\frac{10}{21}d$

#### Page No 162:

#### Question 39:

A ball falls on the ground from a height of 2.0 m and rebounds up to a height of 1.5 m. Find the coefficient of restitution.

#### Answer:

Let the velocity of the ball falling from height *h*_{1} be *u *(when it approaches the ground).

Velocity on the ground, $u=\sqrt{2g{h}_{1}}$

$\Rightarrow u=\sqrt{2\times 9.8\times 2}$

Let the velocity of ball when it separates from the ground be *v. *(Assuming it goes up to height *h*_{2})

$\Rightarrow v=\sqrt{2g{h}_{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2\times 9.8\times 1.5}\phantom{\rule{0ex}{0ex}}$

Let the coefficient of restitution be *e*.

We know, *v* = *eu*

$\Rightarrow e=\frac{\sqrt{2\times 9.8\times 1.5}}{\sqrt{2\times 9.8\times 2}}=\frac{\sqrt{3}}{2}$

Hence, the coefficient of restitution is $\frac{\sqrt{3}}{2}$.

#### Page No 162:

#### Question 40:

In a gamma decay process, the internal energy of a nucleus of mass *M* decreases, a gamma photon of energy *E* and linear momentum *E/c* is emitted and the nucleus recoils. Find the decrease in internal energy.

#### Answer:

Let the nucleus recoils with a velocity *v.*

Applying the law of conservation of linear momentum, we get:

Linear momentum of recoiled nucleus = Linear momentum of gamma photon

*⇒* *mv* = $\frac{E}{c}$

$\therefore $ $v=\frac{E}{mc}$

Kinetic energy of the recoiled nucleus = $\frac{1}{2}M{v}^{2}$

$\Rightarrow \mathrm{K}.\mathrm{E}.=\frac{1}{2}m{\left(\frac{E}{mc}\right)}^{2}=\frac{1}{2}\frac{{E}^{2}}{m{c}^{2}}$

Decrease in the internal energy = photon energy + the kinetic energy of the recoiled nucleus

*⇒ *Decrease in the internal energy = $E+\frac{{E}^{2}}{2m{c}^{2}}$

#### Page No 162:

#### Question 41:

A block of mass 2.0 kg is moving on a frictionless horizontal surface with a velocity of 1.0 m/s towards another block of equal mass kept at rest. The spring constant of the spring fixed at one end is 100 N/m. Find the maximum compression of the spring

Figure

#### Answer:

Given,

Mass of each block, *M*_{A} = *M*_{B} = 2 kg

Initial velocities of block A, *V _{a}* = 1 m/s

Initial velocity of block B,

*V*= 0

_{b}Spring constant of the spring = 100 N/m

Block A strikes the spring with a velocity of 1 m/s.

After the collision, it's velocity decreases continuously. At an instant the whole system (Block A + the compound spring + Block B) moves together with a common velocity

*V*(say).

Using the law of conservation of energy, we get:

$\left(\frac{1}{2}\right){M}_{A}{V}_{A}^{2}+\left(\frac{1}{2}\right){M}_{B}{V}_{B}^{2}=\left(\frac{1}{2}\right){M}_{A}{V}^{2}+\left(\frac{1}{2}\right){M}_{B}{V}^{2}+\left(\frac{1}{2}\right)k{x}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\frac{1}{2}\right)\times 2(1{)}^{2}+0=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\times {v}^{2}+\left(\frac{1}{2}\right){x}^{2}\times 100$

(where

*x*is the maximum compression of the spring)

⇒ 1 − 2

*v*

^{2}= 50

*x*

^{2}...(1)

As there is no external force acting in the horizontal direction, the momentum is conserved.

$\Rightarrow {M}_{A}{V}_{A}+{M}_{B}{V}_{B}=({M}_{A}+{M}_{B})V\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times 1=4\times V\phantom{\rule{0ex}{0ex}}\Rightarrow V=\left(\frac{1}{2}\right)\mathrm{m}/\mathrm{s}...\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{Susbstituting}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{V}\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}1=2\times \left(\frac{1}{4}\right)+50{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{4}=50{x}^{2}\mathit{}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=\frac{1}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{1}{10}\mathrm{m}\phantom{\rule{0ex}{0ex}}\Rightarrow x=10\mathrm{cm}$

#### Page No 162:

#### Question 42:

A bullet of mass 20 g travelling horizontally with a speed of 500 m/s passes through a wooden block of mass 10.0 kg initially at rest on a level surface. The bullet emerges with a speed of 100 m/s and the block slides 20 cm on the surface before coming to rest. Find the friction coefficient between the block and the surface.

Figure

#### Answer:

It is given that:

Mass of bullet, *m* = 20 g =0.02 kg

The initial speed, *v*_{1} = 500 m/s

Mass of block, *M* = 10 kg

The initial speed of block = 0

Final velocity of bullet, *v*_{2}= 100 m/s

Let the final velocity of block when the bullet emerges out = *v*'

Applying conservation of linear momentum,

*mv*_{1} + *M*_{ }× 0 = *mv*_{2} + *Mv**'*

⇒ 0.02 × 500 = 0.02 × 100 + 10 × *v*'

⇒ *v*' = 0.8 m/s

Distance covered by the block, *d = *20 cm = 0.02 m .

Let friction coefficient between the block and the surface = *μ*

Thus, the value of friction force, $F=\mu mg$$F=\mu mg$

Change in K.E. of block = Work done by the friction force

$\Rightarrow \frac{1}{2}\times M\times 0-\frac{1}{2}\times M\times {\left(v\text{'}\right)}^{2}=\mu mgd\phantom{\rule{0ex}{0ex}}\Rightarrow 0-\left(\frac{1}{2}\right)\times 10\times (0.8{)}^{2}=\mu \times 10\times 10\times 0.2\phantom{\rule{0ex}{0ex}}\Rightarrow \mu =0.16$

#### Page No 162:

#### Question 43:

A projectile is fired with a speed *u* at an angle θ above a horizontal field. The coefficient of restitution of collision between the projectile and the field is *e*. How far from the starting point, does the projectile makes its second collision with the field?

#### Answer:

Given:

Initial velocity of the projectile = *u*

Angle of projection of the projectile with respect to ground = θ

When the projectile hits the ground for the first time, the velocity remains same *i.e.* *u*.

The component of velocity parallel to ground, *u* cos θ should remain constant.

However, the vertical component of the projectile undergoes a change after the collision.

If the coefficient of restitution of collision between the projectile and the field is *e,*

The velocity of separation is given by,

⇒ *v* = *eu *sin *θ*

Therefore, for the second projectile motion,

Velocity of projection (*u'*) will be,

$u\text{'}=\sqrt{(u\mathrm{cos}\theta {)}^{2}+(eu\mathrm{sin}\theta {)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Angle}\mathrm{of}\mathrm{projection},\alpha ={\mathrm{tan}}^{-1}\left(\frac{eu\mathrm{sin}\theta}{u\mathrm{cos}\theta}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha ={\mathrm{tan}}^{-1}(e\mathrm{tan}\theta )\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{tan}\alpha =e\mathrm{tan}\theta ...\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{Also},y=x\mathrm{tan}\mathrm{\alpha}-\left(\frac{g{x}^{2}{\mathrm{sec}}^{2}\mathrm{\alpha}}{2u{\text{'}}^{2}}\right)...\left(3\right)\phantom{\rule{0ex}{0ex}}\mathrm{Here},\mathit{}y=0\phantom{\rule{0ex}{0ex}}\therefore \mathrm{tan}\alpha =e\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{sec}}^{2}\mathrm{\alpha}=1+{e}^{2}{\mathrm{tan}}^{2}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}\mathrm{and}u{\text{'}}^{2}={u}^{\mathit{2}}\mathit{}{\mathrm{cos}}^{\mathit{2}}\mathit{}\mathrm{\theta}+{e}^{2}{u}^{2}\mathit{}{\mathrm{sin}}^{2}\mathit{}\mathrm{\theta}$

Putting the above calculated values in equation (3), we get:

$xe\mathrm{tan\theta}=\frac{g{x}^{2}(1+{e}^{2}{\mathrm{tan}}^{2}\mathrm{\theta})}{2{u}^{2}({\mathrm{cos}}^{2}\mathrm{\theta}+{e}^{2}\mathit{}{\mathrm{sin}}^{\mathit{2}}\mathit{}\mathrm{\theta})}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{or},x=\frac{2e{u}^{2}\mathrm{tan}\theta ({\mathrm{cos}}^{2}\mathrm{\theta}+{e}^{2}{\mathrm{sin}}^{2}\mathrm{\theta})}{g(1+{e}^{2}{\mathrm{tan}}^{2}\mathrm{\theta})}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{2e{u}^{2}\mathrm{tan}\theta .{\mathrm{cos}}^{2}\mathrm{\theta}}{g}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{e{u}^{2}\mathrm{sin}2\mathrm{\theta}}{g}$

Thus, from the starting point the projectile makes its second collision with the field at a distance,

$x\text{'}=\frac{{u}^{2}\mathrm{sin}2\mathrm{\theta}}{g}+\frac{e{u}^{2}\mathrm{sin}2\mathrm{\theta}}{g}\phantom{\rule{0ex}{0ex}}\Rightarrow x\text{'}=\frac{{u}^{2}\mathrm{sin}2\mathrm{\theta}}{g}(1+e)$

#### Page No 162:

#### Question 44:

A ball falls on an inclined plane of inclination θ from a height *h* above the point of impact and makes a perfectly elastic collision. Where will it hit the plane again?

#### Answer:

Given:

The angle of inclination of the inclined plane is θ.

A ball falls on the inclined plane from height *h.*

The coefficient of restitution is* e*.

Let the ball strikes the inclined plane at origin with velocity ${v}_{0}=\sqrt{2gh}$.

As the ball elastically rebounds, it recalls with the same velocity *v*_{0} at the same angle *θ* from the normal or *y*-axis.

Let the ball strikes the incline second time at any point P, which is at a distance *l* from the origin along the incline.

From the equation,

$y={v}_{oy}t+\frac{1}{2}{a}_{y}{t}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{we}\mathrm{may}\mathrm{write}:\phantom{\rule{0ex}{0ex}}0={v}_{0}\mathrm{cos}\theta t-\frac{1}{2}g\mathrm{cos}\theta {t}^{2}\phantom{\rule{0ex}{0ex}}$

where *t* is time of motion of the ball in air, as the ball moves from origin to point P.

$\mathrm{As}t\ne 0,\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{t}\mathrm{is}\frac{2{v}_{0}}{g}.$

Now from the equation,

$x={v}_{0}t+\frac{1}{2}{a}_{x}{t}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{we}\mathrm{can}\mathrm{write},\phantom{\rule{0ex}{0ex}}l={v}_{0}\mathrm{sin}\theta t+\frac{1}{2}g\mathrm{sin}\theta {t}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathit{\therefore}\mathit{}l={v}_{0}\mathrm{sin}\theta \left(\frac{2{v}_{\mathit{0}}}{g}\right)+\frac{1}{2}g\mathrm{sin}\theta {\left(\frac{2{v}_{0}}{g}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\frac{4{v}_{\mathit{0}}^{\mathit{2}}\mathit{}\mathrm{sin}\mathit{}\theta}{g}\phantom{\rule{0ex}{0ex}}$

Substituting the value of *v*_{0} in the above equation, we get:

*l* = 8*h** *sin *θ*

Therefore, the ball again hits the plain at a distance, *l* = 8*h** *sin θ

#### Page No 162:

#### Question 45:

Solve the previous problem if the coefficient of restitution is *e*.

$\mathrm{Use}\mathrm{\theta}=45\xb0,e=\frac{3}{4}\mathrm{and}h=5\mathrm{m}.$

#### Answer:

Given:

The angle of inclination of the plane, θ = 45°

A ball falls on the inclined plane from a height (h) of 5 m*.*

The coefficient of restitution,* *$e=\left(\frac{3}{4}\right)$

The velocity with which ball strikes the inclined plane is given as,

$v=\sqrt{2g\times 5}=10\mathrm{m}/\mathrm{s}$

The ball makes an angle β with the horizontal, after the collision.

The horizontal component of velocity, 10cos45° remains unchanged.

However, the velocity in perpendicular direction to the plane after the collision will now be:

${v}_{1}=e\times 10\mathrm{sin}45\xb0\phantom{\rule{0ex}{0ex}}=\left(\frac{\mathit{3}}{\mathit{4}}\right)\times 10\times \frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=(3.75)\sqrt{2}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},{v}_{2}=5\sqrt{2}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}u=\sqrt{{\mathrm{v}}_{2}^{2}+{v}_{\mathit{1}}^{\mathit{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{50+28.125}\phantom{\rule{0ex}{0ex}}=\sqrt{78.125}\phantom{\rule{0ex}{0ex}}=8.83\mathrm{m}/\mathrm{s}$

Angle of reflection from the wall is given as,

$\mathrm{\beta}={\mathrm{tan}}^{-1}\frac{(3.75\sqrt{2})}{5\sqrt{2}}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)=37\xb0$

Angle of projection α = 90 − *(θ + β)*

$\Rightarrow $α = 90° − (45° + 37°) = 8°

Let the distance where the ball falls after the collision be *L.
⇒ x = L *cos

*θ*

Angle of projection (

*α*) = −8°

Now,

*Y*= $x\mathrm{tan}\alpha -\frac{g{x}^{2}se{c}^{2}\alpha}{2{u}^{2}}$

$\Rightarrow $ −L sin θ = L cos θ × tan 8° $-\frac{g}{2}\frac{{\mathrm{L}}^{2}{\mathrm{cos}}^{2}\mathrm{\theta}{\mathrm{sec}}^{2}80\xb0}{(u{)}^{2}}$

$\Rightarrow -\mathrm{sin}45\xb0=\mathrm{cos}45\xb0\times \mathrm{tan}8\xb0-\frac{10{\mathrm{cos}}^{2}45{\mathrm{sec}}^{2}8\xb0}{(8.83{)}^{2}}\phantom{\rule{0ex}{0ex}}$

On solving the above equation, we get:

*L*= 18.5 m

#### Page No 162:

#### Question 46:

A block of mass 200 g is suspended through a vertical spring. The spring is stretched by 1.0 cm when the block is in equilibrium. A particle of mass 120 g is dropped on the block from a height of 45 cm. The particle sticks to the block after the impact. Find the maximum extension of the spring.

Figure

#### Answer:

It is given that:

Mass of block, *M* = 200 g = 0.20 kg

Mass of the particle, *m* = 120 gm = 0.12 kg

Height of the particle, *h* = 45 cm = 0.45 m

According to question, as the block attains equilibrium, the spring is stretched by a distance, *x* = 1.00 cm = 0.01 m.

*i.e.* *M* × g = *K* × *x*

⇒ 0.2 × g = *K* × *x*

⇒ 2 = K × 0.01

⇒ *K* = 200 N/m

The velocity with which the particle *m* strikes *M* is given by,

$u=\sqrt{2gh}\phantom{\rule{0ex}{0ex}}u=\sqrt{2\times 10\times 0.45}\phantom{\rule{0ex}{0ex}}=\sqrt{9}=3\mathrm{m}/\mathrm{s}$

After the collision, let the velocity of the particle and the block be *V*.

According to law of conservation of momentum, we can write:

*mu = (m + M)V*

Solving for *V* , we get:

$V=\frac{0.12\times 3}{0.32}=\frac{9}{8}\mathrm{m}/\mathrm{s}$

Let the spring be stretched through an extra deflection of *δ*.

On applying the law of conservation of energy, we can write:

Initial energy of the system before collision = Final energy of the system

$\Rightarrow \frac{1}{2}m{u}^{2}+\frac{1}{2}K{x}^{2}=\frac{1}{2}\left(m+M\right){V}^{2}+\frac{1}{2}K{\left(x+\delta \right)}^{2}$

Substituting appropriate values in the above equation, we get:

$\left(\frac{1}{2}\right)\times 0.12\times 9+\left(\frac{1}{2}\right)\times 200\times (0.01{)}^{2}=\left(\frac{1}{2}\right)0.32\times \left(\frac{81}{64}\right)+\left(\frac{1}{2}\right)\times 200\times (\delta +0.1{)}^{2}$

On solving the above equation, we get:

*δ* = 0.061

m = 6.1 cm

#### Page No 163:

#### Question 47:

A bullet of mass 25 g is fired horizontally into a ballistic pendulum of mass 5.0 kg and gets embedded in it. If the centre of the pendulum rises by a distance of 10 cm, find the speed of the bullet.

#### Answer:

Given:

Mass of bullet, *m = *25 *g* = 0.025 kg

Mass of ballistic pendulum, *M* = 5 kg

Vertical displacement*, h =* 10 cm = 0.1 m

Let the bullet strikes the pendulum with a velocity *u*.

Let the final velocity be *v*.

Using the law of conservation of linear momentum, we can write:

$mu=(M+m)v\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{m}{(M+m)}u\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{0.25}{5.025}\times u=\frac{u}{201}$

Applying the law of conservation of energy, we get:

$\left(\frac{1}{2}\right)(M+m){v}^{2}=(M+m)gh\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{u}^{2}}{(201{)}^{2}}=2\times 10\times 0.1\phantom{\rule{0ex}{0ex}}\Rightarrow u=201\times \sqrt{2}=280\mathrm{m}/\mathrm{s}$

The bullet strikes the pendulum with a velocity of 280 m/s.

#### Page No 163:

#### Question 48:

A bullet of mass 20 g moving horizontally at a speed of 300 m/s is fired into a wooden block of mass 500 g suspended by a long string. The bullet crosses the block and emerges on the other side. If the centre of mass of the block rises through a height of 20.0 cm, find the speed of the bullet as it emerges from the block.

#### Answer:

Given:

Mass of bullet, *m* = 20 gm = 0.02 kg

Horizontal speed of the bullet, *u* = 300 m/s

Mass of wooden block, *M* = 500 gm = 0.5 kg

Let the bullet emerges out with velocity *v*.

Let the velocity of the block be* v'*.

Using the law of conservation of momentum, we get:

*mu* =* **Mv**'* + *m**v* ...(1)

Now, applying the work-energy principle for the block after the collision, we get:

$0-\left(\frac{1}{2}\right)M\times {\left(v\text{'}\right)}^{2}=-Mgh\phantom{\rule{0ex}{0ex}}\Rightarrow (v\text{'}{)}^{2}=2gh\phantom{\rule{0ex}{0ex}}v\text{'}=\sqrt{2gh}\phantom{\rule{0ex}{0ex}}=\sqrt{20\times 10\times 0.2}=2\mathrm{m}/\mathrm{s}$

On substituting the value of *v'* in equation (1), we get:

$0.02\times 300=0.5\times 2+0.02\times v\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{6-1}{0.02}=\frac{5}{0.02}\phantom{\rule{0ex}{0ex}}\Rightarrow v=250\mathrm{m}/\mathrm{s}$

Hence, the speed of the bullet as it emerges out from the block is 250 m/s.

#### Page No 163:

#### Question 49:

Two mass *m*_{1} and *m*_{2} are connected by a spring of spring constant *k* and are placed on a frictionless horizontal surface. Initially the spring is stretched through a distance *x*_{0} when the system is released from rest. Find the distance moved by the two masses before they again come to rest.

#### Answer:

It is given that two blocks of masses *m*_{1} and *m*_{2} are connected with a spring having spring constant *k.*

Initially the spring is stretched by a distance *x*_{0}.

For the block to come to rest again,

Let the distance travelled by *m*_{1} be *x*_{1}_{ }(towards right), and that travelled by *m*_{2} be _{ }*x*_{2} towards left.

As no external force acts in horizontal direction, we can write:

*m*_{1}*x*_{1} = *m*_{2}*x*_{2} ...(1)

As the energy is conserved in the spring, we get:

$\left(\frac{1}{2}\right)k{x}_{0}^{2}=\left(\frac{1}{2}\right)k({x}_{1}+{x}_{2}-{x}_{0}{)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{0}={x}_{1}+{x}_{2}-{x}_{0}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}+{x}_{2}=2{x}_{0}...(2)$

$\therefore {x}_{1}=2{x}_{0}-{x}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{Putting}\mathrm{this}\mathrm{value}\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}{m}_{1}(2{x}_{0}-{x}_{0})={m}_{2}{x}_{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2{m}_{1}{x}_{0}-{m}_{1}{x}_{2}={m}_{2}{x}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{2}=\frac{2{m}_{2}}{{m}_{1}+{m}_{2}}{x}_{0}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},{x}_{1}=\left(\frac{2{m}_{2}}{{m}_{1}+{m}_{2}}\right){x}_{0}$

#### Page No 163:

#### Question 50:

Two blocks of masses *m*_{1} and *m*_{2} are connected by a spring of spring constant *k*. The block of mass *m*_{2} is given a sharp impulse so that it acquires a velocity *v*_{0} towards right. Find (a) the velocity of the centre of mass, (b) the maximum elongation that the spring will suffer.

Figure

#### Answer:

Given,

Velocity of mass, *m*_{2} = *v*_{0}

Velocity of mass, *m*_{1}_{ }= 0

(*a*) Velocity of centre of mass is given by,

${v}_{cm}=\frac{{m}_{1}{v}_{1}+{m}_{2}{v}_{2}}{{m}_{1}+{m}_{2}}$

$\Rightarrow {v}_{cm}=\frac{{m}_{1}\times 0+{m}_{2}\times {v}_{0}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{cm}=\frac{{m}_{2}{v}_{0}}{{m}_{1}+{m}_{2}}$

(b) Let the maximum elongation in spring be *x.*

The spring attains maximum elongation when velocities of both the blocks become equal to the velocity of centre of mass.

*i.e*. *v*_{1} = *v*_{2} = *v*_{cm}

On applying the law of conservation of energy, we can write:

Change in kinetic energy = Potential energy stored in spring

$\Rightarrow \frac{1}{2}{m}_{2}{v}_{0}^{2}-\frac{1}{2}({m}_{1}+{m}_{2}){\left(\frac{{m}_{2}{v}_{0}}{{m}_{1}+{m}_{2}}\right)}^{2}=\frac{1}{2}k{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{2}{v}_{0}^{2}\left(1-\frac{{m}_{2}}{{m}_{1}+{m}_{2}}\right)=k{x}^{2}$

$\Rightarrow x={v}_{0}{\left[\frac{{m}_{1}{m}_{2}}{\left({m}_{1}+{m}_{2}\right)k}\right]}^{1/2}$

#### Page No 163:

#### Question 51:

Consider the situation of the previous problem. Suppose each of the blocks is pulled by a constant force *F* instead of any impulse. Find the maximum elongation that the spring will suffer and the distance moved by the two blocks in the process.

#### Answer:

It is given that the force on both the blocks is *F.
*

Let

*x*

_{1}and

*x*

_{2}be the extensions of blocks

*m*

_{1}and

*m*

_{2}respectively.

Total work done by the forces on the blocks =

*Fx*

_{1}

*+*

*Fx*

_{2}...(1)

∴ Increase in the potential energy of spring = $\left(\frac{1}{2}\right)K({x}_{1}+{x}_{2}{)}^{2}...(2)$

$\mathrm{Equating}\mathrm{the}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\Rightarrow ({x}_{1}+{x}_{2})=\left(\frac{2\mathrm{F}}{k}\right)\dots \left(3\right)$

As the net external force on the system is zero, the centre of mass does not shift.

$\therefore {m}_{1}{x}_{1}={m}_{2}{x}_{2}...\left(4\right)\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{the}\mathrm{equations}\left(3\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\frac{{m}_{1}}{{m}_{2}}{x}_{1}+{x}_{1}=\frac{2F}{k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}\left(1+\frac{{m}_{1}}{{m}_{2}}\right)=\frac{2F}{k}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}=\frac{2F{m}_{2}}{k({m}_{1}+{m}_{2})}$

_{}

Similarly,

${x}_{2}=\frac{2F{m}_{1}}{k({m}_{1}+{m}_{2})}$

#### Page No 163:

#### Question 52:

Consider the situation of the previous problem. Suppose the block of mass *m*_{1} is pulled by a constant force *F*_{1} and the other block is pulled by a constant force *F*_{2}. Find the maximum elongation that the spring will suffer.

#### Answer:

Given:

Force on block of mass, *m*_{1} = *F*_{1}

Force on block of mass, *m*_{2} = *F*_{2}

Let the acceleration produced in mass *m*_{1} be *a*_{1}.

${a}_{1}=\frac{{F}_{1}-{F}_{2}}{{m}_{1}+{m}_{2}}$

Let the acceleration of mass *m*_{2}_{ }be *a*_{2}.

${a}_{2}=\frac{{F}_{2}-{F}_{1}}{{m}_{1}+{m}_{2}}$

Due to the force *F*_{2}, the mass *m*_{1} experiences a pseudo force.

$\therefore \mathrm{Net}\mathrm{force}\mathrm{on}{m}_{1}={F}_{1}+{m}_{1}{a}_{2}\phantom{\rule{0ex}{0ex}}F\text{'}={F}_{1}+{m}_{1}\times \frac{({F}_{2}-{F}_{1})}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{{m}_{1}{F}_{1}+{m}_{2}{F}_{1}+{m}_{1}{F}_{2}-{m}_{1}{F}_{1}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{{m}_{2}{F}_{1}+{m}_{1}{F}_{2}}{{m}_{1}+{m}_{2}}$

Similarly, mass *m*_{2} experiences a pseudo force due to force *F*_{1}.

$\therefore \mathrm{Net}\mathrm{force}\mathrm{on}{m}_{2}={F}_{2}+{m}_{2}{a}_{1}$

$F"={F}_{2}+{m}_{2}\times \frac{({F}_{1}-{F}_{2})}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{{m}_{1}{F}_{2}+{m}_{2}{F}_{2}+{m}_{2}{F}_{1}-{m}_{2}{F}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{{m}_{1}{F}_{2}+{m}_{2}{F}_{1}}{{m}_{1}+{m}_{2}}$

Let *m*_{1} be displaced by a distance *x*_{1} and m_{2} be displaced by a distance *x*_{2}.

Therefore, the maximum elongation of the spring = *x*_{1} + *x*_{2}

Work done by the blocks = Energy stored in the spring

$\Rightarrow \frac{{m}_{2}{F}_{1}+{m}_{1}{F}_{2}}{{m}_{1}+{m}_{2}}\times {x}_{1}\times \frac{{m}_{2}{F}_{1}+{m}_{1}{F}_{2}}{{m}_{1}+{m}_{2}}\times {x}_{2}=\left(\frac{1}{2}\right)k({x}_{1}+{x}_{2}{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}+{x}_{2}=\frac{2}{k}\left(\frac{{m}_{1}{F}_{2}+{m}_{2}{F}_{1}}{{m}_{1}+{m}_{2}}\right)$

#### Page No 163:

#### Question 53:

Consider a gravity-free hall in which an experimenter of mass 50 kg is resting on a 5 kg pillow, 8 ft above the floor of the hall. He pushes the pillow down so that it starts falling at a speed of 8 ft/s. The pillow makes a perfectly elastic collision with the floor, rebounds and reaches the experimenter's head. Find the time elapsed in the process.

#### Answer:

Given,

Mass of the man, *M*_{m} = 50 kg

Mass of the pillow, *M*_{P} = 5 kg

Velocity of pillow w.r.t. man, ${\overrightarrow{V}}_{pm}$ = 8 ft/s

Let the velocity of man be ${\overrightarrow{V}}_{\mathrm{m}}$.

${\overrightarrow{V}}_{\mathrm{pm}}={\overrightarrow{V}}_{\mathrm{p}}-(-{\overrightarrow{V}}_{\mathrm{m}})\phantom{\rule{0ex}{0ex}}={\overrightarrow{V}}_{\mathrm{p}}+{\overrightarrow{V}}_{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\mathit{}{\overrightarrow{V}}_{\mathrm{p}}\mathit{}={\overrightarrow{V}}_{\mathrm{pm}}-{\overrightarrow{V}}_{\mathrm{m}}$

As no external force acts on the system, the acceleration of centre of mass is zero and the velocity of centre of mass remains constant.

Using the law of conservation of linear momentum:

*M*_{m} × *V*_{m} = *M*_{p} × *V*_{p} ...(*1*)

${M}_{\mathrm{m}}\times {V}_{\mathrm{m}}={M}_{\mathrm{p}}\times ({V}_{\mathrm{pm}}-{V}_{\mathrm{m}})\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 50\times {V}_{\mathrm{m}}=5\times (8-{V}_{\mathrm{m}})\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{\mathrm{m}}=\frac{8}{11}=0.727\mathrm{ft}/\mathrm{s}$

∴ Velocity of pillow, ${\overrightarrow{V}}_{\mathrm{p}}$ = 8 − 0.727 = 7.2 ft/s

The time taken to reach the floor is given by,

$t=\frac{s}{v}=\frac{8}{7.2}=1.1\mathrm{s}$

As the mass of wall is much greater than the mass of the pillow,

velocity of block before the collision = velocity after the collision

⇒ time of ascent = 1.11 s

Hence, total time taken = 1.11 + 1.11 = 2.22 s

#### Page No 163:

#### Question 54:

The track shown is figure is frictionless. The block *B* of mass 2*m* is lying at rest and the block *A* or mass *m* is pushed along the track with some speed. The collision between *A* and *B* is perfectly elastic. With what velocity should the block *A* be started to get the sleeping man awakened?

Figure

#### Answer:

Given:

Mass of the block, A = *m*

Mass of the block, B = 2*m*

Let the initial velocity of block A be *u*_{1} and the final velocity of block A,when it reaches the block B be *v*_{1}.

Using the work-energy theorem for block A, we can write:

Gain in kinetic energy = Loss in potential energy

$\therefore \left(\frac{1}{2}\right)m{v}_{1}^{2}-\left(\frac{1}{2}\right)m{u}_{1}^{2}=mgh\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}^{2}-{u}_{1}^{2}=2\mathrm{gh}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}=\sqrt{2gh+{u}_{1}^{2}}...\left(1\right)$

Let the block B just manages to reach the man's head.

*i.e.* the velocity of block B is zero at that point.

Again, applying the work-energy theorem for block B, we get:

$\left(\frac{1}{2}\right)\times 2m\times (0{)}^{2}-\left(\frac{1}{2}\right)\times 2m\times {v}^{2}=mgh\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{v}=\sqrt{2\mathrm{gh}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{before}\mathrm{the}\mathrm{collision}:\phantom{\rule{0ex}{0ex}}\mathrm{Velocity}\mathrm{of}\mathrm{A},{u}_{\mathrm{A}}={v}_{1}\phantom{\rule{0ex}{0ex}}\mathrm{Velocity}\mathrm{of}\mathrm{B},\mathit{}{u}_{\mathrm{B}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{After}\mathrm{the}\mathrm{collision}:\phantom{\rule{0ex}{0ex}}\mathrm{Velocity}\mathrm{of}\mathrm{A},{v}_{\mathrm{A}}=v(\mathrm{say})\phantom{\rule{0ex}{0ex}}\mathrm{Velocity}\mathrm{of}\mathrm{B},{v}_{\mathrm{B}}=\sqrt{2\mathrm{gh}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

As the collision is elastic, K.E. and momentum are conserved.

$m{v}_{1}+2m\times 0=mv+2m\sqrt{2gh}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}-v=2\sqrt{2gh}...\left(2\right)$

$\Rightarrow \left(\frac{1}{2}\right)m{v}_{1}^{2}+\left(\frac{1}{2}\right)2m\times (0{)}^{2}=\left(\frac{1}{2}\right)m{v}^{2}+\left(\frac{1}{2}\right)2m{\left(\sqrt{2gh}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}^{2}-{v}^{2}=2\times \sqrt{2gh}\times \sqrt{2gh}...\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{equation}(3)\mathrm{by}\mathrm{equation}(2),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}{v}_{1}+v=\sqrt{2gh}...\left(4\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\mathrm{the}\mathrm{equations}\left(4\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}2{v}_{1}=3\sqrt{2gh}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now}\mathrm{using}\mathrm{equation}\left(1\right)\mathrm{to}\mathrm{substitue}\mathrm{the}\mathrm{value}\mathrm{of}{v}_{\mathit{1}},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\sqrt{2gh+{u}^{2}}=\left(\frac{3}{2}\right)\sqrt{2gh}\phantom{\rule{0ex}{0ex}}\Rightarrow 2gh+{u}^{2}=\left(\frac{9}{4}\right)(2gh)\phantom{\rule{0ex}{0ex}}\Rightarrow u=\sqrt{2.5gh}$

Block a should be started with a minimum velocity of $\sqrt{2.5gh}$ to get the sleeping man awakened.

#### Page No 163:

#### Question 55:

A bullet of mass 10 g moving horizontally at a speed of 50√7 m/s strikes a block of mass 490 g kept on a frictionless track as shown in figure. The bullet remains inside the block and the system proceeds towards the semicircular track of radius 0.2 m. Where will the block strike the horizontal part after leaving the semicircular track?

Figure

#### Answer:

Given:

Mass of block = 490 gm

Initial speed of the block = 0

Mass of bullet = 10 gm

Initial speed of the bullet, *v*_{1 = }$50\sqrt{7}\mathrm{m}/\mathrm{s}$

As the bullet gets embedded inside the block, this is an example of a perfectly inelastic collision.

Let the final velocity of the system (block and bullet) be *V*_{A}.

Using the law of conservation of linear momentum, we get:

*$$*

m

_{1}

v

_{1}

+

m

_{2}

× 0 = (

m

_{1}

+

m

_{2}

)

v

_{A}

⇒

×

v

_{A}

∴

v

_{A}

= $\sqrt{7}$ m / s

When the block loses contact at D, the component

*mg*acts on it.

Let the velocity at D be

*v*

_{B}.

$\frac{m({v}_{\mathrm{B}}{)}^{2}}{r}=mg\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}({v}_{\mathrm{B}}{)}^{2}=gr\mathrm{sin}\theta ...(1)$

Using work energy theorem

$\left(\frac{1}{2}\right)m{v}_{\mathrm{B}}^{2}-\left(\frac{1}{2}\right)m{v}_{\mathrm{A}}^{2}=-mg(0.2+0.2\mathrm{sin}\theta )\phantom{\rule{0ex}{0ex}}\left(\frac{1}{2}\right)gr\mathrm{sin}\theta -\left(\frac{1}{2}\right){\left(\sqrt{7}\right)}^{2}=g(0.2+0.2\mathrm{sin}\theta )$

$3.5-\frac{1}{2}+9.8+0.2+\mathrm{sin}\theta =9.8+0.2(1+\mathrm{sin}\theta )\phantom{\rule{0ex}{0ex}}3.5-0.98\mathrm{sin}\theta =1.96+1.96\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}\mathrm{sin}\theta =\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \theta ={30}^{\mathrm{o}}$

∴ Angle of projection = 90° − 30° = 60°

Time of reaching the ground,

$\Rightarrow t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times (0.2+0.2\mathrm{sin}{30}^{\mathrm{o}})}{9.8}}\phantom{\rule{0ex}{0ex}}\Rightarrow t=0.247\mathrm{s}$

Distance travelled in the horizontal direction is given by,

*S*=

*v*cos

*θ*×

*t*

$$

*S*

*g*

*r*

*θ*

*θ*

*×*

*t*

*S*

*S*=

Total distance = (0.2 − 0.2 cos 30° + 0.196)

= 0.22 m

#### Page No 163:

#### Question 56:

Two balls having masses *m* and 2*m* are fastened to two light strings of same length *l*. The other ends of the strings are fixed at *O*. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. (a) Find the velocity of the balls just after their collision. (b) How high will the ball rise after the collision?

Figure

#### Answer:

Given:

Mass of the 1st ball = *m*

Mass of the 2nd ball = 2*m*

When the balls reach the lower end,

Let the velocity of *m* be *v; *and the velocity of 2*m* be *v*_{'}.

Using the work-energy theorem, we can write:

$\left(\frac{1}{2}\right)\times m{v}^{2}-\left(\frac{1}{2}\right)\times m(0{)}^{2}=mgl\phantom{\rule{0ex}{0ex}}\Rightarrow v=\sqrt{2gl}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\mathrm{velocity}\mathrm{of}\mathrm{block}\mathrm{having}\mathrm{mass}2m\mathrm{will}\mathrm{be},\phantom{\rule{0ex}{0ex}}v\text{'}=\sqrt{2gl}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore v=v\text{'}=u(\mathrm{say})\phantom{\rule{0ex}{0ex}}$

Let the velocities of *m* and 2*m* after the collision be *v*_{1} and *v*_{2} respectively.

Using the law of conservation of momentum, we can write:

$m\times u-2mu=m{v}_{1}+2m{v}_{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}+2{v}_{2}=-u...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{applying}\mathrm{collision}\mathrm{formula},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}{v}_{1}-{v}_{2}=\left[\right(u-v\left)\right]=-2u...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Substracting}\mathrm{equation}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}3{v}_{2}=u\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{2}=\frac{u}{3}=\sqrt{\frac{2gl}{3}}\phantom{\rule{0ex}{0ex}}\mathrm{Substituting}\mathrm{this}\mathrm{value}\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}{v}_{1}-{v}_{2}=-2u\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}=-2u+{v}_{2}\phantom{\rule{0ex}{0ex}}=-2u+\left(\frac{u}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{-5}{3}u=\frac{-5}{3}\times \sqrt{2gl}\phantom{\rule{0ex}{0ex}}=\frac{-\sqrt{50gl}}{3}$

(b) Let the heights reached by balls 2*m* and *m* be *h*_{1} and *h* respectively.

Using the work energy principle, we get:

$\left(\frac{1}{2}\right)\times 2m\times (0{)}^{2}-\left(\frac{1}{2}\right)\times 2m\times {\left({v}_{2}\right)}^{2}=-2m\times g\times {h}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}_{1}=\left(\frac{l}{9}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\phantom{\rule{0ex}{0ex}}\left(\frac{1}{2}\right)\times m\times (0{)}^{2}-\left(\frac{1}{2}\right)\times m{v}_{1}^{2}=m\times g\times {h}_{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{1}{2}\right)\times \frac{50gl}{9}=g\times {h}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}_{2}=\frac{25l}{9}$

As *h*_{2} is more than 2*l**,* the velocity at the highest point will not be zero.

Also, the mass *m* will rise by a distance 2*l**.*

#### Page No 163:

#### Question 57:

A uniform chain of mass *M* and length *L* is held vertically in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to test. Assuming that the chain does not form a heap on the floor, calculate the force exerted by it on the floor when a length *x* has reached the floor.

#### Answer:

Given is a uniform chain of mass *M *and length *L*.

Let us consider a small element of chain at a distance *x* from the floor having length '*dx*'.

Therefore, mass of element, $\mathrm{d}m=\frac{M}{L}\mathrm{d}x$ ...(1)

The velocity with which the element strikes the floor is,

$v=\sqrt{2gx}$ ...(2)

∴ The momentum transferred to the floor is,

$P=\left(\mathrm{d}m\right)v=\frac{M}{L}.\mathrm{d}x\sqrt{2gx}$

According to the given condition, the element comes to rest.

Thus, the force (say *F*_{1}) exerted on the floor = Change in momentum

${F}_{1}=\frac{\mathrm{d}P}{\mathrm{d}t}=\frac{M}{L}\times \frac{\mathrm{d}x}{\mathrm{d}t}\sqrt{2gx}\phantom{\rule{0ex}{0ex}}(\mathrm{for}\mathrm{the}\mathrm{chain}\mathrm{element})\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}As\frac{\mathrm{d}x}{\mathrm{d}t}=v,\phantom{\rule{0ex}{0ex}}{F}_{1}=\frac{M}{L}\times {\left(\sqrt{2gx}\right)}^{2}\left[\mathrm{using}\mathrm{equation}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {F}_{1}=\frac{M}{L}2gx=\frac{2Mgx}{L}$

Again, the force exerted due to length *x* of the chain on the floor due to its own weight is given by,

$W=\frac{M}{L}\left(x\right)\times g=\frac{Mgx}{L}$

Thus, the total force exerted is given by.

*F* = *F*_{1} + *W*

where *W* is the weight of chain below the element d*x* up to the floor.

$\Rightarrow F=\frac{2Mgx}{L}+\frac{Mgx}{L}=\frac{3Mgx}{L}$

#### Page No 163:

#### Question 58:

The blocks shown in figure have equal masses. The surface of *A* is smooth but that of *B* has a friction coefficient of 0.10 with the floor. Block *A* is moving at a speed of 10 m/s towards *B* which is kept at rest. Find the distance travelled by *B* if (a) the collision is perfectly elastic and (b) the collision is perfectly inelastic.

Figure

#### Answer:

Given,

Speed of the block A = 10 m/s

The block B is kept at rest.

Coefficient of friction between floor and block B, *μ* = 0.10

Lets *v*_{1} and *v*_{2} be the velocities of A and B after collision respectively.

(a) If the collision is perfectly elastic, linear momentum is conserved.

Using the law of conservation of linear momentum, we can write:

$m{u}_{1}+m{u}_{2}=m{v}_{1}+m{v}_{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 10+0={v}_{1}+{v}_{2}\phantom{\rule{0ex}{0ex}}{v}_{1}+{v}_{2}=10...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know},\phantom{\rule{0ex}{0ex}}\mathrm{Velocity}\mathrm{of}\mathrm{separation}\left(\mathrm{after}\mathrm{collision}\right)=\mathrm{Velocity}\mathrm{of}\mathrm{approach}\left(\mathrm{before}\mathrm{collision}\right)\phantom{\rule{0ex}{0ex}}{v}_{1}-{v}_{2}=-({u}_{1}-{v}_{2})\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}-{v}_{2}=-10...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Substracting}\mathrm{equation}\left(2\right)\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}2{v}_{2}=20\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{2}=10\mathrm{m}/\mathrm{s}$

The deceleration of block B is calculated as follows:

Applying the work energy principle, we get:

$\left(\frac{1}{2}\right)\times m\times (0{)}^{2}-\left(\frac{1}{2}\right)\times m\times {v}^{2}=-m\times a\times {s}_{1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow -\left(\frac{1}{2}\right)\times (10{)}^{2}=-\mu g\times {s}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {s}_{1}=\frac{100}{2\times 1\times 10}=50\mathrm{m}$

(b) If the collision is perfectly inelastic, we can write:

$m\times {u}_{1}+m\times {u}_{2}=(m+m)\times v\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow m\times 10+m\times 0=2m\times v\phantom{\rule{0ex}{0ex}}\Rightarrow v=\left(\frac{10}{2}\right)=5\mathrm{m}/\mathrm{s}$

The two blocks move together, sticking to each other.

∴ Applying the work-energy principle again, we get:

$\left(\frac{1}{2}\right)\times 2m\times (0{)}^{2}-\left(\frac{1}{2}\right)\times 2m\times (v{)}^{2}=2\mathrm{m}\times \mathrm{\mu}g\times {s}_{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{(5{)}^{2}}{0.1\times 10\times 2}={s}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {s}_{2}=12.5\mathrm{m}$

#### Page No 163:

#### Question 59:

The friction coefficient between the horizontal surface and each of the block shown in figure is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest.

Figure

#### Answer:

Given:

Initial velocity of 2 kg block, *v*_{1}_{ }= 1.0 m/s

Initial velocity of the 4 kg block, *v*_{2} = 0

Let the velocity of 2 kg block, just before the collision be *u*_{1}.

Using the work-energy theorem on the block of 2 kg mass:

The separation between two blocks, s = 16 cm = 0.16 m

$\therefore \left(\frac{1}{2}\right)m\times {u}_{1}^{2}-\left(\frac{1}{2}\right)m\times (1{)}^{2}=-\mu \times mg\times s\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {u}_{\mathit{1}}=\sqrt{(1{)}^{2}-2\times 0.20\times 10\times 0.16}\phantom{\rule{0ex}{0ex}}\Rightarrow {u}_{1}=0.6\mathrm{m}/\mathrm{s}$

As the collision is perfectly elastic, linear momentum is conserved.

Let *v*_{1}, *v*_{2} be the velocities of 2 kg and 4 kg blocks, just after collision.

Using the law of conservation of linear momentum, we can write:

${m}_{1}{u}_{1}+{m}_{2}{u}_{2}={m}_{1}{v}_{1}+{m}_{2}{v}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times 0.6+4\times 0=2{v}_{1}+4{v}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2{v}_{1}+4{v}_{2}=1.2...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

For elastic collision,

Velocity of separation (after collision) = Velocity of approach (before collision)

$i\mathit{.}e\mathit{.}{v}_{1}-{v}_{2}=+({u}_{1}-{u}_{2})\phantom{\rule{0ex}{0ex}}=+(0.6-0)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}-{v}_{2}=-0.6...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Substracting}\mathrm{equation}\left(2\right)\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}3{v}_{2}=1.2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{2}=0.4m/s\phantom{\rule{0ex}{0ex}}\therefore {v}_{1}=-0.6+0.4=-0.2\mathrm{m}/\mathrm{s}$

Let the 2 kg block covers a distance of *S*_{1}.

∴ Applying work-energy theorem for this block, when it comes to rest:

$\left(\frac{1}{2}\right)\times 2\times (0{)}^{2}+\left(\frac{1}{2}\right)\times 2\times (0.2{)}^{2}=-2\times 0.2\times 10\times {\mathrm{S}}_{1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{S}}_{1}=1\mathrm{cm}.$

Let the 4 kg block covers a distance of *S*_{2}.

Applying work energy principle for this block:

$\left(\frac{1}{2}\right)\times 4\times (0{)}^{2}-\left(\frac{1}{2}\right)\times 4\times (0.4{)}^{2}=-4\times 0.2\times 10\times {\mathrm{S}}_{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times 0.4\times 0.4=4\times 0.2\times 10\times {\mathrm{S}}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{S}}_{2}=4\mathrm{cm}$

Therefore, the distance between the 2 kg and 4 kg block is given as,

S_{1} + S_{2} = 1 + 4 = 5 cm

#### Page No 163:

#### Question 60:

A block of mass *m* is placed on a triangular block of mass *M*, which in turn is placed on a horizontal surface as shown in figure. Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.

Figure

#### Answer:

According to the question, the surface is frictionless. Thus, the block *m* will slide down the inclined plane of mass *M*.

Acceleration, *a*_{1} = *g *sin α (Relative to the inclined plane)

The horizontal component of acceleration *a*_{1} is given by *a _{x}* =

*g*sin α cos α, for which the block

*M*accelerates towards left.

Let the left acceleration be

*a*

_{2}.

By the concept of centre of mass, we can say that the external force is zero in the horizontal direction.

$m{a}_{x}=(M+m){a}_{2}$

Absolute (resultant) acceleration of

*m*on the plane

*M,*along the direction of the incline will be = $a=g\mathrm{sin}\alpha -{a}_{2}\mathrm{cos}\alpha \phantom{\rule{0ex}{0ex}}$

Let the time taken by the block

*m*to reach the bottom end be

*t.*

Now,

$s=ut+\left(\frac{1}{2}\right)a{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{h}{\mathrm{sin}\alpha}=\left(\frac{1}{2}\right)a{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\sqrt{\frac{2h}{a\mathrm{sin}\alpha}}$

Thus, the velocity of the bigger block after time

*t*will be,

${v}_{m}=u={a}_{2}t\phantom{\rule{0ex}{0ex}}=\frac{mg\mathrm{sin}\alpha \mathrm{cos}\alpha}{M+m}\sqrt{\frac{2h}{a\mathrm{sin}\alpha}}={\left[\frac{2{m}^{2}{g}^{2}h{\mathrm{sin}}^{2}\alpha {\mathrm{cos}}^{2}\alpha}{(M+m{)}^{2}a\mathrm{sin}\alpha}\right]}^{1/2}$

Subtracting the value of

*a*from equation (2), we get:

${v}_{\mathrm{M}}=\left[\frac{2{m}^{2}{g}^{2}h{\mathrm{sin}}^{2}\alpha}{(M+m{)}^{2}\mathrm{sin}\alpha}\times \frac{{\mathrm{cos}}^{2}\alpha}{g\mathrm{sin}\alpha}\frac{(M+m)}{(M+m{\mathrm{sin}}^{2}\alpha )}\right]$

#### Page No 163:

#### Question 61:

Figure shows a small body of mass *m* placed over a larger mass *M* whose surface is horizontal near the smaller mass and gradually curves to become vertical. The smaller mass is pushed on the longer one at a speed *v* and${v}_{m}=u={a}_{2}t\phantom{\rule{0ex}{0ex}}=\frac{mg\mathrm{sin}\alpha \mathrm{cos}\alpha}{\mathrm{M}+m}\sqrt{\frac{2h}{a\mathrm{sin}\alpha}}={\left[\frac{2{m}^{2}{g}^{2}h{\mathrm{sin}}^{2}\alpha c{o}^{2}\alpha}{(\mathrm{M}+m{)}^{2}a\mathrm{sin}\alpha}\right]}^{1/2}$ the system is left to itself. Assume that all the surface are frictionless. (a) Find the speed of the larger block when the smaller block is sliding on the vertical part. (b) Find the speed of the smaller mass when it breaks off the larger mass at height *h*. (c) Find the maximum height (from the ground) that the smaller mass ascends. (d) Show that the smaller mass will again land on the bigger one. Find the distance traversed by the bigger block during the time when the smaller block was in its flight under gravity.

Figure

#### Answer:

,

According to the question, mass *m* is given with a speed *v* over the larger mass *M*.

(a) When the smaller block travels on the vertical part, let the velocity of the bigger block be *v*_{1}, towards left.

From the law of conservation of momentum (in the horizontal direction), we get:

$mv=(\mathrm{M}+m){v}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}=\frac{mv}{M+m}$

(b) When the smaller block breaks off, let its resultant velocity be *v*_{2}.

Using the law a of conservation of energy, we get:

$\left(\frac{1}{2}\right)m{v}^{2}=\left(\frac{1}{2}\right)\mathrm{M}{v}_{1}^{2}+\left(\frac{1}{2}\right)m{v}_{2}^{2}+\mathrm{mgh}\phantom{\rule{0ex}{0ex}}$

(c) Vertical component of the velocity *v*_{2} of mass *m* is given by,

${{v}_{y}}^{2}={{v}_{2}}^{2}-{{v}_{1}}^{2}\phantom{\rule{0ex}{0ex}}$

To determine the maximum height (from the ground),

Let us assume that the body rises to a height *h*_{1} over and above *h*.

$\mathrm{Now},\left(\frac{1}{2}\right)m{v}_{y}^{2}=mg{h}_{1}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow H=\frac{M{v}^{2}}{(M+m)2g}$

(d) Because the smaller mass also has a horizontal component of velocity *V*_{1} at the time it breaks off from M (that has a velocity *v*_{1}), the block *m* will again land on the block M.

The time of flight of block *m* after it breaks off is calculated as:

During the upward motion (BC),

$0={v}_{y}-gt\phantom{\rule{0ex}{0ex}}$

Thus, the time for which the smaller block is in flight is given by,

$T=2{t}_{1}\phantom{\rule{0ex}{0ex}}=\frac{2}{g}\left[\frac{M{v}^{2}-2(M+m)gh}{(M+m)}\right]$

The distance travelled by the bigger block during this time is,

$S={v}_{1}T$

#### Page No 164:

#### Question 62:

A small block of superdense material has a mass of 3 × 10^{24}kg. It is situated at a height *h* (much smaller than the earth's radius) from where it falls on the earth's surface. Find its speed when its height from the earth's surface has reduce to to *h*/2. The mass of the earth is 6 × 10^{24}kg.

#### Answer:

It is given that *h* is much lesser than the radius of the earth.

Mass of the earth, *M*_{e} = 6 × 10^{24} kg

Mass of the block, *M _{b}* = 3 × 10

^{24}kg

Let

*V*

_{e}be the velocity of earth and

*V*

_{b}be the velocity of the block.

Let the earth and the block be attracted by gravitational force.

Thus, according to the conservation law of energy, the change in the gravitational potential energy will be the K.E. of block.

$G{M}_{e}{M}_{b}\left(\frac{1}{R+\left({\displaystyle \frac{h}{2}}\right)}-\frac{1}{R+h}\right)=\left(\frac{1}{2}\right){M}_{e}\times {V}_{e}^{2}+\left(\frac{1}{2}\right){\mathrm{M}}_{b}\times {\mathrm{V}}_{b}^{2}$ ...(1)

As only the internal force acts in this system, the momentum is conserved.

*M*

_{e}

*V*

_{e}=

*M*

_{b}

*V*

_{b}

$\Rightarrow {V}_{e}=\frac{{M}_{b}{V}_{b}}{{M}_{e}}...\left(ii\right)\phantom{\rule{0ex}{0ex}}$

#### Page No 164:

#### Question 63:

A body of mass *m* makes an elastic collision with another identical body at rest. Show that if the collision is not head-on, the bodies go at right angle to each other after the collision.

#### Answer:

According to the question, the collision of the two bodies of mass *m *is not head-on. Thus, the two bodies move in different directions.

Let the velocity vectors of the two bodies after collision be *v*_{1} and *v*_{2}.

As the collision in the question is elastic, momentum is conserved.

On applying the law of conservation of momentum in X-direction, we get:

$m{u}_{1}+m\times 0=m{v}_{1}\mathrm{cos}\alpha +m{v}_{2}\mathrm{cos}\mathrm{\beta}\mathit{.}\mathit{.}\mathit{.}\mathit{\left(}i\mathit{\right)}\phantom{\rule{0ex}{0ex}}$

On applying the law of conservation of momentum in Y-direction, we get:

$0=m{v}_{1}\mathrm{sin}\alpha -m{v}_{2}\mathrm{sin}\mathrm{\beta}...\left(ii\right)\phantom{\rule{0ex}{0ex}}$

#### Page No 164:

#### Question 64:

A small particle travelling with a velocity *v* collides elastically with a spherical body of equal mass and of radius *r* initially kept at rest. The centre of this spherical body is located a distance ρ(< *r*) away from the direction of motion of the particle. Find the final velocities of the two particles.

Figure

#### Answer:

It is given that the mass of both the bodies (small particle and spherical body) is same.

Let the velocity of the particle be *v*.

We break the particle velocity into two components: *v *cosα (normal to the sphere) and *v *sinα (tangential to the sphere).

From figure,

cosα$=\sqrt{{r}^{2}-{\rho}^{2}}$ and sinα $=\frac{\rho}{r}$

Here, the collision occurs due to the component *v *cosα.

After an elastic collision, bodies of same mass exchange their respective velocities.

Thus, the spherical body will have a velocity *v *cosα, while the particle will not have any component of velocity in this direction.

However, the tangential velocity of the particle *v *sinα will remain unaffected.

Thus, we have:

Velocity of the sphere = *v *cosα$=\frac{v}{r}\sqrt{{r}^{2}-{\rho}^{2}}$

And,

Velocity of the particle = *v *sinα $=\frac{v\rho}{r}$

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