# Hc Verma i Solutions for Class 11 Science Physics Chapter 5 - Newton's Laws Of Motion

Hc Verma i Solutions for Class 11 Science Physics Chapter 5 Newton's Laws Of Motion are provided here with simple step-by-step explanations. These solutions for Newton's Laws Of Motion are extremely popular among class 11 Science students for Physics Newton's Laws Of Motion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma i Book of class 11 Science Physics Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Hc Verma i Solutions. All Hc Verma i Solutions for class 11 Science Physics are prepared by experts and are 100% accurate.

#### Page No 76:

#### Question 1:

The apparent weight of an object increases in an elevator while accelerating upward. A person sells peanuts using a beam balance in an elevator. Will he gain more if the elevator is accelerating up?

#### Answer:

No, the accelerating elevator will affect the weight of both sides of the beam balance. So, the net effect of the accelerating elevator cancels out, and we get the actual mass.

#### Page No 76:

#### Question 2:

A boy puts a heavy box of mass *M* on his head and jumps from the top of a multi-storied building to the ground. What is the force exerted by the box on the boy's head during his free fall? Does the force greatly increase during the period he balances himself after striking the ground?

#### Answer:

During free fall:

Acceleration of the boy = Acceleration of mass *M* =* g *

Acceleration of mass *M* w.r.t. boy, *a* = 0

So, the force exerted by the box on the boy's head = *M × a* = 0

Yes, the force greatly increases during the period he balances himself after striking the ground because of the weight of the box.

#### Page No 76:

#### Question 3:

A person drops a coin. Describe the path of the coin as seen by the person if he is in (a) a car moving at constant velocity and (b) in a free falling elevator.

#### Answer:

(a) In the car, the path of the coin will be vertically downward because the only force acting on the coin is gravity in the downward direction.

(b) In a free falling elevator, the coin as well as the person will be in a condition of weightlessness. So, the coin will remain stationary w.r.t. the person.

#### Page No 76:

#### Question 4:

Is it possible for a particle to describe a curved path if no force acts on it? Does your answer depend on the frame of reference chosen to view the particle?

#### Answer:

If no force acts on the particle it cannot change its direction. So, it is not possible for a particle to describe a curved path if no force acts on it.

Yes, the answer depends on the frame of reference chosen to view the particle if the frame of reference describes a curved path.

#### Page No 76:

#### Question 5:

You are travelling in a car. The driver suddenly applies the brakes and you are pushed forward. Why does this happen?

#### Answer:

We are pushed forward because of the inertia of motion, as our body opposes the sudden change.

#### Page No 76:

#### Question 6:

It is sometimes heard that the inertial frame of reference is only an ideal concept and no such inertial frame actually exists. Comment.

#### Answer:

We can't find a body whose acceleration is zero with respect to all other bodies in the universe because every body in the universe is moving with respect to other bodies.As we live on earth which itself is accelerates due to its revolution around the sun and spinning about its own axis, so whatever observations and measurements ,we make , are w.r.t to earth which itself is not an inertial frame.Similarly all other planets are also in motion around the sun so tdeally no inertial frame is possible.

#### Page No 76:

#### Question 7:

An object is placed far away from all the objects that can exert force on it. A frame of reference is constructed by taking the origin and axes fixed in this object. Will the frame be necessarily inertial?

#### Answer:

Yes, if the force on the object is zero, its acceleration w.r.t. all the other objects will we zero. So, the frame will necessarily be an inertial frame.

#### Page No 76:

#### Question 8:

The figure shows a light spring balance connected to two blocks of mass 20 kg each. The graduations in the balance measure the tension in the spring. (a) What is the reading of the balance? (b) Will the reading change if the balance is heavy, say 2.0 kg? (c) What will happen if the spring is light but the blocks have unequal masses?

Figure

#### Answer:

(a)

The reading of the balance = Tension in the string

And tension in the string = 20*g*

So, the reading of the balance = 20*g** =* 200 N

(b) If the balance is heavy, the reading will not change because the weight of spring balance does not affect the tension in the string.

(c) If the blocks have unequal masses, the spring balance will accelerate towards the heavy block with an acceleration *a. *Then the reading will be equal to the tension in the string.

Suppose *m*_{1}_{ > }*m*_{2}.

Then tension in the string,

$T=\frac{2{m}_{1}{m}_{2}g}{{m}_{1}+{m}_{2}}$

#### Page No 77:

#### Question 9:

The acceleration of a particle is zero, as measured from an inertial frame of reference. Can we conclude that no force acts on the particle?

#### Answer:

No. The acceleration of the particle can also be zero if the vector sum of all the forces is zero, i.e. no net force acts on the particle.

#### Page No 77:

#### Question 10:

Suppose you are running fast in a field and suddenly find a snake in front of you. You stop quickly. Which force is responsible for your deceleration?

#### Answer:

The force of friction acting between my feet and ground is responsible for my deceleration.

#### Page No 77:

#### Question 11:

If you jump barefoot on a hard surface, your legs are injured. But they are not injured if you jump on a soft surface like sand or pillow. Why?

#### Answer:

In both the cases, change in momentum is same but the time interval during which momentum changes to zero is less in the first case. So, by $F=\frac{dP}{dt}$ , force in the first case will be more. That's why we are injured when we jump barefoot on a hard surface.

#### Page No 77:

#### Question 12:

According to Newton's third law, each team pulls the opposite team with equal force in a tug of war. Then, why does one team win and the other lose?

#### Answer:

The forces on the rope must be equal and opposite, according to Newton's third law. But not all the forces acting on each team are equal. The friction between one team and the ground does not depend on the other team and can be larger on one side than on the other. In addition, the grips on the rope need not be equal and opposite. Thus, the net force acting on each team from all sources need not be equal.

#### Page No 77:

#### Question 13:

A spy jumps from an airplane with his parachute. The spy accelerates downward for some time when the parachute opens. The acceleration is suddenly checked and the spy slowly falls to the ground. Explain the action of the parachute in checking the acceleration.

#### Answer:

Air applies a velocity-dependent force on the parachute in upward direction when the parachute opens. This force opposes the gravitational force acting on the spy. Hence, the net force in the downward direction decreases and the spy decelerates.

#### Page No 77:

#### Question 14:

Consider a book lying on a table. The weight of the book and the normal force by the table on the book are equal in magnitude and opposite in direction. Is this an example of Newton's third law?

#### Answer:

Yes, this is an example of Newton's third law of motion, which sates that every action has an equal and opposite reaction.

#### Page No 77:

#### Question 15:

Two blocks of unequal masses are tied by a spring. The blocks are pulled stretching the spring slightly and the system is released on a frictionless horizontal platform. Are the forces due to the spring on the two blocks equal and opposite? If yes, is it an example of Newton's third law?

#### Answer:

Yes, the forces due to the spring on the two blocks are equal and opposite.

But it's not an example of Newton's third law because there are three objects (2 blocks + 1 spring). Spring force on one block and force by the same block on the spring is an action-reaction pair.

#### Page No 77:

#### Question 16:

When a train starts, the head of a standing passenger seems to be pushed backward. Analyse the situation from the ground frame. Does it really go backward? Coming back to the train frame, how do you explain the backward movement of the head on the basis of Newton's laws?

#### Answer:

No, w.r.t. the ground frame, the person's head is not really pushed backward.

As the train moves, the lower portion of the passenger's body starts moving with the train, but the upper portion tries to be in rest according to Newton's first law and hence, the passenger seems to be pushed backward.

#### Page No 77:

#### Question 17:

A plumb bob is hung from the ceiling of a train compartment. If the train moves with an acceleration '*a*' along a straight horizontal track , the string supporting the bob makes an angle tan^{−1} (*a*/*g*) with the normal to the ceiling. Suppose the train moves on an inclined straight track with uniform velocity. If the angle of incline is tan^{−1} (*a*/*g*), the string again makes the same angle with the normal to the ceiling. Can a person sitting inside the compartment tell by looking at the plumb line whether the train is accelerating on a horizontal straight track or moving on an incline? If yes, how? If not, then suggest a method to do so.

#### Answer:

No, a person sitting inside the compartment can't tell just by looking at the plumb line whether the train is accelerating on a horizontal straight track or moving on an incline.

When the train is accelerating along the horizontal, the tension in the string is $m\sqrt{{g}^{2}+{a}^{2}}$; when it is moving on the inclined plane, the tension is* mg*. So, by measuring the tension in the string we can differentiate between the two cases.

#### Page No 77:

#### Question 1:

A body of weight *w*_{1} is suspended from the ceiling of a room by a chain of weight w_{2}. The ceiling pulls the chain by a force

(a) *w*_{1}

(b) w_{2}

(c) *w*_{1}* + *w_{2}

(d) $\frac{{w}_{1}+{w}_{2}}{2}$

#### Answer:

(c) *w*_{1}* + *w_{2}

From the free-body diagram,

(*w*_{1} + *w*_{2}) – *N *= 0

*N = **w*_{1}* + **w*_{2}

The ceiling pulls the chain by a force (*w*_{1} + *w*_{2}).

#### Page No 77:

#### Question 2:

When a horse pulls a cart, the force that helps the horse to move forward is the force exerted by

(a) the cart on the horse

(b) the ground on the horse

(c) the ground on the cart

(d) the horse on the ground

#### Answer:

(b) the ground on the horse

The horse pushes the ground in the backward direction and, in turn, the ground pushes the horse in the forward direction, according to Newton's third law of motion.

#### Page No 77:

#### Question 3:

A car accelerates on a horizontal road due to the force exerted by

(a) the engine of the car

(b) the driver of the car

(c) the earth

(d) the road

#### Answer:

(d) the road

The car pushes the ground in the backward direction and according to the third law of motion, reaction force of the ground in the forward direction acts on the car.

#### Page No 77:

#### Question 4:

A block of mass 10 kg is suspended from two light spring balances, as shown in the figure.

Figure

(a) Both the scales will read 10 kg.

(b) Both the scales will read 5 kg.

(c) The upper scale will read 10 kg and the lower zero.

(d) The readings may be anything but their sums will be 10 kg.

#### Answer:

(a) Both the scales will read 10 kg.

From the free-body diagram,

*K*_{1}*x*_{1} = *mg* = 10$\times $9.8 = 98 N

*K*_{2}*x*_{2} = *K*_{1}*x*_{1}

So, *K*_{1}*x*_{1} = *K*_{2}*x*_{2} = 98 N

Therefore, both the spring balances will read the same mass, i.e. 10 kg.

#### Page No 77:

#### Question 5:

A block of mass *m* is placed on a smooth inclined plane of inclination θ with the horizontal. The force exerted by the plane on the block has a magnitude

(a) *mg*

(b) *mg*/cosθ

(c) *mg* cosθ

(d) *mg* tanθ

#### Answer:

(c) *mg* cosθ

From the free-body diagram,

*N = mg *cosθ

Normal force exerted by the plane on the block is *mg* cosθ.

#### Page No 77:

#### Question 6:

A block of mass *m* is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude

(a) *mg*

(b) *mg*/cosθ

(c) *mg* cosθ

(d) *mg* tanθ

#### Answer:

(b) *mg*/cosθ

Free-body Diagram of the Small Block of Mass '*m*'

The block is at equilibrium w.r.t. to wedge. Therefore,

*mg* sin*θ* = *ma *cosθ

*⇒ a = **g*tan*θ*

Normal reaction on the block is

*N = mg *cos*θ** *+* **ma *sin*θ*

Putting the value of *a,* we get:

*N = mg *cos*θ** + mg *tan*θ*sin*θ*

*$N=mg\mathrm{cos}\theta +mg\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}\mathrm{sin}\theta N=\frac{mg}{\mathrm{cos}\theta}$*

#### Page No 77:

#### Question 7:

Neglect the effect of rotation of the earth. Suppose the earth suddenly stops attracting objects placed near its surface. A person standing on the surface of the earth will

(a) fly up

(b) slip along the surface

(c) fly along a tangent to the earth's surface

(d) remain standing

#### Answer:

(d) remain standing

If the earth suddenly stops attracting objects placed near its surface, the net force on the person will become zero and according to the first law of motion, the person will remain standing.

#### Page No 77:

#### Question 8:

Three rigid rods are joined to form an equilateral triangle *ABC* of side 1 m. Three particles carrying charges 20 μC each are attached to the vertices of the triangle. The whole system is at rest in an inertial frame. The magnitude of the resultant force on the charged particle at A is

(a) zero

(b) 3.6 N

(c) 3.6√3 N

(d) 7.2 N

#### Answer:

(a) zero

Using, *F*_{net}_{ }=* ma*,

*a* = 0 *⇒ **F*_{net}_{ = 0}

As the whole system is at rest, the resultant force on the charged particle at A is zero.

#### Page No 78:

#### Question 9:

A force *F*_{1} acts on a particle accelerating it from rest to a velocity *v*. Force *F*_{1} is then replaced by *F*_{2} which decelerates the particle to rest.

(a) *F*_{1} must be equal to *F*_{2}.

(b) *F*_{1} may be equal to *F*_{2}.

(c) *F*_{1} must be unequal to *F*_{2}.

(d) None of these.

#### Answer:

(b) *F*_{1} may be equal to *F*_{2}.

Any force applied in the direction opposite the motion of the particle decelerates it to rest.

#### Page No 78:

#### Question 10:

Two objects A and B are thrown upward simultaneously with the same speed. The mass of A is greater than that of B. Suppose the air exerts a constant and equal force of resistance on the two bodies.

(a) The two bodies will reach the same height.

(b) *A* will go higher than *B.*

*B*will go higher than

*A*.

(d) Any of the above three may happen depending on the speed with which the objects are thrown.

#### Answer:

(b) *A* will go higher than *B*.

Let the air exert a constant resistance force = F (in downward direction).

Acceleration of particle A in downward direction due to air resistance, *a*_{A} = *F*/*m*_{A}.

Acceleration of particle B in downward direction due to air resistance,* **a*_{B} = *F*/*m*_{B}.

*m*_{A} > *m*_{B}

*a*_{A} < *a*_{B}

$S=ut+\frac{1}{2}a{t}^{2}$

$\mathrm{So},{H}_{A}=ut-\frac{1}{2}({a}_{A}+g){t}^{2}$

${H}_{B}=ut-\frac{1}{2}({a}_{B}+g){t}^{2}$

${H}_{A}>{H}_{B}$

Therefore, *A* will go higher than *B. *

#### Page No 78:

#### Question 11:

A smooth wedge* A* is fitted in a chamber hanging from a fixed ceiling near the earth's surface. A block* B* placed at the top of the wedge takes time *T* to slide down the length of the wedge. If the block is placed at the top of the wedge and the cable supporting the chamber is broken at the same instant, the block will

(a) take a time longer than *T* to slide down the wedge

(b) take a time shorter than *T* to slide down the wedge

(c) remain at the top of the wedge

(d) jump off the wedge

#### Answer:

(c) remain at the top of the wedge

Downward gravitational force will be balanced by the upward pseudo force (because of the motion of the wedge in downward direction). The block will remain at its position, as both the box and the inclined plane are falling with the same acceleration (*g*).

#### Page No 78:

#### Question 12:

In an imaginary atmosphere, the air exerts a small force *F* on any particle in the direction of the particle's motion. A particle of mass *m* projected upward takes time *t*_{1} in reaching the maximum height and *t*_{2} in the return journey to the original point. Then

(a) *t*_{1} < *t*_{2}

(b) *t*_{1} > *t*_{2}

(c) *t*_{1} = *t*_{2}

(d) the relation between *t*_{1} and *t*_{2} depends on the mass of the particle

#### Answer:

(b) *t*_{1} > *t*_{2}

Let acceleration due to air resistance force be *a. *

Let *H* be maximum height attained by the particle.

Direction of air resistance force is in the direction of motion.

In the upward direction of motion, ${a}_{\mathrm{eff}}=\left|g-a\right|$.

${t}_{1}=\sqrt{\frac{2H}{\left|g-a\right|}}...\left(1\right)$

In the downward direction of motion, ${a}_{\mathrm{eff}}=g+a$.

${t}_{2}=\sqrt{\frac{2H}{g+a}}...\left(2\right)$

So, *t*_{1} > *t*_{2}.

#### Page No 78:

#### Question 13:

A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in time *t*_{1} if the elevator is stationary and in time *t*_{2} if it is moving uniformly. Then

(a) *t*_{1} = *t*_{2}

(b) *t*_{1} < *t*_{2}

(c) *t*_{1} > *t*_{2}

(d) *t*_{1} < *t*_{2} or *t*_{1} > *t*_{2} depending on whether the lift is going up or down.

#### Answer:

(a) *t*_{1} = *t*_{2}

After the coin is dropped, the only force acting on it is gravity, which is same for both the cases.

So *t*_{1} = *t*_{2}.

#### Page No 78:

#### Question 14:

A free ${}^{238}U$ nucleus kept in a train emits an alpha particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes* x* at time *t* after the decay. If the decay takes place while the train is moving at a uniform velocity *v,* the distance between the alpha particle and the recoiling nucleus at a time *t* after the decay, as measured by the passenger, is

(a) *x* + *v t*

(b) *x* − *v t*

(c) *x*

(d) depends on the direction of the train

#### Answer:

(c) *x*

The moving train does not put any extra force on the alpha particle and the recoiling nucleus. So, the distance between the alpha particle and the recoiling nucleus at a time *t* after the decay, as measured by the passenger, will be same as before, i.e.* x*.

#### Page No 78:

#### Question 1:

A reference frame attached to the earth

(a) is an inertial frame by definition

(b) cannot be an inertial frame because the earth is revolving around the sun

(c) is an inertial frame because Newton's laws are applicable in this frame

(d) cannot be an inertial frame because the earth is rotating about its axis

#### Answer:

(b) cannot be an inertial frame because the earth is revolving around the sun

(d) cannot be an inertial frame because the earth is rotating about its axis

A reference frame attached to the earth cannot be an inertial frame because the earth is revolving around the sun and also rotating about its axis.

#### Page No 78:

#### Question 2:

A particle stays at rest as seen in a frame. We can conclude that

(a) the frame is inertial

(b) resultant force on the particle is zero

(c) the frame may be inertial but the resultant force on the particle is zero

(d) the frame may be non-inertial but there is a non-zero resultant force

#### Answer:

(c) the frame may be inertial but the resultant force on the particle is zero

(d) the frame may be non-inertial but there is a non-zero resultant force

According to Newton's second law which says that net force acting on the particle is equal to rate of change of momentum ( or mathematically F = ma), so if a particle is at rest then *F*_{net} = ma = m$\frac{dv}{dt}=m\frac{d\left(0\right)}{dt}=m\times 0=0$.

Now, if the frame is inertial, then the resultant force on the particle is zero.

If the frame is non-inertial,

vector sum of all the forces plus a pseudo force is zero.

i.e. *F*_{net} ≠ 0.

#### Page No 78:

#### Question 3:

A particle is found to be at rest when seen from a frame *S*_{1} and moving with constant velocity when seen from another frame *S*_{2}. Mark out the possible options.

(a) Both the frames are inertial.

(b) Both the frames are non-inertial.

(c) *S*_{1} is inertial and *S*_{2} is non-inertial.

(d) *S*_{1} is non-inertial and *S*_{2} is inertial

#### Answer:

(a) Both the frames are inertial.

(b) Both the frames are non-inertial.

*S*_{1} is moving with constant velocity w.r.t frame *S*_{2}. So, if *S*_{1}_{ }is inertial, then *S*_{2}_{ }will be inertial and if *S*_{1}_{ }is non-inertial, then *S*_{2}_{ }will be non-inertial.

#### Page No 78:

#### Question 4:

The figure shows the displacement of a particle going along the *X*-axis as a function of time. The force acting on the particle is zero in the region

(a) *AB*

(b) *BC*

(c) *CD*

(d) *DE*

Figure

#### Answer:

(a) *AB*

(c) *CD*

Slope of the *x-t* graph gives velocity. In the regions *AB* and *CD,* slope or velocity is constant, i.e. acceleration is zero. Hence, from the second law, force is zero in these regions.

#### Page No 78:

#### Question 5:

The figure shows a heavy block kept on a frictionless surface and being pulled by two ropes of equal mass *m*. At *t* = 0, the force on the left rope is withdrawn but the force on the right end continues to act. Let *F*_{1} and *F*_{2} be the magnitudes of the forces exerted by right rope and the left rope on the block, respectively.

Figure

(a) *F*_{1}_{ }= *F*_{2} = *F* for *t* < 0

(b) *F*_{1}_{ }= *F*_{2} = *F* + *mg* for *t* < 0

(c) *F*_{1}_{ }= *F*, F_{2} = *F* for *t* > 0

(d) *F*_{1} < *F*, F_{2} = *F* for *t* > 0

#### Answer:

(a) *F*_{1}_{ }= *F*_{2} = *F*, for *t* < 0

At *t *< 0, the block is in equilibrium in the horizontal direction.

So, *F*_{1}_{ }= *F*_{2} = *F*

At *t* > 0, *F*_{2}_{ }= 0 and *F*_{1}_{ }= *F*.

#### Page No 79:

#### Question 6:

A particle of mass 50 g moves in a straight line. The variation of speed with time is shown in figure (5−E1). Find the force acting on the particle at* t* = 2, 4 and 6 seconds.

Figure

#### Answer:

Given:

Mass of the particle,* m* = 50 g = 5 × 10^{−2} kg

Slope of the* v-t* graph gives acceleration.

At *t* = 2 s,

Slope = $\frac{15}{3}=5\mathrm{m}/{\mathrm{s}}^{2}$

So, acceleration, *a* = 5 m/s^{2}

*F* = *ma* = 5 × 10^{−2} × 5

⇒ *F* = 0.25 N along the motion.

At *t *= 4 s,

Slope = 0

So, acceleration,* a* = 0

⇒ *F* = 0

At *t* = 6 sec,

Slope = $\frac{-15}{3}=-5\mathrm{m}/{\mathrm{s}}^{2}$

So, acceleration, *a* = − 5 m/s^{2}

*F* = *ma* = − 5 × 10^{−2} × 5

⇒ *F* = − 0.25 N along the motion

or, *F *= 0.25 N opposite the motion.

#### Page No 79:

#### Question 7:

Two blocks A and B of mass *m*_{A} and *m*_{B} , respectively, are kept in contact on a frictionless table. The experimenter pushes block A from behind, so that the blocks accelerate. If block A exerts force F on block B, what is the force exerted by the experimenter on block A?

#### Answer:

Let *F' *= force exerted by the experimenter on block A and *F* be the force exerted by block A on block B.

Let *a* be the acceleration produced in the system.

For block A,

$F\text{'}-F={m}_{A}a$ ...(1)

For block B,

*F* = *m*_{B}*a** ...*(2)

Dividing equation (1) by (2), we get:

$\frac{F\text{'}}{F}-1=\frac{{m}_{A}}{{m}_{B}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow F\text{'}=F\left(1+\frac{{m}_{\mathrm{A}}}{{m}_{\mathrm{B}}}\right)$

∴ Force exerted by the experimenter on block A is $F\left(1+\frac{{m}_{\mathrm{A}}}{{m}_{\mathrm{B}}}\right)$.

#### Page No 79:

#### Question 8:

Raindrops of radius 1 mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head.

#### Answer:

Given:

Radius of a raindrop, *r* = 1 mm = 10^{−3} m

Mass of a raindrop, *m* = 4 *mg* = 4 × 10^{−6} kg

Distance coved by the drop on the head, *s* = 10^{−3} m

Initial speed of the drop, *v* = 0

Final speed of the drop, *u* = 30 m/s

Using $a=\frac{{v}^{2}-{u}^{2}}{2s}$, we get:

$a=\frac{-{\left(30\right)}^{2}}{2\times {10}^{-3}}=-4.5\times {10}^{5}m/{s}^{2}$

Force, *F = ma
⇒ F = *4 × 10

^{−6}× 4.5 × 10

^{5}

= 1.8 N

#### Page No 79:

#### Question 9:

A particle of mass 0.3 kg is subjected to a force F = −*kx* with *k* = 15 N/m. What will be its initial acceleration if it is released from a point *x* = 20 cm?

#### Answer:

Displacement of the particle from the mean position*, x* = 20 cm = 0.2 m

*k* = 15 N/m

Mass of the particle, *m* = 0.3 kg

Acceleration, $a=\frac{\left|\mathit{F}\right|}{m}$

$\Rightarrow a=\frac{kx}{m}=\frac{15\left(0.2\right)}{0.3}=\frac{3}{0.3}=10\mathrm{m}/{\mathrm{s}}^{2}$

So, the initial acceleration when the particle is released from a point *x* = 20 cm is 10 m/s^{2}.

#### Page No 79:

#### Question 1:

A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first seconds. Find the magnitude of F.

#### Answer:

Given:

Mass of the block,* m* = 2 kg,

Distance covered, *S* = 10 m and initial velocity, *u* = 0

Let* a* be the acceleration of the block.

Using, $\mathrm{S}=ut+\frac{1}{2}a{t}^{2}$, we get:

$10=\frac{1}{2}a\left({2}^{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 10=2a\phantom{\rule{0ex}{0ex}}\Rightarrow a=5\mathrm{m}/{\mathrm{s}}^{2}$

∴ Force,* F* = *ma *= 2 × 5 = 10 N

#### Page No 79:

#### Question 2:

A car moving at 40 km/hr is to be stopped by applying brakes in the next 4 m. If the car weighs 2000 kg, what average force must be applied to stop it?

#### Answer:

Given:

Initial speed of the car, *u* = 40 km/hr $=\frac{4000}{3600}=11.11\mathrm{m}/\mathrm{s}$

Final speed of the car, *v* = 0

Mass of the car,* m* = 2000 kg

Distance to be travelled by the car before coming to rest, *s** *= 4m

Acceleration, $a=\frac{{v}^{2}-{u}^{2}}{2\mathrm{s}}$

$\Rightarrow a=\frac{{0}^{2}-{\left(11.11\right)}^{2}}{2\times 4}=\frac{-123.43}{8}=-15.42\mathrm{m}/{\mathrm{s}}^{2}$

∴ Average force to be applied to stop the car, *F* = *ma*

*⇒ F* = 2000 × 15.42 ≈ 3.1 × 10^{4} N

#### Page No 79:

#### Question 3:

In a TV picture tube, electrons are ejected from the cathode with negligible speed and they attain a velocity of 5 × 10^{6} m/s in travelling one centimetre. Assuming straight-line motion, find the constant force exerted on the electrons. The mass of an electron is 9.1 × 10^{−31} kg.

#### Answer:

Initial velocity of the electrons is negligible, i.e. *u* = 0.

Final velocity of the electrons*, v* = 5 × 10^{6} m/s

Distance travelled by the electrons,

*s* = 1 cm = 1 × 10^{−2} m

∴ Acceleration, $a=\frac{{v}^{2}-{u}^{2}}{2\mathrm{S}}$

$\Rightarrow a=\frac{{\left(5\times {10}^{5}\right)}^{2}-0}{2\times 1\times {10}^{-3}}=\frac{25\times {10}^{12}}{2\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}\Rightarrow a=12.5\times {10}^{14}\mathrm{m}/{\mathrm{s}}^{2}$

So, force on the electrons, *F* = *ma*

*⇒* *F* = 9.1 × 10^{−31} × 12.5 × 10^{−14}

= 1.1 × 10^{−15} N

#### Page No 79:

#### Question 4:

A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of mass 0.3 kg is suspended from the first block by another string. Find the tensions in the two strings. Take g = 10 m/s^{2}.

#### Answer:

The free-body diagrams for both the blocks are shown below:

From the free-body diagram of the 0.3 kg block,

*T* = 0.3*g*

⇒ *T*= 0.3 × 10 = 3 N

Now, from the free-body diagram of the 0.2 kg block,

*T*_{1}_{ }= 0.2*g* + *T*

⇒ *T*_{1}= 0.2 × 10 + 3 = 5 N

∴ The tensions in the two strings are 5 N and 3 N, respectively.

#### Page No 79:

#### Question 5:

Two blocks of equal mass *m* are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks.

#### Answer:

Let* a *be the common acceleration of the blocks.

For block 1,

$F-T=ma$ ...(1)

For block 2,

*T* = *ma* ...(2)

Subtracting equation (2) from (1), we get:

$F-2T=0$

$\Rightarrow T=\frac{F}{2}$

#### Page No 79:

#### Question 10:

Both the springs shown in figure (5−E2) are unstretched. If the block is displaced by a distance *x* and released, what will be the initial acceleration?

Figure

#### Answer:

Let the block *m* be displaced towards left by displacement *x*.

∴ *F*_{1} = $-$*k*_{1}*x* (compressed)

*F*_{2} = $-$*k*_{2}*x* (expanded)

$ma={F}_{1}+{F}_{2}$

$\Rightarrow a=\frac{-x\left({k}_{1}+{k}_{2}\right)}{m}$

i.e. $\left({k}_{1}+{k}_{2}\right)\frac{x}{m}$ opposite the displacement or towards the mean position.

#### Page No 79:

#### Question 11:

A small block B is placed on another block A of mass 5 kg and length 20 cm. Initially, the block B is near the right end of block A (Figure 5−E3). A constant horizontal force of 10 N is applied to the block A. All the surfaces are assumed frictionless. Find the time that elapses before block B separates from A.

Figure

#### Answer:

Mass of block A, *m* = 5 kg

*F *= *ma* = 10 N

$\Rightarrow a=\frac{10}{5}=2\mathrm{m}/{\mathrm{s}}^{2}$

As there is no friction between A and B, when block A moves, block B remains at rest in its position.

Initial velocity of A, *u* = 0

Distance covered by A to separate out,

*s* = 0.2 m

Using $s=ut+\frac{1}{2}a{t}^{2}$, we get:

$0.2=0+\frac{1}{2}\times 2{t}^{2}$

⇒ *t*^{2} = 0.2

⇒* t* = 0.44 s ≈ 0.45 s

#### Page No 79:

#### Question 12:

A man has fallen into a ditch of width *d* and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure (5−E4). Show that the force (assumed equal for both the friends) exerted by each friend on the road increases as the man moves up. Find the force when the man is at a depth *h*.

Figure

#### Answer:

(a) At any depth, let the ropes makes an angle *θ* with the vertical.

From the free-body diagram,

*F*cos*θ** + **F*cos*θ** *− *mg *= 0

2*F*cos*θ** = mg*

$\Rightarrow F=\frac{mg}{2cos\theta}$

As the man moves up, *θ* increases, i.e. cos*θ* decreases. Thus,* F* increases.

(b) When the man is at depth *h*,

$\mathrm{cos}\theta =\frac{h}{\sqrt{{\left(d/2\right)}^{2}+{h}^{2}}}$

$F=\frac{mg}{2\mathrm{cos}\theta}\phantom{\rule{0ex}{0ex}}=\frac{mg}{2\left[h/\sqrt{{\left({\displaystyle \frac{d}{2}}\right)}^{2}+{h}^{2}}\right]}\phantom{\rule{0ex}{0ex}}=\frac{mg}{4h}\sqrt{{d}^{2}+4{h}^{2}}$

#### Page No 80:

#### Question 13:

The elevator shown in figure (5−E5) is descending with an acceleration of 2 m/s^{2}. The mass of the block A is 0.5 kg. What force is exerted by the block A on the block B?

Figure

#### Answer:

When the elevator is descending, a pseudo-force acts on it in the upward direction, as shown in the figure.

From the free-body diagram of block A,

$mg-N=ma$

$N=m\left(g-a\right)\phantom{\rule{0ex}{0ex}}\Rightarrow N=0.5\left(10-2\right)=4\mathrm{N}\phantom{\rule{0ex}{0ex}}$

So, the force exerted by the block A on the block B is 4 N.

#### Page No 80:

#### Question 14:

A pendulum bob of mass 50 g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator (a) goes up with acceleration 1.2 m/s^{2}, (b) goes up with deceleration 1.2 m/s^{2}, (c) goes up with uniform velocity, (d) goes down with acceleration 1.2 m/s^{2}, (e) goes down with deceleration 1.2 m/s^{2} and (f) goes down with uniform velocity.

#### Answer:

(a) When the elevator goes up with acceleration 1.2 m/s^{2}:

$T=mg+ma$

*⇒ T* = 0.05 (9.8 + 1.2) = 0.55 N

(b) Goes up with deceleration 1.2 m/s^{2} :

$T=mg+m\left(-a\right)=m\left(g-a\right)$

⇒ *T* = 0.05 (9.8 − 1.2) = 0.43 N

(c) Goes up with uniform velocity:

$T=mg$

⇒ *T* = 0.05 × 9.8 = 0.49 N

(d) Goes down with acceleration 1.2 m/s^{2} :

$T+ma=mg\phantom{\rule{0ex}{0ex}}\Rightarrow T=m\left(g-a\right)$

⇒ *T* = 0.05 (9.8 − 1.2) = 0.43 N

(e) Goes down with deceleration 1.2 m/s^{2} :

$T+m\left(-a\right)=mg\phantom{\rule{0ex}{0ex}}\Rightarrow T=m\left(g+a\right)$

⇒ *T *= 0.05 (9.8 + 1.2) = 0.55 N

(f) Goes down with uniform velocity:

$T=mg$

⇒* T *= 0.05 × 9.8 = 0.49 N

#### Page No 80:

#### Question 15:

A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are 72 kg and 60 kg, respectively. Assuming that the magnitudes of acceleration and deceleration are the same, find (a) the true weight of the person and (b) the magnitude of the acceleration. Take *g* = 9.9 m/s^{2}.

Figure

#### Answer:

Maximum weight will be recorded when the elevator accelerates upwards.

Let *N* be the normal reaction on the person by the weighing machine.

So, from the free-body diagram of the person,

$N=mg+ma$ ...(1)

This is maximum weight, *N* = 72 × 9.9 N

When decelerating upwards, minimum weight will be recorded.

$N\text{'}=mg+m\left(-a\right)$ ...(2)

This is minimum weight, *N*' = 60 × 9.9 N

From equations (1) and (2), we have:

2 *mg *= 1306.8

$\Rightarrow m=\frac{1306.8}{2\times 9.9}=66\mathrm{kg}$

So, the true mass of the man is 66 kg.

And true weight = 66 $\times $ 9.9 = 653.4 N

(b) Using equation (1) to find the acceleration, we get:

*mg* + *ma* = 72 × 9.9

$\Rightarrow a=\frac{72\times 9.9-66\times 9.9}{66}=\frac{9.9\times 6}{66}=\frac{9.9}{11}\phantom{\rule{0ex}{0ex}}\Rightarrow a=0.9\mathrm{m}/{\mathrm{s}}^{2}$

#### Page No 80:

#### Question 16:

Find the reading of the spring balance shown in figure (5−E6). The elevator is going up with an acceleration g/10, the pulley and the string are light and the pulley is smooth.

#### Answer:

Let the left and right blocks be A and B, respectively.

And let the acceleration of the 3 kg mass relative to the elevator be '*a*' in the downward direction.

From the free-body diagram,

${m}_{A}a=T-{m}_{A}g-\frac{{m}_{A}g}{10}...\left(1\right)$

${m}_{B}a={m}_{B}g+\frac{{m}_{B}g}{10}-T...\left(2\right)$

Adding both the equations, we get:

$a\left({m}_{A}+{m}_{B}\right)=\left({m}_{B}-{m}_{A}\right)g+\left({m}_{B}-{m}_{A}\right)\frac{g}{10}$

Putting value of the masses,we get:

$9a=\frac{33g}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{g}=\frac{11}{30}...\left(3\right)$

Now, using equation (1), we get:

$T={m}_{A}\left(a+g+\frac{g}{10}\right)$

The reading of the spring balance = $\frac{2T}{g}=\frac{2}{g}{m}_{A}\left(a+g+\frac{g}{10}\right)$

$\Rightarrow 2\times 1.5\left(\frac{a}{g}+1+\frac{1}{10}\right)=3\left(\frac{11}{30}+1+\frac{1}{10}\right)\phantom{\rule{0ex}{0ex}}=4.4\mathrm{kg}$

#### Page No 80:

#### Question 17:

A block of 2 kg is suspended from a ceiling by a massless spring of spring constant *k* = 100 N/m. What is the elongation of the spring? If another 1 kg is added to the block, what would be the further elongation?

#### Answer:

Given,

mass of the first block, *m* = 2 kg

*k* = 100 N/m

Let elongation in the spring be *x. *

From the free-body diagram,

*kx* = *mg*

$x=\frac{mg}{k}=\frac{2\times 9.8}{100}\phantom{\rule{0ex}{0ex}}=\frac{19.6}{100}=0.196\approx 0.2\mathrm{m}$

Suppose, further elongation, when the 1 kg block is added, is *$\u2206x$*.

Then,$k\left(x+\u2206x\right)=m\text{'}g$

⇒ *k$\u2206x$*= 3*g* − 2*g* = *g*

$\Rightarrow \u2206x=\frac{g}{k}=\frac{9.8}{100}=0.098\approx 0.1\mathrm{m}$

#### Page No 80:

#### Question 18:

Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of 2.0 m/s^{2}. Find the elongation.

#### Answer:

When the ceiling of the elevator is going up with an acceleration 'a', then a pseudo-force acts on the block in the downward direction.

*
a* = 2 m/s

^{2}

From the free-body diagram of the block,

*kx = m*

*g*+

*ma*

⇒

*kx*= 2

*g*+ 2

*a*

= 2 × 9.8 + 2 × 2

= 19.6 + 4

$\Rightarrow x=\frac{23.6}{100}=0.236\approx 0.24\mathrm{m}$

When 1 kg body is added,

total mass = (2 + 1) kg = 3 kg

Let elongation be

*x*'.

∴

*kx'*= 3

*g*+ 3

*a*= 3 × 9.8 + 6

$\Rightarrow x\text{'}=\frac{35.4}{100}\phantom{\rule{0ex}{0ex}}=0.354\approx 0.36\mathrm{m}$

So, further elongation =

*x' − x*

= 0.36 − 0.24 = 0.12 m.

#### Page No 80:

#### Question 19:

The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall to the earth's surface with a constant velocity *v*. How much mass should be removed from the balloon so that it may rise with a constant velocity *v*?

#### Answer:

Let *M* be mass of the balloon.

Let the air resistance force on balloon be *F *.

Given that* F ∝ v*.

⇒ *F _{ }= kv*,

where

*k*= proportionality constant.

When the balloon is moving downward with constant velocity,

*B + kv = Mg*...(i)

$\Rightarrow M=\frac{B+kv}{g}$

Let the mass of the balloon be

*M' so that it can*rise with a constant velocity

*v*in the upward direction.

*B = Mg + kv*

$\Rightarrow M\text{'}=\frac{B+kv}{g}$

∴ Amount of mass that should be removed =

*M − M'.*

$\u2206M=\frac{B+kv}{g}-\frac{B-kv}{g}\phantom{\rule{0ex}{0ex}}=\frac{B+kv-B+kv}{g}\phantom{\rule{0ex}{0ex}}=\frac{2kv}{g}=\frac{2\left(Mg-B\right)}{g}\phantom{\rule{0ex}{0ex}}=2\left\{M-\frac{B}{g}\right\}$

$\u2206M=\frac{B+kv}{g}-\frac{B-kv}{g}\phantom{\rule{0ex}{0ex}}=\frac{B+kv-B+kv}{g}\phantom{\rule{0ex}{0ex}}=\frac{2kv}{g}=\frac{2\left(Mg-B\right)}{g}\phantom{\rule{0ex}{0ex}}=2\left\{M-\frac{B}{g}\right\}$

#### Page No 80:

#### Question 20:

An empty plastic box of mass *m* is found to accelerate up at the rate of *g*/6 when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of *g*/6?

#### Answer:

Let *U* be the upward force of water acting on the plastic box.

Let *m* be the initial mass of the plastic box.

When the empty plastic box is accelerating upward,

$U-mg=\frac{mg}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow U=\frac{7mg}{6}$

$\Rightarrow m=\frac{6U}{7g}....\left(i\right)$

Let M be the final mass of the box after putting some sand in it.

$Mg-U=\frac{Mg}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow Mg-\frac{Mg}{6}=U\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{6U}{5g}....\left(ii\right)$

Mass added$=\frac{6U}{5g}-\frac{6U}{7g}$

$=\frac{6U\left(7-5\right)}{35g}\phantom{\rule{0ex}{0ex}}=\frac{6U\xb72}{35g}$

From equation (i), $m=\frac{6U}{7g}$

∴ Mass added$=\frac{2}{5}m$ .

#### Page No 80:

#### Question 21:

A force $\overrightarrow{F}=\overrightarrow{v}\times \overrightarrow{A}$ is exerted on a particle in addition to the force of gravity, where $\overrightarrow{v}$ is the velocity of the particle and $\overrightarrow{A}$ is a constant vector in the horizontal direction. With what minimum speed, a particle of mass *m* be projected so that it continues to move without being defelected and with a constant velocity?

#### Answer:

For the particle to move without being deflected and with constant velocity, the net force on the particle should be zero.

$\overrightarrow{F}+m\overrightarrow{g}=0$

$\Rightarrow \left(\overrightarrow{v}\times \overrightarrow{A}\right)+\overrightarrow{mg}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\overrightarrow{v}\times \overrightarrow{A}\right)=-\overrightarrow{mg}$

$\left|vA\mathrm{sin}\theta \right|=\left|mg\right|$

$\therefore v=\frac{mg}{A\mathrm{sin}\theta}$

*v* will be minimum when sin*θ* = 1.

⇒ *θ* = 90°

$\therefore {v}_{min}=\frac{mg}{A}$

#### Page No 80:

#### Question 22:

In a simple Atwood machine, two unequal masses *m*_{1} and *m*_{2} are connected by a string going over a clamped light smooth pulley. In a typical arrangement (figure 5−E7), *m*_{1} = 300 g and *m*_{2} = 600 g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds; (b) find the tension in the string; (c) find the force exerted by the clamp on the pulley.

Figure

#### Answer:

The masses of the blocks are *m*_{1} = 0.3 kg and *m*_{2} = 0.6 kg

The free-body diagrams of both the masses are shown below:

For mass *m*_{1},

* T* − *m*_{1}*g* = *m*_{1}*a* ...(i)

For mass *m*_{2},

*m*_{2}*g* − *T*= *m*_{2}*a* ...(ii)

Adding equations (i) and (ii), we get:

*g*(*m*_{2} − *m*_{1}) = *a*(*m*_{1} + *m*_{2})

$\Rightarrow a=g\left(\frac{{m}_{2}-{m}_{1}}{{m}_{1}+{m}_{2}}\right)\phantom{\rule{0ex}{0ex}}=9.8\times \frac{0.6-0.3}{0.6+0.3}\phantom{\rule{0ex}{0ex}}=3.266\mathrm{m}/{\mathrm{s}}^{2}$

(a) *t* = 2 s, *a* = 3.266 ms^{−2}, *u* = 0

So, the distance travelled by the body,

$\mathrm{S}=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}=0+\frac{1}{2}\left(3.266\right){2}^{2}=6.5\mathrm{m}$

(b) From equation (i),

*T* = *m*_{1} (*g *+ *a*)

= 0.3 (3.8 + 3.26) = 3.9 N

(c)The force exerted by the clamp on the pulley,

*F = *2*T** *= 2 × 3.9 = 7.8 N

#### Page No 80:

#### Question 23:

Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment, 2.0 s after the system is set into motion. Find the time that elapses before the string is tight again.

#### Answer:

*a* = 3.26 m/s^{2}, *T* = 3.9 N

After 2 s, velocity of mass *m*_{1}_{,}

*v* = *u* + *at* = 0 + 3.26 × 2

= 6.52 m/s upward

At this time, *m*_{2} is moving 6.52 m/s downward.

At time 2 s, *m*_{2} stops for a moment. But *m*_{1} is moving upward with velocity 6.52 m/s. It will continue to move till final velocity (at highest point) becomes zero.

Here, *v* = 0, *u* = 6.52 m/s

*a* = −*g* = − 9.8 m/s^{2}

*v* = *u* + *at* = 6.52 + (−9.8)*t*

$\Rightarrow t=\frac{6.52}{9.8}\approx \frac{2}{3}\mathrm{sec}$

After this time, the mass* **m*_{1} also starts moving downward.

So, the string becomes tight again after $\frac{2}{3}\mathrm{s}$.

#### Page No 80:

#### Question 24:

Figure (5−E8) shows a uniform rod of length 30 cm and mass 3.0 kg. The strings shown in the figure are pulled by constant forces of 20 N and 32 N. Find the force exerted by the 20 cm part of the rod on the 10 cm part. All the surfaces are smooth and the strings and the pulleys are light.

Figure

#### Answer:

Mass per unit length$=\frac{3}{30}\mathrm{kg}/\mathrm{cm}$ = 0.10 kg/cm

Mass of the 10 cm part, *m*_{1} = 1 kg

Mass of the 20 cm part, *m*_{2} = 2 kg

Let *F* = contact force between them.

From the free-body diagram,

${m}_{1}a=F-20...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}{m}_{2}a=32-F...\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\mathrm{both}\mathrm{the}\mathrm{equation}\text{s},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}a=\frac{12}{{m}_{1}+{m}_{2}}=\frac{12}{3}=4\mathrm{m}/{\mathrm{s}}^{2}$

So, contact force,

*F* = 20 + 1*a*

*F* = 20 + 4 = 24 N

#### Page No 81:

#### Question 25:

Consider the situation shown in figure (5−E9). All the surfaces are frictionless and the string and the pulley are light. Find the magnitude of acceleration of the two blocks.

#### Answer:

Mass of each block is 1 kg, $\mathrm{sin}{\mathrm{\theta}}_{1}=\frac{4}{5}$, $\mathrm{sin}{\mathrm{\theta}}_{2}=\frac{3}{5}$.

The free-body diagrams for both the boxes are shown below:

*m**g*sin*θ*_{1} − *T = ma* ...(i)

* T − **mg*sin*θ*_{2} = *ma *...(ii)

Adding equations (i) and (ii),we get:

*mg*(sin*θ*_{1} *− *sin*θ*_{2}) = 2*ma** *

⇒ 2*a* = *g* (sin*θ*_{1} − sin*θ*_{2})

$\Rightarrow a=\frac{g}{5}\times \frac{1}{2}\phantom{\rule{0ex}{0ex}}=\frac{g}{10}$

#### Page No 81:

#### Question 26:

A constant force F = *m*_{2}*g*/2 is applied on the block of mass *m*_{1} as shown in figure (5−E10). The string and the pulley are light and the surface of the table is smooth. Find the acceleration of *m*_{1}.

Figure

#### Answer:

The free-body diagrams for both the blocks are shown below:

From the free-body diagram of block of mass *m*_{1},

*m*_{1}*a* =* T − F* ...(i)

From the free-body diagram of block of mass *m*_{2},

*m*_{2}*a* = *m*_{2}*g* − T ...(ii)

Adding both the equations, we get:

$a\left({m}_{1}+{m}_{2}\right)={m}_{2}g-\frac{{m}_{2}g}{2}\left[\mathrm{because}F=\frac{{m}_{2}g}{2}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{{m}_{2}g}{2\left({m}_{1}+{m}_{2}\right)}$

So, the acceleration of mass *m*_{1},

$a=\frac{{m}_{2}g}{2\left({m}_{1}+{m}_{2}\right)},\mathrm{towards}\text{the}\mathrm{right}.$

#### Page No 81:

#### Question 27:

In figure (5−E11), *m*_{1} = 5 kg, *m*_{2} = 2 kg and F = 1 N. Find the acceleration of either block. Describe the motion of *m*_{1} if the string breaks but F continues to act.

Figure

#### Answer:

Let the acceleration of the blocks be *a*.

The free-body diagrams for both the blocks are shown below:

From the free-body diagram,

*m*_{1}*a* = *m*_{1}*g* +* F − T ...*(i)

Again, from the free-body diagram,

*m*_{2}*a* = *T − m*_{2}*g* *− F ...*(ii)

Adding equations (i) and (ii), we have:

$a=g\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow a=\frac{3g}{7}=\frac{29.4}{7}\phantom{\rule{0ex}{0ex}}=4.2\mathrm{m}/{\mathrm{s}}^{2}$

Hence, acceleration of the block is 4.2 m/s^{2}.

After the string breaks, *m*_{1} moves downward with force *F* acting downward. Then,

*m*_{1}*a* = *F* + *m*_{1}*g*

5*a* = 1 + 5*g*

$\Rightarrow a=\frac{5g+1}{5}\phantom{\rule{0ex}{0ex}}=g+0.2\mathrm{m}/{\mathrm{s}}^{2}$

#### Page No 81:

#### Question 28:

Let *m*_{1} = 1 kg, *m*_{2} = 2 kg and *m*_{3} = 3 kg in figure (5−E12). Find the accelerations of *m*_{1}, *m*_{2} and *m*_{3}. The string from the upper pulley to *m*_{1} is 20 cm when the system is released from rest. How long will it take before *m*_{1} strikes the pulley?

Figure

#### Answer:

The free-body diagram for mass *m*_{1} is shown below:

(Figure 1)

The free-body diagram for mass *m*_{2} is shown below:

(Figure 2)

The free-body diagram for mass *m*_{3} is shown below:

(Figure 3)

Suppose the block *m*_{1} moves upward with acceleration *a*_{1} and the blocks *m*_{2} and *m*_{3} have relative acceleration *a*_{2} due to the difference of weight between them.

So, the actual acceleration of the blocks *m*_{1}, *m*_{2} and *m*_{3} will be *a*_{1}, (*a*_{1} − *a*_{2}) and (*a*_{1} + *a*_{2}), as shown.

From figure 2, *T* − 1*g* − 1*a*_{1} = 0 ...(i)

From figure 3, $\frac{T}{2}-2g-2\left({a}_{1}-{a}_{2}\right)=0...\left(\mathrm{ii}\right)$

From figure 4, $\frac{T}{2}-3g-3\left({a}_{1}+{a}_{2}\right)=0...\left(\mathrm{iii}\right)$

From equations (i) and (ii), eliminating T, we get:

1*g* + 1*a*_{2} = 4*g* + 4 (*a*_{1} + *a*_{2})

5*a*_{2} − 4*a*_{1} = 3*g* ...(iv)

From equations (ii) and (iii), we get:

2*g* + 2(*a*_{1} − *a*_{2}) = 3*g* − 3 (*a*_{1} − *a*_{2})

5*a*_{1} + *a*_{2} = *g* ...(v)

Solving equations (iv) and (v), we get:

${a}_{1}=\frac{2g}{29}\phantom{\rule{0ex}{0ex}}\mathrm{and}{a}_{2}=g-5{a}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{2}=g-\frac{10g}{29}=\frac{19g}{29}\phantom{\rule{0ex}{0ex}}\mathrm{So},{a}_{1}-{a}_{2}=\frac{2g}{29}-\frac{19g}{29}=-\frac{17g}{29}\phantom{\rule{0ex}{0ex}}\mathrm{and}{a}_{1}+{a}_{2}=\frac{2g}{29}+\frac{19g}{29}=\frac{21g}{29}$

So, accelerations of *m*_{1}, *m*_{2} and *m*_{3} are $\frac{19g}{29}\left(\mathrm{up}\right),\frac{17g}{29}\left(\mathrm{down}\right)\mathrm{and}\frac{21g}{29}\left(\mathrm{down}\right)$ , respectively.

Now, *u* = 0, *s* = 20 cm = 0.2 m

${a}_{2}=\frac{19g}{29}\phantom{\rule{0ex}{0ex}}\therefore s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 0.2=\frac{1}{2}\times \frac{19}{29}g{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow t=0.25\mathrm{s}$

#### Page No 81:

#### Question 29:

In the previous problem, suppose *m*_{2} = 2.0 kg and *m*_{3} = 3.0 kg. What should be the mass *m*, so that it remains at rest?

#### Answer:

For m_{1} to be at rest, *a*_{1}_{ }= 0.

*T *−* **m*_{1}*g* = 0

*T = **m*_{1}*g** *...(i)

For mass *m*_{2},

*T*/2 − 2*g** *= 2*a** *

*T =** *4*a* + 4*g** ** * ...(ii)

For mass *m*_{3},

3*g* –* T*/2= 2*a** *

*T =** *6*g* − 6*a** ** * ...(ii)

From equations (ii) and (iii), we get:

3*T** *– 12*g** *= 12*g* – 2*T** *

*T* = 24g/5= 4.08g

Putting the value of *T* in equation (i), we get:

*m*_{1}_{ }= 4.8kg

#### Page No 81:

#### Question 30:

Calculate the tension in the string shown in figure (5−E13). The pulley and the string are light and all the surfaces are frictionless. Take *g* = 10 m/s^{2}.

Figure

#### Answer:

The free-body diagrams for both the bodies are shown below:

*T* + *ma* =*mg *...(i)

and *T* = *ma* ...(ii)

From equations (i) and (ii), we get:

*ma* + *ma* = m*g*

⇒ 2*ma* = *g*

$\Rightarrow a=\frac{g}{2}=\frac{10}{5}=5\mathrm{m}/{\mathrm{s}}^{2}$

From equation (ii),

*T* = *ma* = 5 N

#### Page No 81:

#### Question 31:

Consider the situation shown in figure (5−E14). Both the pulleys and the string are light and all the surfaces are frictionless. (a) Find the acceleration of the mass M; (b) find the tension in the string; (c) calculate the force exerted by the clamp on the pulley A in the figure.

Figure

#### Answer:

Let the acceleration of mass *M* be* a*.

So, the acceleration of mass 2*M** *will be* $\frac{a}{2}$.*

(a) 2*M*(*a*/2) − 2*T* = 0

⇒* Ma = *2*T*

*T + Ma − Mg *= 0

$\Rightarrow \frac{Ma}{2}+Ma=Mg\phantom{\rule{0ex}{0ex}}\Rightarrow 3Ma=2Mg\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{2g}{3}$

(b) Tension, $T=\frac{Ma}{2}=\frac{M}{2}\times \frac{2g}{3}=\frac{Mg}{3}$

(c) Let *T*' = resultant of tensions

$\therefore T\text{'}=\sqrt{{T}^{2}+{T}^{2}}=\sqrt{2}T\phantom{\rule{0ex}{0ex}}\therefore T\text{'}=\sqrt{2}T=\frac{\sqrt{2}Mg}{3}\phantom{\rule{0ex}{0ex}}\mathrm{Again},\mathrm{tan}\theta =\frac{T}{T}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \theta =45\xb0$

So, the force exerted by the clamp on the pulley is $\frac{\sqrt{2}\mathrm{M}g}{3}$ at an angle of 45° with the horizontal.

#### Page No 81:

#### Question 32:

Find the acceleration of the block of mass M in the situation shown in figure (5−E15). All the surfaces are frictionless and the pulleys and the string are light.

Figure

#### Answer:

The free-body diagram of the system is shown below:

Let acceleration of the block of mass 2*M* be *a*.

So, acceleration of the block of mass *M* will be 2*a**.*

*M*(2*a*) + *Mg*sin*θ** *−* T* = 0

⇒ *T* = 2*Ma* + *Mg*sin*θ* ...(i)

2*T* + 2*Ma* − 2*Mg* = 0

From equation (i),

2(2*Ma* + *Mg*sin*θ*) + 2*Ma* − 2*M*g = 0

4*Ma* + 2*Mg*sin*θ* + 2*Ma* −* Mg* = 0

6*Ma* + 2*Mg*sin30° + 2*Mg* = 0

6*Ma** = Mg*

$\Rightarrow a=\frac{g}{6}$

Hence, the acceleration of mass $M=2a=2\times \frac{g}{6}=\frac{g}{3}\left(\mathrm{up}\mathrm{the}\mathrm{plane}\right).$

#### Page No 82:

#### Question 33:

Find the mass M of the hanging block in figure (5−E16) that will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.

Figure

#### Answer:

The free-body diagram of the system is shown below:

Block ‘*m*’ will have the same acceleration as that of *M',* as it does not slip over *M*'.

From the free body diagrams,

*T + Ma – Mg = *0* ...*(i)

*T – M'a – **R*sin*θ** = *0 ...(ii)

*R*sin*θ** – ma* = 0

*R*cos*θ** – mg *= 0

Eliminating

*T, R*and

*a*from the above equations, we get:

$M=\frac{M\mathit{\text{'}}\mathit{+}m}{\mathrm{cot}\mathrm{\theta}-1}$

#### Page No 82:

#### Question 34:

Find the acceleration of the blocks A and B in the three situations shown in figure (5−E17).

Figure

#### Answer:

(a) 5*a* + *T* − 5*g* = 0

From free-body diagram (1),

*T* = 5*g* − 5*a* .....(i)

Again, $\left(\frac{1}{2}\right)T-4g-8a=0$

⇒ *T* − 8*g* − 16*a* = 0

(1)

From free-body diagram (2),

*T* = 8*g* + 16*a* ......(ii)

From equations (i) and (ii), we get:

5*g* − 5*a* = 8*g* + 16*a*

$\Rightarrow 21a=-3g-a=-\frac{9}{7}$

So, the acceleration of the 5 kg mass is $\frac{9}{7}\mathrm{m}/{\mathrm{s}}^{2}\left(\mathrm{upward}\right)$ and that of the 4 kg mass is $\mathit{2}a\mathit{=}\frac{\mathit{2}\mathit{g}}{\mathit{7}}\mathit{}\left(\mathrm{downward}\right)$.

(b) From free body diagram-3,

$4a-\frac{T}{2}=0$

(2)

⇒ 8*a* − *T* = 0

⇒ *T* = 8*a*

Again, *T *+ 5*a* − 5*g* = 0

From free body diagram-4,

8*a* + 5*a* − 5*g* = s0

⇒ 13*a* − 5*g** *= 0

$\Rightarrow a=\frac{5g}{13}\left(\mathrm{downward}\right)$

Acceleration of mass 2 kg is $2a=\frac{10}{13}\left(g\right)$ and 5 kg is $\frac{5g}{13}$.

(c) *T* + 1*a* − 1*g* = 0

From free body diagram-5

*T* = 1*g* − 1*a* .....(i)

Again, from free body diagram-6,

$\frac{T}{2}-2g-4a=0$

⇒ *T *− 4*g* − 8*a* = 0 .....(ii)

From equation (i)

1*g* − 1*a* − 4*g* − 8*a* = 0

$\Rightarrow a=\frac{g}{3}\left(\mathrm{downward}\right)$

Acceleration of mass 1 kg is $\frac{g}{3}\left(\mathrm{upward}\right)$,

Acceleration of mass 2 kg is $\frac{2g}{3}\left(\mathrm{downward}\right)$.

(3)

#### Page No 82:

#### Question 35:

Find the acceleration of the 500 g block in the figure (5−E18).

Figure

#### Answer:

Given,

*m*_{1} = 100 g = 0.1 kg

*m*_{2} = 500 g = 0.5 kg

*m*_{3} = 50 g = 0.05 kg

The free-body diagram for the system is shown below:

From the free-body diagram of the 500 g block,

*T* + 0.5*a* − 0.5*g* = 0 .....(i)

From the free-body diagram of the 50 g block,

*T*_{1} + 0.05*g* − 0.05*a* = *a* ....(ii)

From the free-body diagram of the 100 g block,

T_{1}_{ }+ 0.1*a* − *T* + 0.5*g* = 0 ....(iii)

From equation (ii),

*T*_{1} = 0.05*g* + 0.05*a* .....(iv)

From equation (i),

*T*_{1} = 0.5*g* − 0.5*a* .....(v)

Equation (iii) becomes

*T*_{1} + 0.1*a* − *T* + 0.05*g* = 0

From equations (iv) and (v), we get:

0.05*g* + 0.05*a* + 0.1*a* − 0.5*g* + 0.5*a* + 0.05*g* = 0

0.65*a* = 0.4 *g*

$\Rightarrow a=\frac{0.4}{0.65}g\phantom{\rule{0ex}{0ex}}=\frac{40}{65}g=\frac{8}{13}g\left(\mathrm{downward}\right)$

So, the acceleration of the 500 gm block is $\frac{8g}{13}$downward.

#### Page No 82:

#### Question 36:

A monkey of mass 15 kg is climbing a rope fixed to a ceiling. If it wishes to go up with an acceleration of 1 m/s^{2}, how much force should it apply on the rope? If the rope is 5 m long and the monkey starts from rest, how much time will it take to reach the ceiling?

#### Answer:

Mass of the monkey, *m* = 15 kg,

Acceleration of the monkey in the upward direction, *a* = 1 m/s^{2}

The free-body diagram of the monkey is shown below:

From the free-body diagram,

*T* − [15*g* + 15(*a*)] = 0

*T* − [15*g* + 15(1)] = 0

⇒ *T* = 5 (10 + 1)

⇒ *T* = 15 × 11 = 165 N

The monkey should apply a force of 165 N to the rope.

Initial velocity, *u* = 0

*s* = 5 m

Using, $s=ut+\frac{1}{2}a{t}^{2}$, we get:

$5=0+\left(\frac{1}{2}\right)\times 1\times {t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}=5\times 2\phantom{\rule{0ex}{0ex}}\Rightarrow t=\sqrt{10}\mathrm{s}$

Hence, the time required to reach the ceiling is $\sqrt{10}\mathrm{s}$.

#### Page No 82:

#### Question 37:

A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end (figure 5−E19). Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If initially both were at rest, their separation will not change as time passes.

Figure

#### Answer:

Suppose the monkey accelerates upward with acceleration *a* and the block accelerates downward with acceleration *a*'.

Let force exerted by the monkey be *F*.

From the free-body diagram of the monkey, we get:

*F*− *mg* − *ma* = 0 ...(i)

*⇒ **F* = *mg* + *ma*

Again, from the free-body diagram of the block,

*F* + *ma*' − *mg* = 0

*mg* + *ma* + *ma'* − *mg* = 0 [From (i)]

*⇒ ma* = −*ma*'

*⇒ a* = −*a*'

If acceleration −*a*' is in downward direction then the acceleration *a*' will be in upward direction.

This implies that the block and the monkey move in the same direction with equal acceleration.

If initially they were at rest (no force is exerted by the monkey), then their separation will not change as time passes because both are moving same direction with equal acceleration.

#### Page No 82:

#### Question 38:

The monkey B, shown in the figure (5−E20), is holding on to the tail of monkey A that is climbing up a rope. The masses of monkeys A and B are 5 kg and 2 kg, respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry monkey B with it? Take g = 10 m/s^{2}.

Figure

#### Answer:

Let the acceleration of monkey A upwards be *a*, so that a maximum tension of 30 N is produced in its tail.

(i)

(ii)

*T* − 5*g* − 30 − 5*a* = 0 ...(i)

30 − 2*g* − 2*a* = 0 ...(ii)

From equations (i) and (ii), we have:

* T* = 105 N (max.)

and *a* = 5 m/s^{2}

So, A can apply a maximum force of 105 N on the rope to carry monkey B with it.

For minimum force, there is no acceleration of A and B.

*T*_{1} = weight of monkey B

⇒ *T*_{1} = 20 N

Rewriting equation (i) for monkey A, we get:

*T* − 5*g* − 20 = 0

⇒ *T* = 70 N

∴ To carry monkey B with it, monkey A should apply a force of magnitude between 70 N and 105 N.

#### Page No 82:

#### Question 39:

Figure (5−E21) shows a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling by a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight recorded on the machine? What force should he exert on the rope to record his correct weight on the machine?

Figure

#### Answer:

(i) Given, mass of the man = 60 kg

Let *W*' = apparent weight of the man in this case

From the free-body diagram of the man,

*W*' + *T* − 60*g* = 0

⇒ *T* = 60*g* − *W*' ...(i)

From the free-body diagram of the box,

*T* − *W*' = 30*g* = 0 ...(ii)

From equation (i), we get:

60*g* − *W*' − *W*' − 30*g* = 0

⇒* W*' = 15*g*

Hence, the weight recorded on the machine is 15 kg.

(ii) To find his actual weight, suppose the force applied by the men on the rope is *T *because of which the box accelerates upward with an acceleration *a*'.Here we need to find ${T}^{\text{'}}$.

Correct weight = *W* = 60*g*

From the free-body diagram of the man,

T' + *W* − 60*g* − 60*a* = 0

⇒ T' − 60*a* = 0

⇒ T' = 60*a** *...(i)

From the free-body diagram of the box,

T' − *W* − 30*g* − 30*a* = 0

⇒ T' − 60*g* − 30*g* − 30*a* = 0

⇒ T' = 30*a* − 900 ...(ii)

From equations (i) and (ii), we get:

T' = 2T' − 1800

T' = 1800 N

So, the man should exert a force of 1800 N on the rope to record his correct weight on the machine.

#### Page No 83:

#### Question 40:

A block A can slide on a frictionless incline of angle θ and length *l*, kept inside an elevator going up with uniform velocity *v* (figure 5−E22). Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline.

#### Answer:

The force on the block which makes the body move down the plane is the component of its weight parallel to the inclined surface.

*F* = *mg *sin*θ*

Acceleration, *g* = sin θ

Initial velocity of block, *u* = 0

Distance to be covered

*s* = *l*

*a* = *g* sin θ

Using, $s=ut+\frac{1}{2}a{t}^{2}$, we get:

$l=0+\frac{1}{2}\left(g\mathrm{sin}\theta \right){t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}=\frac{2l}{g\mathrm{sin}\theta}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Time}\mathrm{taken},t=\sqrt{\frac{2\mathrm{l}}{\mathrm{gsin}\theta}}$

#### Page No 83:

#### Question 41:

A car is speeding up a horizontal road with acceleration *a*. Consider the following situations in the car: (i) A ball suspended from the ceiling by a string is maintaining a constant angle with the vertical. Find this angle. (ii) A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal.

#### Answer:

Let the pendulum (formed by the ball and the string) make angle *θ * with the vertical.

From the free-body diagram,

*T*cos*θ** − mg = 0*

*T*cos*θ** = mg*

$\Rightarrow T=\frac{mg}{\mathrm{cos}\mathrm{\theta}}...\left(\mathrm{i}\right)$

And, *ma* − *T* sin θ = 0

*⇒ ma* = T sin θ

$\Rightarrow T=\frac{ma}{\mathrm{sin}\mathrm{\theta}}....\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\mathrm{\theta}=\frac{a}{g}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\theta}={\mathrm{tan}}^{-1}\frac{a}{g}$

So, the angle formed by the ball with the vertical is ${\mathrm{tan}}^{-1}\left(\frac{a}{g}\right)$ .

(ii) Let the angle of the incline be *θ*.

From the diagram,

⇒* ma* cos θ = *mg* sin θ

$\frac{\mathrm{sin}\mathrm{\theta}}{\mathrm{cos}\mathrm{\theta}}=\frac{a}{g}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\mathrm{\theta}=\frac{a}{g}\mathrm{\theta}={\mathrm{tan}}^{-1}\left(\frac{a}{g}\right)$

So, the angle of incline is ${\mathrm{tan}}^{-1}\left(\frac{a}{g}\right)$.

#### Page No 83:

#### Question 42:

A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s^{2}. Find the displacement of the block during the first 0.2 s after the start. Take *g* = 10 m/s^{2}.

#### Answer:

The free-body diagram of the system is shown below:

The two bodies are separated because the elevator is moving downward with an acceleration of 12 m/s^{2} (>g) and the body moves with acceleration, *g* = 10 m/s^{2} [Freely falling body]

Now, for the block:

*g* = 10 m/s^{2}*, u* = 0*, t* = 0.2 s

So, the distance travelled by the block is given by

$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}=0+\frac{1}{2}10\times {\left(0.2\right)}^{2}=5\times 0.04\phantom{\rule{0ex}{0ex}}=0.2\mathrm{m}=20\mathrm{cm}$

The displacement of the body is 20 cm during the first 0.2 s.

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