Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 6 - Work, Energy And Power

Answer:

The magnetic force on a charged particle q moving with a velocity $\overrightarrow{v}$ , in a region having magnetic field intensity $\overrightarrow{B}$ , is given as:

$\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

On analysing the above formula it becomes clear that the magnetic force will always be perpendicular to the velocity of the charged particle. The direction of displacement is same as that of the velocity, hence the force is also perpendicular to the displacement which makes the work done equal to zero.

As the work done is zero, the change in kinetic energy is also zero. Hence, the kinetic energy of the charged particle does not change due to magnetic force.

Hence, the correct answer is option (b).

Answer:

The force on both the charged particles will be same, as they have same charge and the source of electric field is also same. As the force is same hence the change in the momentum for both of them will be same, and the mass of the proton is approximately 1837 times that of the positron. In the given amount of time the velocity attained by the proton will be lower than that by the positron, hence the distance travelled by the proton will be much lesser than that of the positron. Thus, the work done will be more in the case of a positron, as it moves away a larger distance.

Hence, the correct answer is option (c).

Answer:

When the man stands from the position of squatting, in the begining the force applied by the man is at some angle to the vertical. Thus, the magnitude of the normal component of the contact force by the ground is equal to the weight of the man and friction balances the horizontal component of the force applied by the man. Hence, initially the reaction force is more than mg and when the man stands up straight the reaction force becomes equal to mg.

Hence, the correct answer is option (d).  

Answer:

The displacement of the road due to the cycle is zero. Hence, the work done by the cyclist on the road is zero.

Hence, the correct answer is option (c).

Answer:

As the body is falling under the effect of gravity, which is a conservative force and there are no dissipative forces, as the experiment is taking place in the vacuum. Thus, the total mechanical energy of the system (Sum of Potential and Kinetic energies) remains constant.

Hence, the correct answer is option (c).

Answer:

During inelastic collision between two bodies, total linear momentum of the system remains conserved.

Hence, the correct answer is option (c).

Answer:

It is mentioned in the question that the tracks AB and AC are frictionless. Hence, the conservation of mechanical energy can be applied to both the stones. Both the stones descend through same height h, due to which the change in potential energy for both of them is same, consequently the change in kinetic energy will also be same. Thus, the two stones reach the bottom with the same speed. The acceleration of the stones I and II will be gsinθ₁ and gsinθ₂ , respectively. It can be observed easily that θ₁ < θ₂ thus, the acceleration of stone II is more. The length AC < AB. This means that the time taken by the stone II to reach the ground will be lesser than that by the stone I.

Hence, the correct answer is option (c).

Answer:

The body executing simple harmonic motion is at rest at extreme position whereas, its speed is maximum at the mean position. The position x = x_m represents the extreme position hence, the potential energy at the given position will be maximum that is equal to E and the kinetic energy will be minimum that is equal zero. 

Hence, the correct answer is option (b).

Answer:

In the elastic collision between two identical objects, their velocities are interchanged. Ball 1 collides with the ball 2, initially ball 2 is at rest and ball 1 is moving with the velocity v, so just after the collision their velocities will get interchanged. Hence, the ball 1 will come to rest and the ball 2 will start moving with the velocity v. At the next moment there is collision between ball 2 and ball 3. Thus, ball 3 starts moving with velocity v and ball 2 comes to rest.

Hence, the correct answer is option (b).

Answer:

To calculate the work done we need to find the force, as velocity v = a x^3/2 we can calculate the acceleration (A) as:

$A=v\frac{dv}{dx}=a{x}^{3/2}\frac{d\left(a{x}^{3/2}\right)}{dx}=\frac{3}{2}{a}^2{x}^2$
Thus, the force can be given calculated as:
$F=mA=m\left(\frac{3}{2}{a}^2{x}^2\right)=\frac{3}{2}m{a}^2{x}^2$
The work done can be calculated as:

$$\begin{gathered} W={\int }_0^{2}Fdx={\int }_0^2\frac{3}{2}m{a}^2{x}^{2}dx=\frac{3}{2}m{a}^2{\left(\frac{x^3}{3}\right)}_0^2=\frac{1}{2}m{a}^2\times 8 \\ \Rightarrow W=\frac{1}{2}\left(0.5\right)\times 25\times 8=50 \mathrm{J} \end{gathered}$$

Hence, the correct answer is option (b).
 

Answer:

As the body is moving under the influence of a source of constant power, the instantaneous power is given by:

$P=\overrightarrow{F}.\overrightarrow{v}$
When the force applied by the source on the body is F and the instantaneous velocity of the body is v.

For the simplicity of the calculations we can consider the angle between the force and the velocity to be 0^o.

Thus, P = Fv
As the power is constant, P = Fv = constant
Let the acceleration of the body at the given instant be a.

Thus, mav = constant
             av = constant
Let av = k 

$$\begin{gathered} \left(v\frac{dv}{dx}\right)v=k \\ {v}^{2}dv=kdx \end{gathered}$$

Integrating both sides
$$\begin{gathered} {\int }_0^v{v}^{2}dv=k{\int }_0^{x}dx \\ {\overline{)\frac{v^3}{3}}}_0^v={\overline{)kx}}_0^x \\ {v}^3=3kx \\ \Rightarrow v={\left(3kx\right)}^{1/3} \end{gathered}$$

We can express the velocity in terms of the displacement and time,
$$\begin{gathered} v={\left(3kx\right)}^{1/3}=\frac{dx}{dt} \\ \Rightarrow \frac{dx}{{\left(x\right)}^{1/3}}={\left(3k\right)}^{1/3}dt \end{gathered}$$
Integrating both sides

$$\begin{gathered} {\int }_0^d\frac{dx}{{\left(x\right)}^{1/3}}={\int }_0^t{\left(3k\right)}^{1/3}dt \\ \Rightarrow d^{2/3}=ct \left(\mathrm{Let} \frac{2\times {\left(3\mathrm{k}\right)}^{1/3}}{3} =c \right) \\ \Rightarrow d={\left(ct\right)}^{3/2} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

As the earth is moving in the elliptical orbit, and the sun lies at the one of the foci of that elliptical path the velocity of the earth in the elliptical orbit is minimum when the distance between the sun and the earth is maximum and vice versa is also true.  Hence, the kinetic energy is maximum when the earth is closest to the sun (at perigee) and minimum when the earth is farthest from the sun (at apogee). Kinetic energy cannot be zero or negative. Hence, the graph that suitably describes the given situation is given in option (d).

Hence, the correct answer is option (d).

Answer:

The motion of the simple pendulum in the air is the example of damped oscillations. The energy of the simple pendulum is decreasing exponentially with the time. The expression for the energy of a damped oscillator in terms of time is given as:

 $E=\frac{1}{2}k{A}^2{e}^{-bt/m}$
Where, k = SHM constant
A = Maximum amplitude
m = mass
b = damping constant

Hence, the correct answer is option (c).

Answer:

In this question we are assuming that the given object is moving in uniform circular motion. Thus, we need to calculate the average speed of the object.

Circumference of the path = 2πR = 2⨯π⨯1 = 2π m

Total distance travelled in one minute = 300 â¨¯ 2π = 600 π

Speed of the object = $\frac{600\pi }{60}=10\pi \mathrm{m}/\mathrm{s}$
Thus, the kinetic energy of the object = $\frac{1}{2}\times 5\times {\left(10\pi \right)}^2=250{\pi }^2$ J

Hence, the correct answer is option (a).

Answer:

The potential energy of the raindrop is maximum at the highest point as the raindrop falls its kinetic energy increases and the potential energy decreases. Due to air drag the kinetic energy cannot increases upto a certain level and after that it becomes constant. On the other side, potential energy decreases continously as the drop is falling continously. The graph given in option (b) is most suitable for the given situation.

Hence, the correct answer is option (b).

Answer:

As nothing is mentioned regarding air resistance so we shall ignore that in our calculation. In the absence of air resistance the total mechanical energy of the system remains conserved. 

Assuming the ground level to be datum (reference level) for the potential energy.


Initially, the energy of the shotput is given as:

PE_i + KE_i = 10 ⨯ 10 â¨¯ 1.5 + $\frac{1}{2}\times 10\times 1^2$ = 155 J

Finally, the shotput reaches the ground, hence its energy can be written as:

 PE_f + KE_f = 0 + KE_f  

Applying conservation of mechanical energy 

Thus,  KE_f = 155 J

Hence, the correct answer is option (d).

Answer:

When the iron sphere is inside the water, there are three forces acting on it:

1. Gravity
2. Buoyant force
3. Viscous drag

Gravity tends to increase the speed as well as the kinetic energy of the iron sphere, whereas the buoyant force and the viscous drag act in upward direction so tend to decrease the speed. Gravity and the buoyant forces are constant whereas, the viscous drag increases with increase in the speed. Thus, considering these factors, the kinetic energy of the iron sphere will first increase and after some time will become constant.

Hence, the correct answer is option (b).

Answer:

The batsman has to apply the force that is equal to average force applied by the ball on the bat. According to Newton's third law the force applied by the bat on the ball is equal in magnitude to the force applied by the ball on the bat.

Average force, $F=\frac{∆p}{∆t}$
If we consider the initial direction of motion of the ball as positive, thus the opposite direction will be negative.

$F=\frac{∆p}{∆t}=\frac{-mv-mu}{∆t}$

The collision is completely elastic, so the magnitude of the final velocity will be same as that of the initial. Thus, the magnitude of the force on the bat is:
$F=\frac{2mu}{∆t}$

Answer:

If we apply work-energy theorem in this situation, then we understand that as the initial and final kinetic energy of the man is zero, hence the work done by all the forces on the man is zero. The work done by the gravitational force is −mgL, as the man is moving in the direction opposite to the gravitational force. The work done by the reaction force on the man is zero as the point of application does not move.

Hence, the correct answers are options (b) and  (d).

Answer:

Let KE₁ and KE₂ be the kinetic energies of the bullet just before and just after it hits the target, and let v₁ and v₂ be the respective speeds of the bullet. 

$$\begin{gathered} {\mathrm{KE}}_1=\frac{1}{2}m{v}_1^2 \\ {\mathrm{KE}}_2=\frac{1}{2}m{v}_2^2 \end{gathered}$$
As the kinetic energy of the bullet has been reduced to half its value,

$$\begin{gathered} {\mathrm{KE}}_2=\frac{{\mathrm{KE}}_1}{2} \\ \frac{1}{2}m{v}_2^2=\frac{1}{2}\times \frac{1}{2}m{v}_1^2 \\ \Rightarrow v_2=\frac{v_1}{\sqrt{2}} \end{gathered}$$
As$\frac{1}{\sqrt{2}}>\frac{1}{2}$
Hence, we can say that v₂ is more than half of v₁, thus option (b) is correct. The bullet cannot continue with the same parabolic path as the velocity of the bullet has changed. As the kinetic energy of the bullet is lost, that will increase the internal energy of the target as the temperature of the target will increase.

Hence, the correct options are (b), (d) and (f).

Answer:

In the absence of friction the mechanical energy of the system of the two blocks and the spring will remain conserved, hence the collision can be considered as elastic. The momentum of the system will also remain conserved. The spring is fully compressed when the two blocks are moving with the same velocity. So, argument of conversion of all kinetic energy into potential when the spring is fully compressed is wrong. Hence, options (a) and (b) are not correct.
If the spring is massless, then the velocities of the two blocks will be exchanged so the final state of the block M₁ will be same as that of M₂ (as M₁ = M₂), i.e. rest.
If the surface on which the blocks are moving has friction, the mechanical energy will not be conserved. 

Hence, the correct answers are options (c) and (d).

Answer:

The calculation of the work done depends on the frame of reference. If we observe from the ground frame then it appears that due to friction the block is held at the wedge and is moving. But there is no relative motion between the block and the rough incline plane.

If we observe from the trolley frame, the block appears at rest and hence there is no work done by the frictional force and hence there is no dissipation of energy. 

Answer:

Electrical power is required to ensure that the elevator may not freely fall under the gravity when it is descending. The limit of the number of passengers is there because the elevator is held by a cable and it has certain limit of holding the load, if the load is further increased it will break.

Answer:

As the body is being raised, the work done by the applied force is positive, as the directions of the force and the displacement are same. The work done by the gravitational force is negative, as the directions of the force and the displacement are opposite.

Answer:

As the gravity is acting in vertical direction and the car is moving on a straight horizontal road, the force acting against the gravity will also be in vertical direction. Thus, the force is perpendicular to displacement and hence the work done by the force on the car will be zero.

Answer:

As the body is falling towards earth in air, the air drag acts on the body. The air drag is a non-conservative force and in its presence, the mechanical energy is not conserved. When the object falls, the potential energy will convert into kinetic energy but some part of the energy will also convert into the other forms as heat, sound etc. 

Answer:

The work done in moving the body in the closed loop will be zero only when conservative forces are present. In case the non-conservative forces are present, the initial and final kinetic energies may be different and hence the work done will not be zero.

Answer:

In an elastic collision of two billiard balls, total momentum is conserved. But during the collision, the balls may get deformed which is due to elastic potential energy. It implies that some part of the kinetic energy gets converted into elastic potential energy. Thus, total kinetic energy is not conserved.

Answer:

m = 100 kg, h = 10 m

Power is the rate of doing work.

Here, work done by the crane = mgh = 100 ⨯ 9.8 â¨¯ 10 = 9,800 J

Time taken, t = 20 s

Power, $P=\frac{9800}{20}=490 \mathrm{W}$

Answer:

Average work done by the human heart while it beats once, W' = 0.5 J

Average work done by the human heart in one minute, W = 72 â¨¯ 0.5 J = 36 J

Time, t = 1 minute = 60 s

Power = $\frac{W}{t}=\frac{36}{60}=0.6 \mathrm{watts}$

Answer:

When the work done by a force on a body is zero, the change in kinetic energy will be zero. For example work done by the tension in a string is zero, when the object attached to the string is whirled in a circle. The reason is that the angle between the tension and the displacement will always be perpendicular to each other. Other example that can be given in this context is that of a charged particle moving in uniform magnetic field, the work done by the magnetic force on the charged particle will be zero.

Answer:

According to the work energy theorem, the total work done on the body is equal to the change in kinetic energy. As both the bodies are moving with the same kinetic energy, hence the work required to be done to stop them will be same. As the force applied on them has same magnitude, assuming the angle between the directions of motion of the bodies and the forces applied on them to be same, it can be said that the distance moved by the bodies before coming to rest will be same.

Answer:

To solve this question the first point we need to consider is that the direction of motion of the particle is always along the tangent to the trajectory. Now, when the object is in vertical circular motion then at every instant its velocity will be along the tangent to the path. The object is able to execute the circular motion because there is a force (resultant of Tension and gravity's component) that is acting as the centripetal force. While in vertical circular motion, at point B the direction of the motion of the bob is vertically downwards. If the string is cut at point B, the tension force vanishes and the body will be only under the effect of gravity. In this case the direction of velocity as well as acceleration both are vertically downwards, hence the trajectory of the bob will be straight line and that too vertically downwards. On the same argument if we consider the point C, the direction of motion is horizontal and as the string is cut, the only force acting on it is gravity. Now, velocity is in horizontal direction and the acceleration is in vertically downward direction, hence the trajectory of the bob will be that of the horizontal projectile (parabola) with point C as the highest point. If we consider point X, the direction of motion is along the tangent and that is at certain angle to the horizontal, again if the string is cut the only force acting on the bob will be gravity and due to this the path will be oblique projectile (parabola) but the highest point this time will be higher than point C.

Answer:

Considering the conservation of mechanical energy, the sum of the kinetic and potential energy should remain same. Thus, E₀ is the total energy.  At point B, V(x) < E₀, so kinetic energy should have certain value other than zero. Further, on increasing x the potential energy is decreasing upto zero, so the kinetic energy should increase upto the maximum value upto point  C. From C to D the potential energy is zero, so the kinetic energy should remain maximum. From D to F, the potential energy is increasing linearly, so the kinetic energy should decrease linearly. Hence, the graph of the kinetic energy versus x is as shown below:




Kinetic energy is $K=\frac{1}{2}m{v}^2 \Rightarrow v=\pm \sqrt{\frac{2K}{m}}$ 
Considering the velocity's dependence on kinetic energy, its graph with respect to x can be drawn as:

Answer:

Let the mass of the balls be m, and the velocities of the first and second balls after the collision be v₁ and v₂ respectively.

Applying conservation of momentum

2mv₀ = mv₁ + mv₂

2v₀ = v₁ + v₂

The coefficient of restitution is defined as:
 $$\begin{gathered} e=\frac{\mathrm{velocity} \mathrm{of} \mathrm{separation}}{\mathrm{velocity} \mathrm{of} \mathrm{approach}} \\ e=\frac{v_2-v_1}{2v_0} \\ \Rightarrow v_1=v_0\left(1-e\right), v_2=v_1+2e{v}_0 \end{gathered}$$

As the value of  0 < e < 1, so both the velocities will be positive, hence in the same direction, in our calculation we have taken the forward velocity to be positive, so this proves that velocities of both the balls will be in forward direction.

(b) In any situation the linear momentum will remain conserved. 

Let $\overrightarrow{p},{\overrightarrow{p}}_{1 }\mathrm{and} {\overrightarrow{p}}_2$ be initial momentum and final momenta of the ball 1 and ball 2 respectively.

$\overrightarrow{p}={\overrightarrow{p}}_1+{\overrightarrow{p}}_2$

As the kinetic energy is not conserved

$\frac{p^2}{2m}>\frac{p_1^2}{2m}+\frac{p_2^2}{2m}$

$\Rightarrow p^2>p_1^2+p_2^2$




In the given figure if we apply triangle law of vector addition then we can say that the angle θ must be less than 90^o.
 

Answer:

In order to answer this question we need to understand that the kinetic energy of the particle cannot be negative whereas potential energy can be negative because the value of potential energy depends on the reference level chosen.
If we consider region A, where V > E that means the kinetic energy must be negative so such situation is impossible hence the particle cannot be found in region A.
In region B, V < E here the total energy is more than the potential energy hence the kinetic energy will have positive value, hence the particle can exist in region B.
In region C, K > E, that means the potential energy is negative, this situation is possible hence the particle can be found in the region C.
In region D, V > K, which is also possible in normal situations so the particle can also exist in region D.

Answer:

(a) As the collision is elastic and the two bob are identical, hence the velocities of the bobs will be interchanged. Bob B was at rest initially and bob A was falling, had certain velocity before collision; after collision as the velocities get interchanged, the bob A will come to rest and will not rise at all.

(b) The ball B will attain the velocity same as that of the ball A before collision, which can be calculated as:

$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 1}=4.42 \mathrm{m}/\mathrm{s}$

Answer:

In the given question, the mass of the raindrop, m = 1 g = 0.001 kg,
height from which it is falling, h = 1 km = 1000 m
speed with which it hits the ground, v = 50 m/s

(a) Loss of potential energy = mgh = 0.001 â¨¯ 10 â¨¯ 1000 = 10 J

(b) Gain in kinetic energy = $\frac{1}{2}m{v}^2=\frac{1}{2}\times 0.001\times {\left(50\right)}^2=1.25 \mathrm{J}$

(c) The loss in potential energy is not equal to the gain in kinetic energy, the reason being that a part of the potential energy is  used as the work done against the air drag.

Answer:

(a) In the given problem two pendulums have identical bobs and same lengths, and they collide elastically. When bob A is displaced by 10^∘ as it collides with the bob B it comes to rest and bob B attains the velocity that bob A had before collision. So, both the bobs will perform periodic motion, but one at a time.

(b) Graph for the given situation is as shown below:

Answer:

In the given problem, average terminal velocity of the raindrops, v = 9 m s^–1. 
Density of water, ρ = 1000 kg/m³

100 cm rain in a year means that in one year, every area of 1 m² collects the water upto a height of 100 cm. In the question the area given is also 1 m². So, the volume of the water collected in the given area, V = $\frac{100}{100}\times 1=1 {\mathrm{m}}^3$
Mass of the water collected, M = ρV = 1000 kg

Here we assume that as the water falls on the surface all its kinetic energy gets transferred to the surface. Hence, the energy transferred can be calculated as:

$E=\frac{1}{2}M{v}^2=\frac{1}{2}\times 1000\times {\left(9\right)}^2=4.05\times {10}^4 \mathrm{J}$

Answer:

According to the question 90% of the kinetic energy of the engine is lost due to friction, that means 10% of energy is used to compress the spring.

The kinetic energy of the engine can be written as:

$K=\frac{1}{2}m{v}^2=\frac{1}{2}\times 5\times {10}^4\times {10}^2=2.5\times {10}^6 \mathrm{J}$


Only 10% of the energy is being used to compress the spring, hence the energy stored in the spring is, E = 2.5 â¨¯10⁵J

Let the spring constant of the given spring be k.
Compression in the spring, x = 1 m

 $$\begin{gathered} \frac{1}{2}k{x}^2=2.5\times {10}^5 \\ \Rightarrow x=5\times {10}^5 \mathrm{N}/\mathrm{m} \end{gathered}$$

 

Answer:

Weight of the adult, W = 600 N

Height of each step, h = 0.25 m

In jogging for 6 km, the adult takes 6000 steps.

Energy utilised in jogging 6 km, E = 6000 â¨¯ 600 â¨¯ 0.25 = 9 â¨¯ 10⁵ J.

It is mentioned that the energy utilised is just 10% of the energy intake.
Thus, the energy intake = 10E =  9 â¨¯ 10⁶ J

Answer:

After combustion of 1 litre of the petrol car moves distance, d = 15 km =15000 m

 Heat energy to drive the car by combustion of one litre petrol, E = 0.5 × 3 × 10⁷ J = 1.5 × 10⁷ J

As the car moves with the constant velocity, the work done by the car is utilised in overcoming friction.

Let the frictional force be f.
∴ fd = 1.5 × 10⁷ J, d = 15000 m

$\Rightarrow f=1000 \mathrm{N}$

Answer:

In the given question, 
mass of the block, m = 1 kg
distance travelled along inclined plane, l = 10 m
force applied along inclined plane = 10 N
coefficient of friction, μ = 0.1
angle of inclination of the inclined plane, θ = 30^o 

(a) The work done against gravity = mg â¨¯ (vertical displacement)

vertical displacement =  l sin 30^o 

Thus, work done = 1 â¨¯ 10 â¨¯ 10 â¨¯ 0.5 = 50 J.                                                                                                 [∵ sin 30^o = 0.5 ]

(b) The work done against force of friction = (μmg cos θ )⨯ l = 0.1 â¨¯ 1 â¨¯10 â¨¯ 0.866 â¨¯ 10 = 8.66 J           [∵ cos 30^o​ = 0.866 ]

(c) Increase in potential energy = mg â¨¯ (vertical displacement)
                                                   = 1 â¨¯ 10 â¨¯ 10 â¨¯ 0.5 = 50 J.    

(d) To calculate the increase in kinetic energy we are required to calculate the speed of the block as it moves by a distance of 10 m.

Net force on block = $F-mg\sin {30}^{\mathrm{o}}-\mu mg\cos {30}^{\mathrm{o}}$
                               = $10-10\left(0.5\right)-0.1\left(10\right)0.87=10-5.87=4.13 \mathrm{N}$
Thus, acceleration of the block, a = 4.13 m/s² 

 $$\begin{gathered} v^2-u^2=2al \\ {v}^2=2\times 4.13\times 10=82.6 {(\mathrm{m}/\mathrm{s})}^2 \end{gathered}$$

Initial kinetic energy = 0

Hence, change in kinetic energy = $\frac{1}{2}m{v}^2=\frac{1}{2}\times 1\times 82.6=41.3 \mathrm{J}$
(e) Work done by applied force = F.l = 10 â¨¯ 10 = 100  J

Answer:

(a) Mechanical energy is conserved for balls 1 and 3. In case of ball 1, friction is enough that it rolls without slipping and mechanical energy remains conserved. In case of ball 2, as it slips its frictional force acts which is a non-conservative force so the mechanical energy is not conserved. The ball 3 has negligible friction, so even if it slips, no frictional force acts on it. The forces acting on ball 3 are gravity and the normal reaction. The normal reaction does not do any work in this case, and gravity is a conservative force so the mechanical energy remains conserved in this case.


(b) Only ball 3 can reach D.

For reaching D any ball is required to cross C. Ball 1 possesses friction so rolls without slipping, when it falls from A, the potential energy is converted into rotational and translational kinetic energies. After B, it starts rising its translational kinetic energy gets converted into potential energy but all of its rotational kinetic energy cannot be converted into potential energy. Thus, it is unable to reach C and cannot reach D. For ball 2 it is clear that because energy will get dissipated so it cannot cross C.  Ball 3 neither roll nor loses energy, so it has exact amount of energy required to cross C and thus it reaches D.

(c) None of the balls reach back A.

In case of the ball 2 it is clear because it loses energy, so it cannot reach A. Ball 1 is in rotational motion, as it 1 reaches B it will have rotation in the opposite sense due to which kinetic friction starts acting on it due to which it cannot reach A. 

Answer:

Calculation of kinetic energy of the rocket at time t and t + Δt :

At time t: ${\mathrm{KE}}_t=\frac{1}{2}M{v}^2$

For time t + Δt, 

It is mentioned that the rocket loses Δm amount of mass due to burning of fuel and it gains a velocity of Δv and the gas released has a relative velocity equal to u.

With respect to ground the velocities of the rockets and the emitted gases can be written as:

v_rocket = v + Δv

vgas = v − u

Thus, the kinetic energy of the system of the rocket and the evolved gases can be written as:

$$\begin{gathered} K{E}_{t+∆t}=\frac{1}{2}\left(M-∆m\right)\times {\left(v+∆v\right)}^2+\frac{1}{2}\left(∆m\right)\times {\left(v-u\right)}^2 \\ =\frac{1}{2}M{v}^2+\frac{1}{2}M∆v^2+Mv∆v-\frac{1}{2}∆m∆v^2-\frac{1}{2}∆m{v}^2-∆mv∆v+\frac{1}{2}∆m{v}^2+\frac{1}{2}∆m{u}^2-∆mvu \end{gathered}$$

Ignoring the terms including âˆ†v² and âˆ†mv∆v, the expression for the kinetic energy becomes :

${\mathrm{KE}}_{t+∆t}=\frac{1}{2}M{v}^2+Mv∆v+\frac{1}{2}∆m{u}^2-∆mvu$

We will apply the work-energy principle 

Change in kinetic energy = Work done 
$$\begin{gathered} {\mathrm{KE}}_{t+∆t}-{\mathrm{KE}}_t=\left(\frac{1}{2}M{v}^2+Mv∆v+\frac{1}{2}∆m{u}^2-∆mvu \right)-\frac{1}{2}M{v}^2=W \\ =\left(\frac{1}{2}∆m{u}^2 +v\left(M∆v-∆mu\right)\right)=W \end{gathered}$$ 
In the above equation (M∆v − âˆ†mu) = 0 because there is no external force on the system of the rocket and the evolved gas. 

$M\frac{dv}{dt}-\frac{dm}{dt}u=0$

Thus, the work done $=\left(\frac{1}{2}\right)∆m u^2$

Answer:

In the given question, 
Initial side length of the cube, l = 0.01 m
Area of cross section of the cube, ​A = l² = 0.0001 m
Young’s modulus, Y= 2 × 10¹¹ N/m²

Let ΔL be the change in side length due to collision


Kinetic energy of both boxes, K.E. = $2\times \frac{1}{2}m{v}^2=5\times {10}^{-4 }\mathrm{J}$
As the two blocks stop on collision, all their kinetic energies get converted into potential energy.

$\mathrm{P}.\mathrm{E}.=2\times \frac{1}{2}k{(\mathrm{∆}l)}^{\mathit{2}}$

Where, k is the compression constant which can be calculated as follows:

By Hooke's law 
$$\begin{gathered} \frac{F}{A}=Y\frac{∆l}{l} \\ \Rightarrow F=\left(\frac{YA}{l}\right)∆l ...\left(1\right) \end{gathered}$$

On comparing equation (1) with F = kΔl we get $k=\left(\frac{YA}{l}\right)=\left(\frac{Y{l}^2}{l}\right)=Yl$

Thus, the potential energy can be written as:

$\mathrm{P}.\mathrm{E}.=Yl{\mathit{(}\mathit{∆}\mathit{l}\mathit{)}}^{\mathit{2}}$
As all the kinetic energy gets converted into potential energy,

P.E. = K.E.
$$\begin{gathered} Yl{(\mathit{∆}l)}^{\mathit{2}}=\mathrm{K}.\mathrm{E}. \\ \Rightarrow \mathit{∆}l\mathit{=}\sqrt{\frac{\mathrm{K}.\mathrm{E}\mathit{.}}{Yl}}\mathit{=}\sqrt{\frac{5\times {10}^{-4}}{2\times {10}^{11}\times 0.1}}=1.58\times {10}^{-7 }\mathrm{m} \end{gathered}$$

Answer:

The first thing to understand here is the reason of rise of the balloon. It rises due to the buoyant force of the surrounding air. 

Let the mass of the balloon be m, volume of the balloon be V and the density of the helium gas inside the balloon be ρ_He and that of the air be ρ_air.

As the balloon rises it displaces the air of volume V (same as its own volume)

Buoyant force on the balloon, F_B = Vρ_airg
Net force on the balloon = Vρ_airg − Vρ_Heg
Let a be the acceleration of the balloon in the above direction

Applying Newton's second law,

 Vρ_airg − Vρ_Heg = ma                  ...(1)
$$\begin{gathered} V{\rho }_{\mathrm{air}}g-V{\rho }_{\mathrm{He}}g=m\frac{dv}{dt} \\ \left(V{\rho }_{\mathrm{air}}g-V{\rho }_{\mathrm{He}}g\right)dt=mdv \end{gathered}$$

The velocity (v) of the balloon at time t can be calculated by integrating with respect to t

$$\begin{gathered} \int \left(V{\rho }_{\mathrm{air}}g-V{\rho }_{\mathrm{He}}g\right)dt=\int mdv \\ \left(V{\rho }_{\mathrm{air}}g-V{\rho }_{\mathrm{He}}g\right)t=mv ...\left(2\right) \end{gathered}$$

 Above we have got the momentum of the balloon, we can find the kinetic energy using the formula, $K=\frac{p^2}{2m}$

Thus, the kinetic energy of the balloon can be calculated as:

$\frac{1}{2}m{v}^2=\frac{{\left[\left(V{\rho }_{\mathrm{air}}g-V{\rho }_{\mathrm{He}}g\right)t\right]}^2}{2m}=\frac{1}{2m}{V}^2\left({\rho }_{\mathrm{air}}-{\rho }_{\mathrm{He}}\right)g^2{t}^2$

Let the balloon rises upto a height of h 

Applying equation $s=ut+\frac{1}{2}a{t}^2$

As the intial velocity is zero, and the acceleration of the balloon from equation (1) can be written as : $a=\frac{V\left({\rho }_{\mathrm{air}}-{\rho }_{\mathrm{He}}\right)g}{m}$
Here, $h=\frac{1}{2}a{t}^2=\frac{1}{2}\frac{V\left({\rho }_{\mathrm{air}}-{\rho }_{\mathrm{He}}\right)}{m}g{t}^2$             ...(3)

Rewriting kinetic energy of the balloon in terms of the velocity calculated above in equation (2)
$\frac{1}{2}m{v}^2=\frac{1}{2}m{\left[\frac{\left(V{\rho }_{\mathrm{air}}g-V{\rho }_{\mathrm{He}}g\right)t}{m}\right]}^2=\frac{1}{2m}{V}^2{\left({\rho }_{\mathrm{air}}-{\rho }_{\mathrm{He}}\right)}^2{g}^2{t}^2$

Rearranging the terms

$\frac{1}{2}m{v}^2=\left[V\left({\rho }_{\mathrm{air}}-{\rho }_{\mathrm{He}}\right)g\right]\times \left[\frac{V\left({\rho }_{\mathrm{air}}-{\rho }_{\mathrm{He}}\right)g{t}^2}{2m}\right]$

From equation (3)

$$\begin{gathered} \frac{1}{2}m{v}^2=\left[V\left({\rho }_{\mathrm{air}}-{\rho }_{\mathrm{He}}\right)g\right]\times h \\ \frac{1}{2}m{v}^2=V{\rho }_{\mathrm{air}}gh-V{\rho }_{\mathrm{He}}gh \\ \Rightarrow \frac{1}{2}m{v}^2+V{\rho }_{\mathrm{He}}gh=V{\rho }_{\mathrm{air}}gh \end{gathered}$$

The terms in the left hand side are the kinetic and potential energies of the balloon and the term in the right hand side is the change in the potential energy of the air.

From the above equation we can understand that as the balloon goes up, an equal volume of the air is displaced and has to come down. Thus, the increase in the mechanical energy (potential + kinetic) of the balloon is compensated by the decrease in the potential energy of the displaced air that comes down. So we can conclude that if the effects of the dissipative forces are ignored then the mechanical energy of the system of the balloon and the air remains conserved.