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Physics 1 Marks Question

Work, Energy, and Power · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 109 questions.

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109 questions · 2 auto-graded MCQ + 107 self-marked written.

Question 11 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer
Negative: The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.
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Question 21 Mark
Underline the correct alternative: In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/ total linear momentum/ total energy of the system of two bodies.
Answer
Total linear momentum: The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.
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Question 31 Mark
Answer carefully, with reasons: If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note: we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Answer
Elastic. Explanation: In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.
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Question 41 Mark
Underline the correct alternative: The rate of change of total momentum of a many-particle system is proportional to the external force/ sum of the internal forces on the system.
Answer
External force: If their is no internal force work on body then the external force is require to attend change in linear momentum.
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MCQ 51 Mark
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time $t$ is proportional to:
  • A
    $\text{t}^{\frac{1}{2}}$
  • B
    $\text{t}$
  • ✓
    $\text{t}^\frac{3}{2}$
  • D
    $\text{t}^2$
Answer
Correct option: C.
$\text{t}^\frac{3}{2}$
As power,$P =$ force $\times$ veclocity
$\text{P}=\big[\text{MLT}^{-2}\big]\big[\text{LT}^{-1}\big]=\big[\text{ML}^2\text{T}^{-3}\big]$
As, $\text{P}=\big[\text{ML}^2\text{T}^{-3}\big]$
$= \text{Constant}$
$\therefore\text{ L}^2\text{T}^3=\text{Constant}$
Or, $\frac{\text{L}^2}{\text{T}^3}=\text{Constant}$
$\therefore\text{ L}^2\propto\text{T}^3$
Or, $\text{L}\propto \text{T}^{\frac{3}{2}}$
Hence, right choice is $(iii) \text{t}^{\frac{3}{2}}$
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Question 61 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Work done by friction in 10 s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$ Total force = F - f = 7 - 1.96 = 5.04N
Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Work done by friction = -fx = -1.96 × 125 = -245J
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Question 71 Mark
Underline the correct alternative: Work done by a body against friction always results in a loss of its kinetic/potential energy.
Answer
Kinetic energy: The work done against the direction of friction decrease the velocity of a body. So, there is a loss of kinetic energy of the body.
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Question 81 Mark
State if the following statements is true or false. Give reasons for your answer: Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
Answer
False. Explanation: The external forces on the body may change the total energy of the body.
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MCQ 91 Mark
A body is initially at rest. It undergoes one$-$dimensional motion with constant acceleration. The power delivered to it at time $t$ is proportional to:
  • A
    $\text{t}^{\frac{1}{2}}$
  • ✓
    $\text{t}$
  • C
    $\text{t}^\frac{3}{2}$
  • D
    $\text{t}^2$
Answer
Correct option: B.
$\text{t}$
From,
$V=u+a t$
$V=0+a t=a t$
As power, $\mathrm{P}=\mathrm{F} \times \mathrm{V}$
$\therefore P=(m a) \times \text { at }=m a^2 t$
As $m$ and a are constants, therefore, $\mathrm{P} \propto \mathrm{t}$
Hence, right choice is $(ii)\  t $.
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Question 101 Mark
Answer carefully, with reasons: In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
Answer
No. Explanation: K.E. is not conserved during the given elastic collision, K.E. before and after collision is the same. Infact, during collision, K.E. of the balls gets converted into potential energy.
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Question 111 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
Answer
Positive: Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.
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Question 121 Mark
Answer the following: Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
Answer
The gravitational force is a conservative force, hence, work done by the gravitational force over one complete (closed) orbit of comet is zero.
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Question 131 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Work done by the net force on the body in 10s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$
Total force = F - f = 7 - 1.96 = 5.04N Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Net work done = (875 - 245) = 630J
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Question 141 Mark
Answer carefully, with reasons: What are the answers to (a) and (b) for an inelastic collision?
Answer
No; Yes. Explanation: In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision. The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.
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Question 151 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by friction on a body sliding down an inclined plane.
Answer
Negative: Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.
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Question 161 Mark
Answer carefully, with reasons: Is the total linear momentum conserved during the short time of an elastic collision of two balls?
Answer
Yes. Explanation: In an elastic collision, the total linear momentum of the system always remains conserved.
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Question 171 Mark
State if the following statements is true or false. Give reasons for your answer: In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Answer
True. Explanation: In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.
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Question 181 Mark
State if the following statements is true or false. Give reasons for your answer: In an elastic collision of two bodies, the momentum and energy of each body is conserved.
Answer
False. Explanation: In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.
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Question 191 Mark
State if the following statements is true or false. Give reasons for your answer: Work done in the motion of a body over a closed loop is zero for every force in nature.
Answer
False. Explanation: Work done in the motion of a body over a closed loop is zero for a conservation force only.
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Question 201 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
Answer
Positive: In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.
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Question 211 Mark
A bolt of mass 0.3kg falls from the ceiling of an elevator moving down with an uniform speed of $7ms^{-1}$. It hits the floor of the elevator (length of the elevator = 3m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
Answer
Mass of the bolt, m= 0.3kg Speed of the elevator = 7m/s Height, h = 3m Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy. Heat produced = Loss of potential energy = mgh = 0.3 × 9.8 × 3 = 8.82J The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.
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Question 221 Mark
Answer the following: An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
Answer
As an artificial satellite gradually loses its energy due to dissipation against atmospheric resistance, its potential energy decreases rapidly. As a result, kinetic energy of satellite slightly increases i.e. its speed increases progressively.
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Question 231 Mark
A $1kg$ block situated on a rough incline is connected to a spring of spring constant $100Nm^{-1}$ as shown in. The block is released from rest with the spring in the unstretched position. The block moves 10cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.
Answer
Mass of the block, $m = 1kg$ Spring constant, $k = 100Nm^{-1}$ Displacement in the block, $X = 10cm = 0.1m$ The given situation can be shown as in the following figure.At equilibrium:

 Normal reaction, $\text{R}=\text{mg}\cos37^{\circ}$ Frictional force, $\text{f}=\mu\text{R}=\text{mg}\sin37^{\circ}$ Where, $\mu$ is the coefficient of friction Net force acting on the block $ =\text{mg}\sin 37^{\circ}-\text{f}$ $=\text{mg}\sin37^{\circ}-\mu\text{m}\cos37^{\circ}$ $=\text{mg}(\sin37^{\circ}-\mu\cos37^{\circ})$ At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e., $=\text{mg}(\sin37^{\circ}-\mu\cos37^{\circ})\text{x}=\Big(\frac{1}{2}\Big)\text{kx}^2$ $1\times9.8(\sin37^{\circ}-\mu\cos37^{\circ})=\Big(\frac{1}{2}\Big)\times100\times(0.1)$ $0.602-\mu\times0.799=0.510$ $\therefore\mu=\frac{0.092}{0.799}=0.115$
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Question 241 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by gravitational force in the above case.
Answer
Negative: In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.
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Question 251 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Change in kinetic energy of the body in 10s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$ Total force = F - f = 7 - 1.96 = 5.04N
Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Change in kinetic energy = Net work done = 630J
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Question 261 Mark
Answer the following: The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
Answer
Heat energy required for burning of casing of rocket comes from the rocket itself. As a result of work done against friction the kinetic energy of rocket continuously deceases and this work against friction reappears as heat energy.
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Question 271 Mark
Underline the correct alternative: When a conservative force does positive work on a body, the potential energy of the body increases/ decreases/ remains unaltered.
Answer
Decreases: When we work on the body then i may be possible it become in motion so if we work on body, then kinetic energy increase and potential energy decrease.
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Question 281 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Work done by the applied force in 10s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$ Total force = F - f = 7 - 1.96 = 5.04N
Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Work done by F is W = Fx = 7 × 125 J = 875J
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Question 291 Mark
A man raises a mass m to a height 'h' and then shifts it horizontally by a length x'. What is the work done against the force of gravity?
Answer
Work done against force of gravity is mgh since no work is done in the horizontal displacement.
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Question 301 Mark
What is the source of kinetic energy of the bullet coming out of the bullet of a rifle?
Answer
The source of kinetic energy of bullet is the potential energy of the compressed spring in the loaded rifle.
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MCQ 311 Mark
A force $\Big(\frac{\text{mv}^2}{\text{r}^2}\Big)$ is acting on a body of mass m moving with a velocity $v$ in a circle of radius $r$. What is the work done by the force in moving the body over half the circumference of the circle?
  • A
    $\frac{\text{mv}^2}{\text{r}}\times\pi\text{r}$
  • B
    $\text{Zero}$
  • C
    $\frac{\text{mv}^2}{\text{r}^2}$
  • D
    $\frac{\pi\text{r}^2}{\text{mv}^2}$
Answer
Work done is zero because force is acting at $90^\circ$ to the direction of motion $\text{w}=\text{Fs}\cos90^\circ=0$
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Question 321 Mark
Can mechanical energy be negative?
Answer
As E = K + UExplanation:
E is negative if U is negative and is numerically greater than K.
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Question 331 Mark
Work done by weight of a body moving horizontally is ­________.
Answer
Work done by weight of a body moving horizontally is ­zero .
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Question 341 Mark
Calculate the power of an electric engine which can lift 20 tonne of coal per hour from a mine 180m deep.
Answer
$\text{Power}=\frac{\text{Work done}}{\text{Time taken}}=\frac{\text{mgh}}{\text{t}}$ $=\frac{20\times1000\times9.8\times180}{60\times60}$ $=9800\text{W}$ $=\text{W}=9.8\text{kW}$
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Question 351 Mark
State work-energy theorem.
Answer
According to work-energy theorem, the work done on a body is equal to the change in K.E in it.
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Question 361 Mark
Answer carefully, with reasons: In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
Answer
No. Explanation: K.E. is not conserved during the given elastic collision, K.E. before and after collision is the same. Infact, during collision, K.E. of the balls gets converted into potential energy.
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Question 371 Mark
When force acting on a body is perpendicular to direction of motion then work done is ­­­­­­­­­________.
Answer
When force acting on a body is perpendicular to direction of motion then work done is ­­­­­­­­­Zero.
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Question 381 Mark
Give the characteristics of inelastic collision.
Answer
  1. Kinetic energy does not remain conserved.
  2. Linear momentaum of the system remains conserved.
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Question 401 Mark
Answer with reason: In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e., when they are in contact)?
Answer
Total K.E. is not conserved because a part of K.E. is used in deforming the balls during that short intevral.
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Question 411 Mark
Write down characteristics of elastic collision.
Answer
  1. Kinetic energy of the system remains conserved.
  2. Linear momentaum of the system remains conserved.
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Question 421 Mark
Is it possible to exert a force which does work on a body without changing its kinetic energy. If, so give example.
Answer
Yes, When spring is compressed or when a body is pulled with a constant velocity on a rough horizontal surface.
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Question 431 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
Answer
Positive: Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.
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Question 441 Mark
Answer the following: Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
Answer
The gravitational force is a conservative force, hence, work done by the gravitational force over one complete (closed) orbit of comet is zero.
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Question 461 Mark
Does the work done in raising a box on to a platform depends upon how fast it is raised up? If not why?
Answer
No. Work done against gravity depends only on the initial and final points. The change in energy is same as the work done.
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Question 471 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Work done by the net force on the body in 10s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$
Total force = F - f = 7 - 1.96 = 5.04N Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Net work done = (875 - 245) = 630J
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Question 481 Mark
Answer carefully, with reasons: What are the answers to (a) and (b) for an inelastic collision?
Answer
No; Yes. Explanation: In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision. The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.
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Question 491 Mark
Two bodies of masses $M_1$ and $M_2$ have equal kinetic energy. The ratio of their linear momenta is ______.
Answer
Two bodies of masses $M_1$ and $M_2$ have equal kinetic energy. The ratio of their linear momenta is $\sqrt{\frac{\text{M}_1}{\text{M}_2}}$.
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Question 501 Mark
Why is the work done by centripetal force zero?
Answer
Because centripetal force is perpendicular to the displacement of the body.
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Question 511 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by friction on a body sliding down an inclined plane.
Answer
Negative: Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.
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Question 521 Mark
What type of energy stored in:
  1. The wound spring of a watch.
  2. A stretched bow while ready to project an arrow?
Answer
In both cases, energy is stored in the form of elastic potential energy due to change in configuration.
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Question 531 Mark
Can acceleration be produced without doing any work? Give example.
Answer
Yes, for uniform circular motion, no work done but a centripetal acceleration is present.
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Question 541 Mark
When work done by a force is independent of the path of motion but depends only on the initial and final position of the body displaced. Such force is known as_____________.
Answer
When work done by a force is independent of the path of motion but depends only on the initial and final position of the body displaced. Such force is known as conservative force.
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Question 551 Mark
Energy possessed by a body by virtue of its position or configuration is known as ­________.
Answer
Energy possessed by a body by virtue of its position or configuration is known as ­potential energy .
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Question 561 Mark
Is the total linear momentum conserved during the short time of an elastic collision of two balls?
Answer
During the short interval of an elastic collision, total linear momentum is conserved.
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Question 571 Mark
Answer carefully, with reasons: Is the total linear momentum conserved during the short time of an elastic collision of two balls?
Answer
Yes. Explanation: In an elastic collision, the total linear momentum of the system always remains conserved.
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Question 581 Mark
What is 1 eV?
Answer
Kinetic energy acquired by an electron to pass through a potential difference of 1volt.
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Question 591 Mark
A box is pushed through 4.0m across a floor offering 100N resistance. How much work is done by the resisting force?
Answer

$\text{F}=100\text{N}$
$\text{S}=4\text{m},\theta=0^\circ$
$\omega=\vec{\text{F}}.\vec{\text{S}}$
$=100\times4=400\text{J}$
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Question 601 Mark
A truck and a car are moving with the same kinetic energy on a straight road. Their engines are simultaneously switched off. Which one will stop at a lesser distance?
Answer
$\text{S}=\frac{\text{K.E}}{\text{Reatarding force}}.$In this case retarding force is the force of friction (f); e, $\text{f}=\mu\text{R}$.Since the weight of truck is more, R will be more in case of truck. Therefore, f will be more. Obviously, S would be less in case of truck so it will come to rest at lesser distance.
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Question 621 Mark
What is one electron-volt?
Answer
$1 \mathrm{eV}=1.6 \times 10^{-19}$ joule. It is the energy required to transfer electron (charge $=1.6 \times 10^{-19 \circ} \mathrm{C}$ ) through a P.D. of 1 volt.
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Question 631 Mark
Commercial unit of power kWh is equal to ­________.
Answer
Commercial unit of power kWh is equal to $3.6 \times 10^6J$.
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Question 651 Mark
In an elastic collision, relative velocity of approach between two bodies is always ______to relative velocity of separation.
Answer
In an elastic collision, relative velocity of approach between two bodies is always equal to relative velocity of separation.
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Question 661 Mark
Is it necessary to do work to maintain a constant velocity with a body on a rough surface?
Answer
Yes, to overcome the frictional force of rough surface.
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Question 671 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
Answer
Positive: In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.
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Question 681 Mark
Ten identical balls are placed in contact on a smooth surface. If an eleventh identical ball moving with a speed 'u' collides on the first, what will you notice?
Answer
When the eleventh ball hits the first ball, the tenth ball will start moving with the same speed due to transfer of momentum of equal proportion.
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Question 691 Mark
A coolie with load on his head is walking on a horizontal plateform. What is work done against gravity?
Answer
Zero.
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Question 701 Mark
A mass is moving in a circular path with constant speed. What is the work done in $\frac{3}{4}\text{th}$ of a rotation?
Answer
Since centripetal force and displacement are perpendicular, work done is zero.
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Question 711 Mark
Write the work energy theorem.
Answer
Work done by a force is the change in kinetic energy associated with the body.
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Question 721 Mark
Work-energy theorem states that work done by the net force acting on a body is equal to the change in­________ of the body.
Answer
Work-energy theorem states that work done by the net force acting on a body is equal to the change in­ kinetic energy of the body.
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Question 731 Mark
Is it possible to have a collision in which all kinetic energy is lost? Give example.
Answer
Consider two particles having momentum $m_1u_1$ and $m_2u_2$ colliding perfectly inelastically to form single mass ($m_1 + m_2$) moving with velocity v. For K.E. to be completely lost $\frac{1}{2}(\text{m}_1+\text{m}_2)\text{v}^2=0 \text{ or}\text{ v}=0$ $\text{ i.e}\frac{\text{m}_1\text{u}_1+\text{m}_2\text{u}_2}{\text{m}_1+\text{m}_2}=0$ $\text{m}_1\text{u}_1=-\text{m}_2\text{u}_2$ i.e., the two particles should have equal and opposite momentum for K.E to be completely lost.
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Question 741 Mark
A bolt of mass 0.3kg falls from the ceiling of an elevator moving down with an uniform speed of $7ms^{-1}$. It hits the floor of the elevator (length of the elevator = 3m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
Answer
Mass of the bolt, m= 0.3kg Speed of the elevator = 7m/s Height, h = 3m Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy. Heat produced = Loss of potential energy = mgh = 0.3 × 9.8 × 3 = 8.82J The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.
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Question 751 Mark
Two equal masses, one at rest and another moving undergo elastic oblique collision. If one mass goes at an angle $\frac{\pi}{3}$with its original direction of motion, what is the direction of motion of the other?
Answer
The direction of the other mass is at an angle$\Big(\frac{\pi}{3}-\frac{\pi}{3}\Big),\text{i.e.},\frac{\pi}{6}\text{ or }30^\circ.$
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Question 761 Mark
Is work done by a non conservative force always negative? Comment.
Answer
No. For example, work done by a non-conservative force like friction is zero, so long as the body does not start moving. Again when friction causes motion, work done by friction is positive.
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Question 771 Mark
A light body and a heavy body have the same momentum. Which one will have greater kinetic energy?
Answer
As $\text{K}=\frac{\text{p}^2}{2\text{m}}$Since p is same for both bodies $\text{K}\propto\frac{1}{\text{m}},$i.e., lighter body has more K.E. than the heavier body.
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Question 781 Mark
Answer the following: An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
Answer
As an artificial satellite gradually loses its energy due to dissipation against atmospheric resistance, its potential energy decreases rapidly. As a result, kinetic energy of satellite slightly increases i.e. its speed increases progressively.
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Question 791 Mark
A $1kg$ block situated on a rough incline is connected to a spring of spring constant $100Nm^{-1}$ as shown in. The block is released from rest with the spring in the unstretched position. The block moves $10cm$ down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.
Answer
Mass of the block, $m = 1kg$ Spring constant, $k = 100Nm^{-1}$ Displacement in the block, $X = 10cm = 0.1m$ The given situation can be shown as in the following figure.At equilibrium:

 Normal reaction, $\text{R}=\text{mg}\cos37^{\circ}$ Frictional force, $\text{f}=\mu\text{R}=\text{mg}\sin37^{\circ}$ Where, $\mu$ is the coefficient of friction Net force acting on the block $ =\text{mg}\sin 37^{\circ}-\text{f}$ $=\text{mg}\sin37^{\circ}-\mu\text{m}\cos37^{\circ}$ $=\text{mg}(\sin37^{\circ}-\mu\cos37^{\circ})$ At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e., $=\text{mg}(\sin37^{\circ}-\mu\cos37^{\circ})\text{x}=\Big(\frac{1}{2}\Big)\text{kx}^2$ $1\times9.8(\sin37^{\circ}-\mu\cos37^{\circ})=\Big(\frac{1}{2}\Big)\times100\times(0.1)$ $0.602-\mu\times0.799=0.510$ $\therefore\mu=\frac{0.092}{0.799}=0.115$
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Question 801 Mark
A body falling from a height of 10m rebounds from a hard floor. It loses 20% of its energy in impact. What is the height to which it would rise after the impact?
Answer
$\frac{80}{100}\text{mgh}=\text{mgh}'$ $\text{h}'=\frac{4}{5}\text{h}=\frac{4}{5}\times10\text{m}=8\text{cm}$
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Question 811 Mark
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: Work done by gravitational force in the above case.
Answer
Negative: In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.
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Question 821 Mark
Two blocks of masses $10 kg$ and $20 kg$ moving at speeds of $10 \mathrm{~ms}^{-1}$ and $20 \mathrm{~ms}^{-1}$ respectively in opposite directions, approach each other and collide. If the collision is completely inelastic, find the thermal energy developed in the process.
Answer
Given, Mass of the first block, $m_1=10 \mathrm{~kg}$ Mass of the second block, $m_2=20 \mathrm{~kg}$ Initial velocity of the first block, $u_1=$
$10 \mathrm{~m} / \mathrm{s}$ Initial velocity of the second block, $u_2=20 \mathrm{~m} / \mathrm{s}$ Let the velocity of the blocks after collision be v . Applying conservation of momentum, we get $\mathrm{m}_2 \mathrm{u}_2-\mathrm{m}_1 \mathrm{u}_1=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{v} \Rightarrow 20 \times 20-10 \times 10=(10+20) \mathrm{v} \Rightarrow 400-100=30 \mathrm{v}$
$\Rightarrow 300=30 \mathrm{v} \Rightarrow \mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ Initial kinetic energy is given by, $\mathrm{K}_{\mathrm{i}}=\frac{1}{2} \mathrm{~m}_1 \mathrm{u}_1^2+\frac{1}{2} \mathrm{~m}_2 \mathrm{u}_2^2$
$\mathrm{K}_{\mathrm{i}}=\frac{1}{2} \times 10 \times(10)^2+\frac{1}{2} \times 20 \times(20)^2 \mathrm{~K}_{\mathrm{i}}=500+4000=4500$ Final kinetic energy is given by, $\mathrm{K}_{\mathrm{f}}=\frac{1}{2}\left(\mathrm{~m}_1+\mathrm{m}_2\right) \mathrm{v}^2 \mathrm{~K}_{\mathrm{f}}=\frac{1}{2}(10+20)(10)^2 \mathrm{~K}_{\mathrm{f}}=\left(\frac{30}{2}\right) \times 100=1500$
$\therefore$ Total change in $\mathrm{KE}=4500 \mathrm{~J}-1500 \mathrm{~J}=$ 3000 J Thermal energy developed in the process $=3000 \mathrm{~J}$
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Question 831 Mark
Moutain roads rarely go straight up but wind up gradually. Why?
Answer
Frictional force f is given by $\text{f}=\mu\text{mg}\cos\theta$. If the roads go straight up the angle of slope $\theta$ would be large and frictional force will be less and vehicles may slip.
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Question 841 Mark
What is the work done by earth's gravitational force in keeping the moon in its orbit?
Answer
Force and displacement are always perpendicular. So W = 0.
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Question 861 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Change in kinetic energy of the body in 10s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$ Total force = F - f = 7 - 1.96 = 5.04N
Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Change in kinetic energy = Net work done = 630J
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Question 871 Mark
What happens when two identical objects moving in mutually opposite directions suffer an elastic collision?
Answer
The objects mutually exchange their velocities as a consequence of collision.
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Question 881 Mark
What is meant by mass energy equivalence? Discuss its significance in Physics.
Answer
Energy and non-mass are interconvertible. In Physics, it gives explanation for the energy released in nuclear reactions.
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Question 891 Mark
In what type of collision, maximum kinetic energy is transferred?
Answer
Maximum kinetic energy is transferred when two bodies of equal mass collide.
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Question 901 Mark
When an air bubble rises in water, what happens to its potential energy?
Answer
Potential energy of air bubble decreases because work is done by upthrust on the bubble.
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Question 911 Mark
A spring is stretched. Is the work done by the stretching force positive or negative?
Answer
Positive work. Both the force and the displacement are in the same direction.
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Question 921 Mark
Area under the force-displacement curve is the measure of ______.
Answer
Area under the force-displacement curve is the measure of work done .
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Question 931 Mark
What is the source of the kinetic energy of the falling rain drop?
Answer
Potential energy stored in the rain drops.
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Question 941 Mark
Is work done a scalar or a vector?
Answer
Work done by a force for a certain displacement is a scalar quantity.
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Question 951 Mark
State the law of Conservation of energy.
Answer
Energy can neither be created nor destroyed, but only can be transformed from one form to another.
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Question 961 Mark
Why is it not necessary for a body following another body, to stop to avoid collision?
Answer
If relative velocity of object is zero, then collision can be stop.
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Question 971 Mark
In a game of tug of war, one team is slowly giving way to the other. Which team is doing positive work and which team negative?
Answer
The winning team (i.e. the team which is pulling the other team towards itself) is doing positive work and the losing team (i.e. the team slowly giving way to the other) is doing negative work.
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Question 981 Mark
The earth moving round the sun in circular orbit is acted upon by a force and hence work must be done on the earth by the force. Explain.
Answer
It is wrong. Since force and displacement are perpendicular to each other work done is zero.
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Question 1001 Mark
Answer the following: The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
Answer
Heat energy required for burning of casing of rocket comes from the rocket itself. As a result of work done against friction the kinetic energy of rocket continuously deceases and this work against friction reappears as heat energy.
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Question 1011 Mark
What is the nature of force involved in the winding of a watch?
Answer
As the energy is recoverable therefore the force is conservative force.
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Question 1021 Mark
It is possible to have a situation when E - U < 0?
Answer
No. Since, E = K + U.
(E - U) represents K which cannot be negative.
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Question 1041 Mark
Underline the correct alternative: When a conservative force does positive work on a body, the potential energy of the body increases/ decreases/ remains unaltered.
Answer
Decreases: When we work on the body then i may be possible it become in motion so if we work on body, then kinetic energy increase and potential energy decrease.
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Question 1051 Mark
Which physical terms remain conserved in an inelastic collision?
Answer
In an inelastic collision, total momentum as well as total energy remain conserved.
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Question 1061 Mark
Can kinetic energy of a body be negative? Can potential energy be negative?
Answer
Kinetic energy of a body cannot be negative because it is given by $\frac{1}{2}\text{mv}^2$ and is always positive whether v is -ve or +ve. The gravitational potential energy of a body may be negative or positive.
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Question 1071 Mark
Is it practically possible to have situation where (E - U) < 0?
Answer
No, it is practically not possible because (E - U) represents KE, which cannot be negative.
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Question 1081 Mark
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Work done by the applied force in 10s.
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$ Total force = F - f = 7 - 1.96 = 5.04N
Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Work done by F is W = Fx = 7 × 125 J = 875J
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Question 1091 Mark
An elevator can carry a maximum load of $1800 kg$ (elevator + passengers) is moving up with a constant speed of $2 m s ^{-1}$. The frictional force opposing the motion is $4000 N$. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.
Answer
The downward force on the elevator is
$
F=m g+F_f=(1800 \times 10)+4000=22000 N
$
The motor must supply enough power to balance this force. Hence,
$
P= F . v =22000 \times 2=44000 W =59 hp
$
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