1 Marks Question

Question 11 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 21 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 31 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 41 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 51 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 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 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 71 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 81 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 91 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 101 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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Question 111 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 121 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 131 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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Question 141 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 151 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 161 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 171 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 181 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 191 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 201 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 211 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 221 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 231 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 241 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 251 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 261 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 271 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:

$\text{t}^{\frac{1}{2}}$
$\text{t}$
$\text{t}^\frac{3}{2}$
$\text{t}^2$

Answer

Solution:

From,

V = u + at

V = 0 + at = at

As power, P = F × V

$\therefore$ P = (ma) × at = ma²t

As m and a are constants, therefore, $\text{P}\propto\text{t}$

Hence, right choice is (ii) t.

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Question 281 Mark

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to:

$\text{t}^{\frac{1}{2}}$
$\text{t}$
$\text{t}^\frac{3}{2}$
$\text{t}^2$

Answer

$\text{t}^\frac{3}{2}$

Solution:

As power, P = force × veclocity

$\text{P}=\big[\text{MLT}^{-2}\big]\big[\text{LT}^{-1}\big]=\big[\text{ML}^2\text{T}^{-3}\big]$

As, $\text{P}=\big[\text{ML}^2\text{T}^{-3}\big]$

= 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 291 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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Question 301 Mark

Is it possible to have a collision in which all kinetic energy is lost? Give example.

Answer

Consider two particles having momentum m₁u₁ and m₂u₂ colliding perfectly inelastically to form single mass (m₁ + m₂) 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 311 Mark

Answer

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Question 321 Mark

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 331 Mark

It is possible to have a situation when E - U < 0?

Answer

No. Since, E = K + U.
341 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 351 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 361 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 371 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 381 Mark

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 391 Mark

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 401 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 411 Mark

How many ergs make one joule?

Answer

One joule = 10= erg.

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Question 421 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

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Question 431 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 441 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 451 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 461 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 471 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

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Question 481 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 491 Mark

What type of energy stored in:

The wound spring of a watch.
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 501 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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