
Q1 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: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest. Ans: (a) 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.
(b) 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.
(c) 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.
(d) 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.
(e) 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.
Q2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results. Ans: Mass of the body, m= 2 kg
Applied force, F = 7 N
Coefficient of kinetic friction, µ= 0.1
Initial velocity, u= 0
Time, t = 10 s
The acceleration produced in the body by the applied force is given by Newton's second law of motion as:
a' = F / m = 7/2 = 3.5 m/s^{2}
Frictional force is given as:
f = µmg
= 0.1 × 2 × 9.8 = 1.96 N
The acceleration produced by the frictional force:
a'' =  1.96 / 2 = 0.98 m/s^{2}
Total acceleration of the body:
a = a' + a''
= 3.5 + (0.98) = 2.52 m/s^{2}
The distance travelled by the body is given by the equation of motion:
s = ut + ½ at^{2}
= 0 + ½ x 2.52 x 10^{2}
= 126 m
(a) Work done by the applied force, Wa= F × s = 7 ×126 = 882 J
(b) Work done by the frictional force, W_{f} = F× s= 1.96 ×126 = 247 J
(c) Net force = 7 + (1.96) = 5.04 N Work done by the net force, W_{net}= 5.04 ×126 = 635 J
(d) From the first equation of motion, final velocity can be calculated as:
v = u + at = 0 + (25.2)^{2} ×10 = 25.2 m/s
Change in kinetic energy = 1/2 mv^{2}  1/2 mu^{2}
= ½ x 2(v^{2}  u^{2}) = (25.2)^{2}  0^{2} = 635 j
Q6 Underline the correct alternative: (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic/potential energy. (c) The rate of change of total momentum of a manyparticle system is proportional to the external force/sum of the internal forces on the system. (d) 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. Ans: (a) Decreases
(b) Kinetic energy
(c) External force
(d) Total linear momentum
Explanation: (a) A conservative force does a positive work on a body when it displaces the body in the direction of force. As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body.
(b) The work done against the direction of friction reduces the velocity of a body. Hence, there is a loss of kinetic energy of the body.
(c) Internal forces, irrespective of their direction, cannot produce any change in the total momentum of a body. Hence, the total momentum of a many particle system is proportional to the external forces acting on the system.
(d) The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.
Q7 State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. Ans: (a) False (b) False (c) False (d) True
Explanation: (a) In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.
(b) Although internal forces are balanced, they cause no work to be done on a body. It is the external forces that have the ability to do work. Hence, external forces are able to change the energy of a system.
(c) The work done in the motion of a body over a closed loop is zero for a conservation force only.
(d) 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.
Q8 Answer carefully, with reasons: (a) 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)? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls? (c) What are the answers to (a) and (b) for an inelastic collision? (d) 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). Ans: (a) No
In an elastic collision, the total initial kinetic energy of the balls will be equal to the total final kinetic energy of the balls. This kinetic energy is not conserved at the instant the two balls are in contact with each other. In fact, at the time of collision, the kinetic energy of the balls will get converted into potential energy.
(b) Yes
In an elastic collision, the total linear momentum of the system always remains conserved.
(c) No; Yes
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.
(d) Elastic
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.
Q9 A body is initially at rest. It undergoes onedimensional motion with constant acceleration. The power delivered to it at time t is proportional to (i) t½ (ii) t (iii) t3/2 (iv) t2 Ans: (ii) t
Mass of the body = m
Acceleration of the body = a
Using Newton's second law of motion, the force experienced by the body is given by the equation:
F = ma
Both m and a are constants.Hence, force F will also be a constant.
F = ma = Constant … (i)
For velocity v, acceleration is given as,
a = dv / dt = constant
dv = Constant x dt
v = αt ....... (ii)
where α is another constant
Power is given by the relation:
P = F.v
Using equations (i) and (iii), we have:
P ∝ t
Hence, power is directly proportional to time.
Q14 A molecule in a gas container hits a horizontal wall with speed 200 ms1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic? Ans: Yes;
Collision is elastic The momentum of the gas molecule remains conserved whether the collision is elastic or inelastic.
The gas molecule moves with a velocity of 200 m/s and strikes the stationary wall of the container, rebounding with the same speed.
It shows that the rebound velocity of the wall remains zero. Hence, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.
Q15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump? Ans: Volume of the tank, V= 30 m^{3}
Time of operation, t = 15 min = 15 × 60 = 900 s
Height of the tank, h= 40 m
Efficiency of the pump, n = 30%
Density of water, p = 10^{3 }kg/m3
Mass of water, m= pV= 30 ×10^{3} kg
Output power can be obtained as:
P_{0} = work done / Time = mgh / t
= 30 x 10^{3} x 9.8 x 40 / 900 = 13.067 x 10^{3 }W
For input power Pi,, efficiency n is given by the relation:
n = P_{0} / P_{i} = 30%
P_{i} = 13.067 / 30 x 100 x 10^{3}
= 43.6 kW
Q19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s1. What is the speed of the trolley after the entire sand bag is empty? Ans: The sand bag is placed on a trolley that is moving with a uniform speed of 27 km/h. The external forces acting on the system of the sandbag and the trolley is zero. When the sand starts leaking from the bag, there will be no change in the velocity of the trolley. This is because the leaking action does not produce any external force on the system. This is in accordance with Newton's first law of motion. Hence, the speed of the trolley will remain 27 km/h.
Q21 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?(b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind's energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m3. What is the electrical power produced? Ans: Area of the circle swept by the windmill = A
Velocity of the wind = v
Density of air = p
(a) Volume of the wind flowing through the windmill per sec = Av
Mass of the wind flowing through the windmill per sec = pAv
Mass m, of the wind flowing through the windmill in time t = pAvt
(b) Kinetic energy of air = ½ mv2
= ½ (pAvt)v2 = ½ pAv^{3}t
(c) Area of the circle swept by the windmill = A = 30 m^{2}
Velocity of the wind = v= 36 km/h
Density of air, p = 1.2 kg m^{3}
Electric energy produced = 25% of the wind energy
= 25/100 x Kinetic energy of air
= 1/8 pAv^{3}t
Electrical Power = Electical energy / time
= 1/8 pAv^{3}t / t = 1/8 pAv^{3}
= 1/8 x 1.2 x 30 x 10^{3}
= 4.5 x 10^{3} w
= 4.5 kW
Q22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 x 107 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up? Ans: (a) Mass of the weight, m = 10 kg
Height to which the person lifts the weight, h = 0.5 m
Number of times the weight is lifted, n = 1000
∴Work done against gravitational force:
= n(mgh)
= 1000 x 10 x 9.8 x 0.5
= 49 x 10^{3}J = 49kJ
(b) Energy equivalent of 1 kg of fat = 3.8 × 107 J
Efficiency rate = 20%
Mechanical energy supplied by the person's body:
= 20/100 x 3.8 x 10^{7} J
= 1/5 x 3.8 x 10^{7} J
Equivalent mass of fat lost by the dieter:
= 1 / (1/5 x 3.8 x 10^{7}) x 49 x 10^{3}
= 245 / 3.8 x 10^{4}
= 6.45 x 10^{3} kg
Q23 A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house. Ans: (a) 200 m^{2}
(a) Power used by the family, P = 8 kW = 8 × 10^{3} W
Solar energy received per square metre = 200 W
Efficiency of conversion from solar to electricity energy = 20 %
Area required to generate the desired electricity = A
As per the information given in the question, we have:
8 x 103 = 20% x (A × 200)
= 20/100 x A x 200
∴ A = 8 x 10^{3} / 40 = 200 m^{2}
(b) The area of a solar plate required to generate 8 kW of electricity is almost equivalent to the area of the roof of a building having dimensions 14 m × 14 m.
Q24 A bullet of mass 0.012 kg and horizontal speed 70 ms1strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block. Ans: Mass of the bullet, m= 0.012 kg
Initial speed of the bullet, u_{b}= 70 m/s
Mass of the wooden block, M= 0.4 kg
Initial speed of the wooden block, u_{B}= 0
Final speed of the system of the bullet and the block = v
Applying the law of conservation of momentum:
mu_{b} + mu_{B} = (m + M) v
0.012 × 70 + 0.4 × 0 = (0.012 + 0.4)v
∴ v = 0.84 / 0.412 = 2.04 m/s
For the system of the bullet and the wooden block:
Mass of the system, m' = 0.412 kg
Velocity of the system = 2.04 m/s
Height up to which the system rises = h
Applying the law of conservation of energy to this system:
Potential energy at the highest point = Kinetic energy at the lowest point
m'gh = ½ m'v^{2}
∴ h = ½ v^{2} / g
= ½ (2.04)^{2} / 9.8
= 0.2123 m
The wooden block will rise to a height of 0.2123 m.
Heat produced = Kinetic energy of the bullet  Kinetic energy of the system
½ mu^{2}  ½ m'v^{2}
= 29.4  0.857 = 28.54 J
Q27 A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary? Ans: Mass of the bolt, m= 0.3 kg
Speed of the elevator = 7 m/s
Height, h = 3 m
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 x 9.8 x 3
= 8.82 J
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.