Question-1
Solution:
Air resistance is not constant in the top half of the fall. Hence the formulae for uniform acceleration cannot be applied. Energy equation is therefore applied].
We know that work done = Force × distance = F × S = mgh joule.
W = mgh joule and mass (m) = volume × density = × ρMass of the water drop m = × π (2 ×10^{-3})^{3× }10^{3} kg = 33.49 × 10^{-6}kg
.^{.}. Potential energy at the top = mgh = 33.49 × 10^{-6 × }9.8 × 500 J = 0.164 J
Work done by gravitational force = Kinetic energy gained by the drop + energy lost by air resistance.
LHS = loss in potential energy during the respective falls.
Work done by gravitational force during the first half is
w_{1 } = Loss of potential energy during the first half of the fall
= mg (h/2) = (32/3) × π × 10^{-6} × 9.8 × 250 J
= 10.666 × 3.14 × 10^{-6} × 9.8 ×250 joules
= 0.082 joules.
Since the distance covered in the second half journey is also 250 m, therefore work done is given by w_{2} = 3.28 ´ 10^{-4} 250 = 0.082 Joules Work done by the resistive force = Change in kinetic energy - work done by Gravitational force
Change in kinetic energy = ½ mv^{2 }- 0
= ½ × 33.49 × 10^{-6 × }(10 × 10) joule
= 0.0016745 joules.
Work done by gravitational force = mgh joule = 33.49 × 10^{-6} × 9.8 × 500 J = 0.164101 joules
Work done by resistive force = 0.0016745 - 0.164101 J = - 0.1624265 J.
Question-2
Solution:
For elastic collision,
(a) Momentum is conserved
(b) Total energy is conserved
(c) Mechanical energy is not converted into any other form such as sound, beat, light.(d) Forces involved during the interaction are of conservative nature.The potential energy of a system of two bodies varies inversely proportional to the distance between them i.e. as 1/r. So all the potential energy curve except the one shown in fig. (vi) cannot describe elastic collision.
Question-3
(i) work done by a man in lifting a bucket out of a well by means of a rope tied to a bucket
(ii) work done by gravitational force in the above case.
(iii) work done by friction on a body sliding down an inclined plane,
(iv) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(v) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Solution:
Work done is considered positive when it is in the same direction as that of the applied force.
(i) Movement of the bucket and the direction of application of the force are both upwards. Hence the work done is positive.
(ii) Work done is -ve because the gravitational force acts in the downward direction while backed moves upward.
(iii) Work done is -ve because the frictional force always acts in the opposite direction of the motion of the body.
(iv) Work is done against the friction force which is in the negative direction. The applied force moves the body in the same direction as its own. Hence it is positive.
(v) Resistive force acts against the direction of vibration of pendulum and hence negative.
Question-4
(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 the results.
Solution:
Reaction of the surface on the body = 2× 9.8 N = 19.6 N
Forward force applied on the body F = 7N
Dynamic friction force against the motion = reaction dynamic friction coefficient
Frictional force = f_{r} = mR = μmg
W_{1}= 19.6 x (-0.1) = 1.96 N
F' = F - f_{r}
Net force = 7N - 1.96 N = 5.04 N
Acceleration produced in the body a = (5.04 N) / (2 Kg)= 2.52 m/s^{2}
(a) Distance traveled in 10 s =
s = 126 m
Work done by applied force = 7N x 126 m = 882 J
(b) Work done by friction force W_{2}= cos180^{o}
=1.96 N x 126m x (-1)
= -247 J
acts in a direction opposite to that of motion of the body.
(c) Work done by net force W_{3} = F's
= 5.04 N x 126 m = 635 J
(d) Change in kinetic energy K_{e}=
= (i.e v = at = 2.52 x 10 = 25.2 ms^{-1})
= 635 J
Change in kinetic energy = work done by net force.
Question-5
Solution:
[Magnitudes of potential energy and kinetic energy should be equal to the total energy. Situations which do not conform to this, are not possible].
a. Not possible in the region x>a. Minimum total energy = 0.
b. Not possible in the region x<a, and x>b. Minimum total energy = -V_{1}.
c. Not possible in the region -b/2<x>-a/2. Minimum total energy = -V_{1}.
Question-6
Solution:
Potential energy = V(x) = ½ k x^{2} = ½ ½ x^{2} (given k= ½ N/m)
Total energy = potential energy + kinetic energy = 1 J (given condition)
When the particle turns back, the velocity and hence the kinetic energy should be zero.
⇒ 1 J = ¼ x^{2} + 0 i.e. x ± 2 m
[In the given sketch O is the turning point, and A and B both represent the centre point where the velocity is maximum].
Question-7
a. The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for the burning obtained? The rocket or the atmosphere.
b. 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 comet is zero. Why?
c. 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 earth?
d. In the figure (i), the man walks 2 m carrying a mass of 15 kg on his hands. In figure (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley and a mass of 15 kg hangs at its other end. In which case is the work done greater?
Solution:
a. Rocket generates heat to burn itself by friction with the atmosphere.
b. The gravitational forces that act on comets are conservative forces. For conservative forces the work done = - (total energy change). Over a complete cycle, the potential energy and the kinetic energy remain same. (Work done during a cycle involving conservative forces is calculated by considering the initial and final situations only and the path and other changes in between do not affect the calculation of work done)
c. When the satellite comes closer and closer to earth, the potential energy decreases because of the decrease in height. Since the total energy is to remain constant, the kinetic energy increases and so the speed. Of course, there is a small mount of frictional loss in energy, but the gain by velocity is more. Hence its speed continues to increase.
d. Though the person carries the weight while walking, there is no relative movement with reference to gravitational force. Hence the only work done is the frictional force during walking. In the second case the a force of 15g N is applied and moved to lift the stone against gravity. Hence the work done is more in this case.
Question-8
(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 loss of its kinetic / potential energy.
(c) The rate of change of momentum of a many-particle system is proportional to the external force/sum of the internal forces of 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.
Solution:
(a) Decreases (e.g. When a mass falls freely under gravity, positive work is done by the gravity and the potential energy decreases.
(b) Kinetic energy (e.g. When a mass slides down an inclined plane against friction, the velocity of the mass decreases. The potential energy, which is dependent on the height and mass of the body, remains unchanged.
(c) External forces. (e.g. People walking inside a moving railway carriage do not constitute to the momentum of the carriage whereas an engine towing the carriage does.)
(d) Total linear momentum does not change in an inelastic collision. The total energy also conserved if we consider all the affected components of the system (e.g. If the heat generated is also considered as within the system, the total energy will be conserved).
Question-9
(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.
Solution:
(a) False. Only the total energy and momentum of both the bodies together are conserved.
(b) False. The external forces affect the energy of the system.
(c) False. It is true only for conservative forces.
(d) True if no other external or internally generated force influence the system.
Question-10
(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 elastic collision of the two balls?
(c) What are the answers to (a) and (b) for an inelastic collision?
(d) If the potential energy of the two billiard balls depends only on separation distance between the 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).
Solution:
(a) No. In the short time of collision the kinetic energy is stored as elastic strain energy and reconverted to kinetic energy.
(b) Yes. By elastic deformation and also by joint movement together in some cases.
(c) Linear momentum is conserved during the collision. Kinetic energy is not conserved even after impact.
(d) Elastic. Whatever deformation caused during the first half of collision, should recover during the second half. If the potential energy depends only on the centre distance, whatever energy stored should be recovered. There will be no loss. Hence the collision will be elastic.
Question-11
(i) t^{1/2} (ii) t (iii) t^{3/2} (iv) t^{2}
Solution:
(ii) t
Question-12
(i) t^{1/2} (ii) t (iii) t^{3/2} (iv) t^{2 }
Solution:
(iii) t^{3/2}
Question-13
L L L
where i, j, k are unit vectors along the x, y and z - axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis.
Solution:
Work done is defined as the component of force along the direction of motion and displacement.
The body moves 4 m along z - axis only
.^{.}.
_{Work done, W =} _{Nm}
Question-14
Solution:
Kinetic energy = E = ½ m v^{2}
V = (2 E / m) ^{½}
Kinetic energy of electron E_{e} = 10 Kev = 10^{4} x 1 ev
= 10^{4} x 1.6 x 10^{-19} J
= 1.6 x 10^{-15} J.
Kinetic energy of proton, E_{p} = 100 kev = 10^{5} x 1 ev
= 10^{5} x 1.6 x 10^{-19} J
= 1.6 x 10^{-14} J.
E_{e }= (1/2 ) m_{e} V_{e}^{2} E_{p} = (1/2) m_{p} V_{p}^{2}
V_{e } = √(2 E_{e} / m_{e}) = = 5.93 X 10^{7} ms ^{-1}
V_{p} = √(2 E_{p} / m_{p}) = = 4.38 x 10^{6} ms^{-1}
The required ratio V_{e}/V_{p} = 5.93 X 10^{7 } = 13.5
^{ }4.38 x 10^{6}
Question-15
Solution:
Momentum is conserved for all systems during collision. Kinetic energy is conserved in elastic collisions only. In the given case, the momentum of the molecule and the wall system is conserved. Due to the momentum of the incoming molecule, the wall deflects and rebounds with the same momentum. This momentum is then transferred to the ball, which rebounds with an equal momentum. At any moment, the total momentum of the ball and the wall system is constant. However, it is to be noticed that the recoil momentum of the wall produces very little velocity, because of the very large mass of the wall compared to that of the ball.
Coefficient of resolution = e = - (Relative velocity of the ball along the common normal with respect to the wall after collision.) / (Relative velocity of the ball along the common normalwith respect to the wall before collision)
Therefore e = -
e = 1
If e = 1, then it is perfectly elastic and if e < 1, then it is inelastic.
[The above events are concerned with the component of the momentum normal to the wall. The component resolved along the surface of the wall remains same throughout the event]
Question-16
Solution:
Power required to lift water for 1 s = 30 / (15 × 60) × 9.8 × 40 J
Since this output will be only 30% of input,
The input power required for 1 s = {30 / (15 × 60) × 9.8 × 40} / 0.3 J = 43,600 W
The pump consumes 43.6 kW of power.
Question-17
Solution:
The momentum and the kinetic energy of the moving balls after collision in the above three cases are given below:
Case |
Momentum |
Kinetic energy |
Case 1 |
2 × ( m v/2 ) = mv |
2 × {½ m (v/2)^{2} } = ¼ (m v^{2}) |
Case 2 |
1 × (m v ) = mv |
1 × ½ m (v)^{2} = ½ (m v^{2}) |
Case 3 |
3 × ( m v/3) = mv |
3 × {1/2 m (v/3)^{2} = 1/6 (mv^{2}) |
Question-18
Solution:
Let the velocity of A while hitting B be v, and let us assume that A moves with a velocity v_{1} and B moves with a velocity v_{2} after impact.
Equating momentum m v_{1} + m v_{2} = m v
v_{1} + v_{2} = v
Equating kinetic energies ½ m v_{1}^{2} + ½ m v_{2}2 = ½ m v^{2}
v_{1}^{2 }+ v_{2}^{2} = v^{2}
= (v_{1} + v_{2})^{2}
For the above situation v_{1} should be zero and v_{2} should be = v.
For an elastic collision, A will stop fully and B will move with a velocity v.
Question-19
Solution:
There is a loss of 5% of the initial energy when the bob moves from A to B
Hence 0.95 × Potential energy of the bob at A = Kinetic energy of the bob at B
95% of this energy is converted into kinetic energy
Þ 0.95 (m × 9.8 × 1.5) = ½ m v^{2}
⇒ v^{2} = 27.93
⇒ v = 5.3 m/s.
Question-20
Solution:
Since the trolley is moving on a frictionless track with uniform speed, there is no external force acting on it. The sand falling from the bag falls in the trolley itself and there is no change in total mass of the moving bodies. Hence the speed of 27 km / h will continue to remain constant.
Question-21
Solution:
W = v = ax^{3/2}, dv = 3/2ax^{1/2 }dx
= max^{3/2 }_{3/2 ax}^{1/2 }dx
= 3/2 ma^{2} x^{2 }dx
= 1/2 x 0.5 x 25 x 8
= 50J.
Question-22
Solution:
(a) Density of air = ρ
Time taken = t
Velocity of air = v
Mass of the air = density x volume
= ρAvt
(b) Kinetic energy = mv^{2}
= × ρAvt × v^{2}
= × ρAv^{3 }t
(c) Velocity, v = 36 km/h
= = 10 m/s
ρ = 1.2 kg m^{-3}
A = 30 m^{3}
Work done = × 1.2× 30× 10× 10× 10
= 18000 J
Efficiency = 25%
Therefore work done = 18000 × J
Question-23
Solution:
(a) Mass m = 10 kg
Distance moved d = 0.5 m
Number of times lifted n = 1000
Therefore the work done against the gravitational force = mgh = 10 ×9.8 ×0.5 ×1000 = 49000 J
(b) 3.8 ×10^{7} J of energy is given by 1 kg fat
49000 J of energy is given by = = = 1.289 ×10^{-3} kg
Efficiency = 20%
Fat used up by the dieter = 1.289 ×10^{-3} × = 6.45 ×10^{-3} kg.
Question-24
Solution:
Power used P = 8 kW
(a) 200 W solar energy is incident on 1 square metre, 20% of this energy can be converted to useful energy
Hence = 40 W
40 W solar energy is incident on 1 square metre
Therefore 8 ×1000 W solar energy is incident on
= 200 m^{2}
(b) It is comparable to the roof of a large house of dimension 14m ×14 m.
Question-25
Solution:
[Momentum is conserved in all collisions. The bullet hits the block and imparts momentum conserving the momentum, by which the kinetic energy of the block with the embedded bullet can be found out. This kinetic energy will raise the block, but will not account for the initial kinetic energy of the bullet. The difference is given out as heat during impact.]
Velocity of the block with the embedded bullet after impact
= (mass of the bullet × Velocity of the bullet) / (mass of block + bullet)
= (0.012 × 70) / (0.412) m/s
= 2.04 m/s
Kinetic energy of the block + bullet =½ mv^{2 }= ½ × (0.4 + 0.012) × ( 2.04)^{2}
= 0.857 J
This gets converted into potential energy of the combination. If h is the height of rise, then
Height of the (block + bullet) will rise = 0.857 / {(0.4 + 0.012) × 9.8} = 0.212 m
Initial kinetic energy of bullet = ½ (0.012) (70)^{2} = 29.4 J
Heat generated = Initial kinetic energy of bullet - Kinetic energy of the block + bullet
= 29.4 - 0.857 = 28 .5 J.
Question-26
Solution:
[The gravitational force on the mass m = mg and acts vertically down. This can be resolved into two components mg sin θ down the plane, and mg cos θ normal to the plane. The normal component is balanced by the reaction of the plane on the body. The component down the plane accelerates the body downwards]
For a mass 'm' at a height 'h' on the plane inclined at θto the horizontal, the acceleration down the plane = mg sin θ
Acceleration produced in the 1^{st} stone a_{1 }= g sinθ1 and a_{2 }= g sinθ2
Velocity at bottom = v = u + at = 0 + (m g sin θ )x t , where t = time to reach the bottom.
Considering the energy equation, m g h = ½ m v^{2}
v = (2 g h)^{1/2}
Potential energy at energy = kinetic energy at B or C
Since v is independent of the slope and mass, both the stones will reach the bottom at the same speed.
Considering the time to reach the bottom, if t_{1} is the time for the stone in the gradual slope and t_{2} for the stone in steep slope, and θ_{ 1} and and and and and θ_{ 2 }their inclination angles,
t_{1} = v / (g sin θ_{1}) and t_{2} = v / (g sinθ_{2})
θ_{2 }> θ_{1} , sin θ_{ 2} > sin θ_{ 1 }, t_{2} < t_{1}
Stone on the steeper slope will take less time to reach the bottom.
mgh = mv^{2
} = v
Therefore v =^{ }= 14 m/s
v1 = v2 = 14 m/s
Therefore t1 = 2.86 s
t2 = 1.65 s
Hence t2 < t1.
Question-27
Solution:
Mass of the block m = 1 kg
Distance moved x = 10 cm = 0.1 m
Spring constant k = 100 Nm^{-1}
Force of friction F = kx = 100 ×10 = 1000 N
Coefficient of friction μ =
=
= 102.
Question-28
Solution:
[Since the bolt is falling in the elevator itself, we have to consider the relative velocity only. The relative acceleration is g only, since the elevator does not have any acceleration]
Relative to the elevator, let the velocity of the bolt hitting the floor = v m/s
v^{2} = 0^{2} + 2 × 9.8 m/s^{2} × 3m = 58.8 (m/s)^{2}
Kinetic energy of the bolt with reference to the elevator = ½ m v^{2}
= ½ × 0.3 × 58.8 = 8.82 J
Since the relative velocity w. r. to the elevator after impact, the total kinetic energy of the system is zero. The bolt does not have a potential energy w. r. to while on floor, the only way the kinetic energy is to be dissipated is through heat generated.
Heat generated is 8.82 J.
Question-29
Solution:
[When the child starts moving on the platform, an impact is given to the trolley by the reaction of the starting force. Afterwards the child does not impart any force on the trolley since the movement is at constant speed of 4 m/s. The impact force will accelerate the trolley with the child.]
Initial speed of the trolley = 36 km/h
= 10 m/s
= u m/s say.
Speed of the child w. r. to the trolley = 4 m/s
final speed of the trolley = v say
Mass of the trolley m_{1} = 200 kg
Mass of the child m_{2} = 20 kg
Consider the speeds and momentum with reference to the ground.
Speed of the child w. r. to the ground = (v - 4) m/s
Equating the momentum before and after impact,
(200 +20) 10 = {20 (v - 4)} + 200 v
v = 10.36m/s = The final speed of the trolley
Time taken by the child to run the 10 m length of the trolley = 10/4 s = 2.5 s
Distance covered by the trolley during this 2.5 s = 10.36 × 2.5 m = 25.9 m.