6.1. 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.

Answer

(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.

6.2. 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.

Answer

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 ms-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 ms-2
Total
acceleration of the body: a’ + a”
= 3.5 + (-0.98) = 2.52 ms-2
The
distance travelled by the body is given by
the equation of motion:
s = ut + (1/2)at2
= 0 + (1/2) × 2.52 × (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, Wf
=
F
× s
= –1.96 ×
126 = –247 J

(c) Net
force = 7 + (–1.96) = 5.04 N
Work
done by the net force, Wnet= 5.04 ×126
= 635 J

(d) From
the first equation of motion, final velocity can be calculated as:
v
= u + at
= 0 + 2.52 ×
10 = 25.2 m/s
Change
in kinetic energy = (1/2) mv2 – (1/2) mu2
= (1/2) × 2(v2u2) = (25.2)2 – 02 = 635 J

6.3. Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

 

Answer

Total energy of a system is given by the relation:
E = P.E. + K. E.
∴ K.E. = E – P.E.
Kinetic energy of a body is a positive quantity. It cannot be negative. Therefore, the particle will not exist in a region where K.E. becomes negative.

(i) For x > a, P.E. (V0) > E
∴ K.E. becomes negative. Hence, the object cannot exist in the region x > a.
(ii) For x < a and > b, P.E. (V0) > E.
∴ K.E. becomes negative. Hence the object cannot be present in the region x < a and x > b.
(c) x > a and x < b;  –V1

In the given case, the condition regarding the positivity of K.E. is satisfied only in the region between x > a and x < b.
The minimum potential energy in this case is –V1. Therfore, K.E.  = E – (–V1) = E + V1. Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than –V1. So, the minimum total energy the particle must have is –V1.

(d) b/2 <  x <  a/2 ; a/2 < x < b/2 ; –V1

In the given case, the potential energy (V0) of the particle becomes greater than the total energy (E) for –b/2 < x < b/2 and –a/2 <  x < a/2. Therefore, the particle will not exist in these regions.

The minimum potential energy in this case is –V1. Therfore, K.E.  = E – (–V1) = E + V1. Therefore, for the positivity of the kinetic energy, the totaol energy of the particle must be greater than –V1. So, the minimum total energy the particle must have is –V1.

Page No: 135

6.4. The
potential energy function for a particle executing linear simple
harmonic motion is given by V(x)
=kx2/2,
where k
is the force constant of the
oscillator. For k
=
0.5 N m–1,
the graph of V(x)
versus x
is shown in Fig. 6.12. Show that
a particle of total energy 1 J moving under this potential must ‘turn
back’ when it reaches x
= ± 2 m.

Answer

Total
energy of the particle, E
= 1 J

Force
constant, k
= 0.5 N m–1
Kinetic
energy of the particle, K = (1/2)mv2
According
to the conservation law:
E
= V + K
1 = (1/2)kx2 + (1/2)mv2
At
the moment of ‘turn back’,
velocity (and hence K)
becomes zero.
∴ 1 = (1/2)kx2
(1/2) × 0.5x2 = 1
x2 = 4
x = ±2
Hence,
the particle turns back when it reaches x
= ± 2 m.

6.5. Answer the following:
(a) 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?

(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 of the 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 the earth?

(d) In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on
his hands. In Fig. 6.13(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?

Answer
 

(a) Rocket

The burning of the casing of a rocket in flight (due to friction) results in the reduction of the mass of the rocket.
According to the conservation of energy:
Total energy = Potential energy + Kinetic energy
= mgh + (1/2)mv2
The reduction in the rocket’s mass causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket.

(b) Gravitational force is a conservative
force. Since the work done by a conservative force over a closed path is
zero, the work done by the gravitational force over every complete
orbit of a comet is zero.

(c) When an artificial satellite, orbiting
around earth, moves closer to earth, its potential energy decreases
because of the reduction in the height. Since the total energy of the
system remains constant, the reduction in P.E. results in an increase in
K.E. Hence, the velocity of the satellite increases. However, due to
atmospheric friction, the total energy of the satellite decreases by a
small amount.

(d)  Work done in fig 6.13 (i)
Mass, m = 15 kg
Displacement, s = 2 m
Work done, W = Fs Cosθ
Where, θ = Angle between force and displacement
= mgs Cosθ = 15 × 2 × 9.8 Cos 900
= 0

Work done in fig 6.13 (ii)
Mass, m = 15 kg
Displacement, s = 2 m
Here, the direction of the force applied on the rope and the direction of the displacement of the rope are same.
Therefore, the angle between them, θ = 0°
Since cos 0° = 1
Work done, W = Fs cosθ = mgs
= 15 × 9.8 × 2 = 294 J

Hence, more work is done in the Fig. 6.13 (ii).

6.6. 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 many-particle 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.

Answer


(a) Decreases
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) Kinetic energy
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) External force
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) Total linear momentum
The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.

6.7. 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.

Answer

(a) False
In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.

(b) False
The external forces on the body may cahnge the total energy of the body.

(c) False
The work done in the motion of a body over a closed loop is zero for a conservation force only.

(d) True
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.

6.8. 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).

Answer

(a) No
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.

(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.

Page No: 136

6.9. 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

(i) t1/2   (ii) t   (iii) t3/2   (iv) t2

Answer

From,
v = u + at
v = 0 + at = at
As power, P = F × v
P = (ma× at = ma2t
As m and a are constants, therefore, P t
Hence, right choice is (ii) t


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

(i) t1/2   (ii) t   (iii) t3/2   (iv) t2


Answer

As power, P = force × velocity
P = [MLT-2] [LT-1] = [ML2T-3]
As, P = [ML2T-3]
= constant
L2T-3 = constant
or, L2/T3 = constant
L2T3
or, LT3/2

Hence, right choice is (iii) t3/2


6.11. A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by

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?

Answer

 

 

 

 

 

 

6.12. An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second
with 100 keV. Which is faster, the electron or the proton? Obtain theratio of their speeds. (electron mass = 9.11 × 10–31kg, proton mass = 1.67 × 10–27 
kg, 1 eV = 1.60 × 10–19 J).

Answer

Electron
is faster; Ratio of speeds is 13.54 : 1

Mass
of the electron, me
= 9.11 × 10–31
kg
Mass
of the proton, mp
= 1.67 × 10
27
kg
Kinetic
energy of the electron, EKe
= 10 keV = 104
eV
=
104 ×
1.60 × 10–19

=
1.60 × 10–15
J
Kinetic
energy of the proton, EKp
= 100 keV = 105
eV = 1.60 × 10–14
J
For the velocity of an electron ve, its kinetic energy is given by the relation:
EKe = (1/2) mve2
∴ ve = (2EKe / m)1/2
= (2 × 1.60 × 10-15 / 9.11 × 10-31)1/2  =  5.93 × 107 m/s
For the velocity of a proton vp, its kinetic energy is given by the relation:
EKp = (1/2)mvp2
vp = (2 × 1.6 × 10-14 / 1.67 × 10-27 )1/2  =  4.38 × 106 m/s
Hence,
the electron is moving faster than the
proton.
The ratio
of their speeds
ve / vp = 5.93 × 107 / 4.38 × 106  =  13.54 : 1

6.13. A
rain drop of radius 2 mm falls from a height of 500 m above the
ground. It falls with decreasing acceleration (due to viscous
resistance of the air) until at half its original height, it attains
its maximum (terminal) speed, and moves with uniform speed
thereafter. What is the work done by the gravitational force on the
drop in the first and second half of its journey? What is the work
done by the resistive force in the entire journey if its speed on
reaching the ground is 10 m s–1?

Answer

Radius
of the rain drop, r
= 2 mm = 2 ×
10–3
m

Volume
of the rain drop, V = (4/3)πr3
= (4/3) × 3.14 × (2 × 10-3)3 m-3
Density
of water, ρ
= 103
kg m–3

Mass
of the rain drop, m
= ρV
= (4/3) × 3.14 × (2 × 10-3)3 × 103 kg
Gravitational
force, F =
mg
= (4/3) × 3.14 × (2 × 10-3)3 × 103 × 9.8  N
The
work done by the gravitational force on the
drop in the first half of its journey:
WI
= Fs
= (4/3) × 3.14 × (2 × 10-3)3 × 103 × 9.8 × 250  =  0.082 J
This
amount of work is equal to the work done by
the gravitational force on the drop in the second half of its
journey, i.e., WII,
= 0.082 J
As
per the law of conservation of energy, if
no resistive force is present, then the total energy of the rain drop
will remain the same.
∴Total
energy at the top:
ET
= mgh + 0

= (4/3) × 3.14 × (2 × 10-3)3 × 103 × 9.8 × 500 × 10-5
= 0.164 J
Due
to the presence of a resistive force, the
drop hits the ground with a velocity of 10 m/s.
∴Total
energy at the ground:
EG = (1/2) mv2 + 0
= (1/2) × (4/3) × 3.14 × (2 × 10-3)3 × 103 × 9.8 × (10)2
= 1.675 × 10-3 J
∴Resistive
force = EG
ET
= –0.162 J

6.14. A
molecule in a gas container hits a horizontal wall with speed 200 m
s–1
and angle 30° with the normal,
and rebounds with the same speed. Is momentum conserved in the
collision? Is the collision elastic or inelastic?

Answer

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.

6.15. 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?

Answer

Volume
of the tank, V
= 30 m3

Time
of operation, t
= 15 min = 15 ×
60 = 900 s
Height
of the tank, h
= 40 m
Efficiency
of the pump, η = 30 %
Density
of water, ρ
= 103
kg/m3
Mass
of water, m
= ρV
= 30 ×
103 kg
Output
power can be obtained as:
P0 = Work done / Time  = mgh / t
= 30 × 103 × 9.8 × 40 / 900  =  13.067 × 103 W
For
input power Pi,,
efficiency η, is given by the relation: 

η = P0 / Pi  =  30%
Pi = 13.067 × 100 × 103  / 30
= 0.436 × 105 W  =  43.6 kW

6.16. Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following figure is a possible result after collision?

Answer

It
can be observed that the total momentum
before and after collision in each case is constant.
For
an elastic collision, the total kinetic
energy of a system remains conserved before and after collision.
For
mass of each ball bearing m,
we can write:
Total
kinetic energy of the system before
collision:
= (1/2)mV2 + (1/2)(2m) × 02
= (1/2)mV2


Case
(i)

Total
kinetic energy of the system after
collision:
= (1/2) m × 0 + (1/2) (2m) (V/2)2
= (1/4)mV2
Hence,
the kinetic energy of the system is not
conserved in case (i).

Case
(ii)

Total
kinetic energy of the system after
collision:
= (1/2)(2m) × 0 + (1/2)mV2
= (1/2) mV2
Hence, the
kinetic energy of the system is conserved in case (ii).

Case
(iii)

Total
kinetic energy of the system after
collision:
= (1/2)(3m)(V/3)2
= (1/6)mV2
Hence,
the kinetic energy of the system is not conserved in case (iii).

Hence, Case II is the only possibility.

Page No: 137

6.17. The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Answer
The bob A will not rise because when two bodies of same mass undergo an elastic collision, their velocities are interchanged. After collision, ball A will come to rest and the ball B would move with the velocity of A, Fig. 6.15. Thus the bob A will not rise after the collision.

6.18. he bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Answer

Length of
the pendulum, l = 1.5 m
Mass of
the bob = m
Energy
dissipated = 5%
According
to the law of conservation of energy, the total energy of the system
remains constant.
At the
horizontal position:

Potential
energy of the bob, EP = mgl
Kinetic
energy of the bob, EK = 0
Total
energy = mgl (i)
At the
lowermost point (mean position):

Potential
energy of the bob, EP = 0
Kinetic
energy of the bob, EK = (1/2)mv2
Total energy Ex = (1/2)mv2    ….(ii)
As the bob
moves from the horizontal position to the lowermost point, 5% of its
energy gets dissipated.
The total
energy at the lowermost point is equal to 95% of the total energy at
the horizontal point, i.e.,
(1/2)mv2 = (95/100) mgl
∴  v = (2 × 95 × 1.5 × 9.8 / 100)1/2
= 5.28 m/s

6.19. 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 s–1. What is the speed of the trolley after the entire sand bag is empty?

Answer

As the trolley carrying the sand bag is moving uniformly, therefore, external force on the system = 0.
When the sand leaks out, it does not lead to the application of any external force on the trolley. Hence, the speed of the trolley shall not change.

6.20. A
body of mass 0.5 kg travels in a straight line with velocity v = ax3/2 where a = 5 m1/2 s-1. What is the work done by the net force during its displacement from x
= 0 to x
= 2 m?

Answer

Mass
of the body, m
= 0.5 kg

Velocity
of the body is governed by the equation, v = ax3/2 and a = 5 m1/2 s-1
Initial
velocity, u (at
x
= 0) = 0
Final
velocity v (at
x
= 2 m) = 102 m/s
Work
done, W =
Change in kinetic energy
= (1/2) m (v2u2)
= (1/2) × 0.5 [ (102)2 – 02]
= (1/2) × 0.5 × 10 × 10 × 2
= 50 J

6.21. 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 m–3.
What is the electrical power produced?

Answer

Area
of the circle swept by the windmill = A

Velocity
of the wind = v
Density
of air =
ρ



(a) Volume
of the wind flowing through the windmill per sec = Av
Mass
of the
wind flowing through the windmill per sec = ρAv
Mass
m,
of the wind flowing through the windmill in time t
= ρAvt


(b) Kinetic
energy of air = (1/2) mv2
= (1/2) (ρAvt)v2 = (1/2)ρAv3t


(c) Area
of the circle swept by the windmill = A
= 30 m2
Velocity
of the wind = v
= 36 km/h
Density
of air, ρ
= 1.2 kg m–3
Electric
energy produced = 25% of the wind energy
= (25/100) × Kinetic energy of air
= (1/8) ρ A v3t
Electrical power = Electrical energy / Time
= (1/8) ρ A v3t / t
= (1/8) ρ A v3
= (1/8) × 1.2 × 30 × (10)3
= 4.5 kW

6.22. 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 ×
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?

Answer

(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 × 10 × 9.8 × 0.5  =  49 kJ

(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) × 3.8 × 107 J
= (1/5) × 3.8 × 107 J
Equivalent mass of fat lost by the dieter:
= [ 1 / (1/5) × 3.8 × 107 ]× 49 × 103
= (245 / 3.8) × 10-4 = 6.45 × 10-3 kg

6.23. 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.

Answer

(a) Power used by the family, P = 8 kW = 8 × 103 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 × 103 = 20% × (A × 200)
= (20 /100) × A × 200
A = 8 × 103 / 40  =  200 m2

(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. (≈ 200).


Additional Excercises


6.24. A
bullet of mass 0.012 kg and horizontal speed 70 m
s–1
strikes 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.

Answer

Mass
of the bullet, m
= 0.012 kg

Initial
speed of the bullet, ub
= 70 m/s
Mass
of the wooden block, M
= 0.4 kg
Initial
speed of the wooden block, uB
= 0
Final
speed of the system of the bullet and the block = ν
Applying
the law of conservation of momentum:
mub + MuB = (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
mgh = (1/2)mv2
h = (1/2)(v2 / g)
= (1/2) × (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
= (1/2) mu2 – (1/2) mv2 = (1/2) × 0.012 × (70)2 – (1/2) × 0.412 × (2.04)2
= 29.4 – 0.857 = 28.54 J

6.25. Two
inclined frictionless tracks, one gradual and the other steep meet at
A from where two stones are allowed to slide down from rest, one on
each track (Fig. 6.16). Will the stones reach the bottom at the same
time? Will they reach there with the same speed? Explain. Given θ1
= 30°,
θ2
= 60°,
and h =
10 m, what are the speeds and times taken by the two stones?

Answer

The given situation can be shown as in the following figure:

AB and AC are two smooth planes inclined to the horizontal at ∠θ1 and ∠θ2 respectively. As height of both the planes is the same, therefore, noth the stones will reach the bottom with same speed.
As P.E. at O = K.E. at A = K.E. at B
mgh = 1/2 mv12 = 1/2 mv22
v1 = v2
As it is clear from fig. above, accleration of the two blocks are a1 = g sin θ1 and a2 = g sin θ2
As θ2 > θ1
a2 > a1
From v = u + at = 0 + at
or, t = v/a
As t ∝ 1/a, and a2 > a1
t2 < t1
i.e., Second stone will take lesser time and reach the bottom earlier than the first stone.

Page No: 138

6.26. A
1 kg block situated on a rough incline is connected to a spring of
spring constant 100 N
m–1
as shown in Fig. 6.17. The block is released from rest with the
spring in the unstretched position. The block moves 10 cm 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
= 1 kg
Spring
constant, k
= 100 N m–1
Displacement
in the block, x
= 10 cm = 0.1 m
The
given situation can be shown as in the
following figure.

At
equilibrium:

Normal
reaction, R
= mg cos
37°
Frictional
force, f = μ R = mg Sin 370
Where,
μ
is the coefficient of friction
Net
force acting on the block = mg
sin
37° – f
= mgsin 37° – μmgcos 37°
= mg(sin
37° – μcos 37°)
At
equilibrium, the work done by the block is equal to the potential
energy of the spring, i.e.,
mg(sin
37° – μcos 37°)x = (1/2)kx2
1 × 9.8 (Sin 370 – μcos 37°) = (1/2) × 100 × (0.1)
0.602 – μ × 0.799 = 0.510
∴ μ = 0.092 / 0.799  =  0.115

6.27. A
bolt of mass 0.3 kg falls from the ceiling of an elevator moving down
with an
uniform speed of 7 m s–1.
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?

Answer

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 × 9.8 × 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.

6.28. A
trolley of mass 200 kg moves with a uniform speed of 36 km/h on a
frictionless track. A child of mass 20 kg runs on the trolley from
one end to the other (10 m away) with a speed of 4 m
s–1
relative to the trolley in a direction opposite to the its motion,
and jumps out of the trolley. What is the final speed of the trolley?
How much has the trolley moved from the time the child begins to run?

Answer

Mass
of the trolley, M
= 200 kg
Speed
of the trolley, v
= 36 km/h = 10 m/s
Mass
of the boy, m
= 20 kg
Initial
momentum of the system of the boy and the trolley
= (M
+ m)v

= (200 +
20) × 10
= 2200 kg
m/s
Let
v‘ be the
final velocity of the trolley with respect to the ground.
Final
velocity of the boy with respect to the ground = v’ – 4
Final momentum = Mv’ + m(v’ – 4)
= 200v’ + 20v’ – 80
=
220v‘ –
80
As
per the law of conservation of momentum:

Initial
momentum = Final momentum
2200
= 220v‘ –
80
∴ v’ = 2280 / 220 = 10.36 m/s
Length
of the trolley, l
= 10 m
Speed
of the boy, v
= 4 m/s
Time
taken by the boy to run, t = 10/4 = 2.5 s
∴ Distance
moved by the trolley = v
× t =
10.36 × 2.5 = 25.9 m

6.29. Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

Answer

The potential energy of
a system of two masses is inversely proportional to the separation
between them. In the given case, the potential energy of the system
of the two balls will decrease as they come closer to each other. It
will become zero (i.e., V(r) = 0) when the two balls touch
each other, i.e., at r = 2R, where R is the radius of each
billiard ball. The potential energy curves given in figures (i),
(ii), (iii), (iv), and (vi) do not satisfy these two conditions.
Hence, they do not describe the elastic collisions between them.

6.30. Show
that the two-body decay of this type must necessarily give an
electron of fixed energy and, therefore, cannot account for the
observed continuous energy distribution in the β-decay
of a neutron or a nucleus (Fig. 6.19).

[Note:
The simple result of
this exercise was one among the several arguments advanced by W.
Pauli to predict the existence of a third particle in the decay
products of β-decay.
This particle is known as neutrino. We now know that it is a particle
of intrinsic spin ½ (like e,
p or
n),
but is neutral, and either massless or having an extremely small mass
(compared to the mass of electron) and which interacts very weakly
with matter. The correct decay process of neutron is: n

p + e+ ν]

Answer

The
decay process of free neutron at rest is given as:
n

p + e
From
Einstein’s mass-energy relation, we have the
energy of electron as Δmc2
Where,
Δm
= Mass defect = Mass of neutron – (Mass of proton + Mass of
electron)
c = Speed
of light

Δm
and c are constants. Hence, the given two-body decay is unable to
explain the continuous energy distribution in the β-decay
of a neutron or a nucleus. The presence of neutrino νon the LHS of the decay correctly
explains the continuous energy distribution.