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?

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 P_i,, efficiency η, is given by the relation:
η = P₀ / P_i  =  30%
P_i = 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?

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.

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.

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, E_P = mgl
Kinetic energy of the bob, E_K = 0
Total energy = mgl … (i)
At the lowermost point (mean position):
Potential energy of the bob, E_P = 0
Kinetic energy of the bob, E_K = (1/2)mv²
Total energy E_x = (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

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?

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.

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.

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 = ν
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 = (1/2)m‘v²
∴ 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) m‘v² = (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?

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.

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.

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?

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?

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.

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.