Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 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  R = mg sin 37°

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.,

Question 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.

Question 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 

Final momentum 

= 200v‘ + 20v‘ – 80

= 220v‘ – 80

As per the law of conservation of momentum:

Initial momentum = Final momentum

2200 = 220v‘ – 80

Length of the trolley, l = 10 m

Speed of the boy, v” = 4 m/s

Time taken by the boy to run, 

∴Distance moved by the trolley = v” × t = 10.36 × 2.5 = 25.9 m

Question 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:

(i), (ii), (iii), (iv), and (vi)

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.

Question 6.30:

Consider the decay of a free neutron at rest: n → p+ e–

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:

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.