# NCERT Class 11 Physics Solutions: Chapter 6 – Work, Energy, and Power Part 11 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Question 6.25:

Two inclined frictionless tracks, one gradual and the other steep meet at from where two stones can slide down from rest, one on each track showing in Figure. Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given and , what are the speeds and times taken by the two stones?

Will the stones reach the bottom at the same time: No

Because of the stone moving down the steep plane will reach the bottom first

Will they reach there with the same speed: Yes

Because of the stone will reach the bottom with the same speed

m/s

The given situation can be shown in the following figure:

Here, the initial height (AD) for both the stones is the same (h) . Hence, both will have the same potential energy at point A.

As per the law of conservation of energy, the kinetic energy of the stones at points B and C will also be the same, i.e.. ,

Where,

Mass of each ston

Speed of each stone at points and

Hence, both stones will reach the bottom with the same speed, .

For stone I:

Net force acting on this stone is given by:

For stone II:

Using the first equation of motion, the time of slide can be obtained as:

For stone I:

For stone II:

Hence, the stone moving down the steep plane will reach the bottom first.

The speed of each stone at points and is given by the relation obtained from the law of conservation of energy.

m/s

The times are given as:

Question 6.26:

A block situated on a rough incline is connected to a spring of spring constant as shown in Figure. The block is released from rest with the spring in the unstretched position. The block moves 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,

Spring constant,

Displacement in the block,

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

At equilibrium:

Normal reaction,

Frictional force,

Where, is the coefficient of friction

Net force acting on the block

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 falls from the ceiling of an elevator moving down with a uniform speed of . It hits the floor of the elevator (length of the elevator ) 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,

Speed of the elevator m/s

Height,

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

The heat produced will remain the same even if the lift is stationary. This is because the relative velocity of the bolt with respect to the lift will remain zero.

Question 6.28:

A trolley of mass moves with a uniform speed of on a frictionless track. A child of mass runs on the trolley from one end to the other a speed of 4 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,

Speed of the trolley, v = 36 km ⁄ h = 10 m/s

Mass of the boy,

Initial momentum of the system of the boy and the trolley

m/s

Let be the final velocity of the trolley with respect to the ground.

Final velocity of the boy with respect to the ground

Final momentum

As per the law of conservation of momentum:

Initial momentum Final momentum

m/s

Length of the trolley,

Speed of the boy, m/s

Time taken by the boy to run,

Distance moved by the trolley

Question 6.29:

Which of the following potential energy curves in Figure cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centers of the balls.

Explanation:

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.. , ) when the two balls touch each other, i.e.. , at , where R is the radius of each billiard ball. The potential energy curves given in figures 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:

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 is shown in figure.

[Note: The simple result of this exercise was one among the several arguments advanced by . 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 , 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: ]

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

From Einstein՚s mass-energy relation, we have the energy of electron as

Where,

Mass defect Mass of neutron (Mass of proton Mass of electron)

Speed of light

and 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 on the of the decay correctly explains the continuous energy distribution.

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