NCERT Class 11 Physics Solutions: Chapter 5 – Laws of Motion Part 6

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Question 5.15:

Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to

( I ) A,

( ii ) B along the direction of string.

What is the tension in the string in each case?

Answer:

Horizontal force,

Mass of body A,

Mass of body B,

Total mass of the system,

Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as:

When force F is applied on body A:

Figure shown the When force F is applied on body A.

Figure 5.15 (A)

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The equation of motion can be written as:

When force F is applied on body B:

Figure show the When force F is applied on body B.

Figure 5.15(B)

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The equation of motion can be written as:

Question 5.16:

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Answer:

The given system of two masses and a pulley can be represented as shown in the following figure:

The given system of two masses and a pulley can be represent …

Figure 5.16

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Smaller mass,

Larger mass,

Tension in the string = T

Mass , owing to its weight, moves downward with acceleration a, and mass moves upward.

Applying Newton’s second law of motion to the system of each mass:

For mass,

The equation of motion can be written as:

For mass ,

The equation of motion can be written as:

Adding equations ( I ) and ( ii ), we get:

Therefore, the acceleration of the masses is

Substituting the value of a in equation ( ii ), we get:

Therefore, the tension in the string is 96 N.

Question 5.17:

A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Answer:

Let m, , and , be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest.

Initial momentum of the system (parent nucleus)

Let and be the respective velocities of the daughter nuclei having masses and .

Total linear momentum of the system after disintegration

According to the law of conservation of momentum:

Total initial momentum = Total final momentum

Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.