NCERT Class 11 Physics Solutions: Chapter 5 – Laws of Motion Part 4

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Question 5.11:

A truck starts from rest and accelerates uniformly at . At , a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the

(a) Velocity, and

(b) Acceleration of the stone at ? (Neglect air resistance.)

Answer:

(a) Velocity: , at an angle of 26.57° with the motion of the truck.

Explanation:

Initial velocity of the truck,

Acceleration,

Time,

As per the first equation of motion, final velocity is given as:

The final velocity of the truck and hence, of the stone is 20 m/s.

At , the horizontal component of velocity, in the absence of air resistance, remains unchanged, i.e.,

The vertical component of velocity of the stone is given by the first equation of motion as:

Where,

and

The resultant velocity (v) of the stone is given as:

Figure shown the The resultant velocity (v)

Figure 5.11

Loading Image

Let be the angle made by the resultant velocity with the horizontal component of velocity,

(b) Acceleration of the stone at :

When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is and it acts vertically downward.

Question 5.12:

A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is What is the trajectory of the bob if the string is cut when the bob is

(a) At one of its extreme positions.

(b) At its mean position.

Answer :

(a) At one of its extreme positions: Vertically downward

At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.

(b) At its mean position: Parabolic path

At the mean position, the velocity of the bob is The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.