# Laws of Motion

Question 545

## A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 s, the average force acting on the wall is 48 N 24 N 12 N 96 N

Solution

B.

24 N

The vector  represents the momentum of the object before the collision, and the vector  that after the collision. The vector  represents the change in momentum of the object .

As the magnitudes of   are equal, the components of  along the wall are equal and in the same direction, while those perpendicular to the wall are equal and opposite. Thus, the change in momentum is due only to the change in direction of the perpendicular components.
Hence,

Its time rate will appear in the form of average force acting on the wall.

Question 546

## The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc is

Solution

C.

The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is

where M is the mass of disc and R its radius. According to theorem of parallel axis, MI of circular disc about an axis touching the disc at it diameter and normal to the disc is

Question 547

## A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is )

Solution

A.

The moment of inertia of the uniform rod about an axis through the end and perpendicular to length is

where m is mass of rod and l its length.
Torque  acting on centre of gravity of rod is given by

or

Question 548

Solution

D.

448 Hz