Question

Ball A of mass 0.55 kg has a velocity of 0.65 m/s east. It strikes a stationary ball, also of mass 0.55 kg. Ball A deflects off ball B at an angle of 37° north of A's original path. Ball B moves in a line 90° right of the final path of A. Find the momentum (in kg m/s) of Ball A after the collision.

Answer 1

Assuming the east direction as positive x-axis and north direction as the positive y-axis.

The initial momentum of the system in the x-direction is

The initial momentum of the system in the y-direction is zero because there is no component of the velocity in the y-direction. i.e

NOW,

After the collision, the ball A deflects an angle of 37 north of east. And ball B move an angle of 90 right of ball A. Which means the ball B moves an angle of (90-37=53) south of east.

So, the final momentum of the system in the x-direction is

And the final momentum of the system in the y-direction is

NOW,

Applying conservation of momentum in the y-direction

Again applying conservation of momentum in the x-direction

Putting the value of v2

The momentum of the ball A is given by

So, the final answer is 0.2855 kg.m/s.