Question

A small ball falling vertically downward with constant velocity 4 m/s strikes elastically a massive inclined cart moving with velocity 4 m/s horizontally as shown.The velocity of the rebound of the ball is



(A) 4√2m/s (B) 4√3m/s (C) 4 m/s (D) 4√5m/s

Answer 1

Given:

Velocity ( ball ) = 4 m/s

Velocity ( cart ) = 4 m/s

To find:

The velocity of the rebound of the ball

Solution:

By law,

When the ball rebounds, it experiences the same amount of momentum in the opposite direction.

Hence,

Change in momentum = 2 * p

Relative velocity ( ball ),

-V along x-axis directed to the right

-V along y-axis directed up

After rebound,

V along x-axis directed to the left

V along y-axis directed down

Therefore,

Velocity ( rebounded ball ) = √ ( 2 V )^2 + V^2

V √5

Velocity ( rebounded ball ) = 4√5 m / s

Hence, the velocity of the rebounded ball is 4√5 m / s

The answer is (d) 4√5 m / s

Answer 2

An object having momentum strikes an elastically massive object maybe a wall rebounds with the same value of momentum in the opposite direction.

The ball rebounds with a velocity v along x- axis and velocity v along y axis in the cart frame. The ball moves with a velocity of 2v along x and y axis. Thus the velocity of the rebounded ball is

√2v^2 + v^2 = v√5 = 4√5 ( given velocity is 4 m/s)