Question

A 0.25 kg ball traveling at an initial velocity of -20 m/s hits a stationary wall and rebounds back with a final velocity of 20 m/s. What is the impulse imparted by the wall?

Answer 1

The impulse imparted by the wall is -10 kg·m/s

Step-by-step explanation:

The impulse imparted by the wall can be found using the principle of conservation of momentum. The impulse is equal to the change in momentum of the ball.

Since the ball rebounded with the same speed but in the opposite direction, the change in momentum is equal to twice the initial momentum.

The initial momentum of the ball can be calculated using the formula p = mv, where p is the momentum, m is the mass, and v is the velocity. Plugging in the values, we get:

p = (0.25 kg)(-20 m/s)

= -5 kg·m/s

Therefore, the impulse imparted by the wall is equal to twice the initial momentum, which is:

Impulse = 2(-5 kg·m/s)

= -10 kg·m/s