Question

A ball of mass 0.10 kg moving at a speed of 3.0 m/s collides with a wall and bounces directly back with the same speed. If the ball was in contact with the wall for 0.01 seconds, what is the magnitude of the average force exerted on the ball by the wall?

Answer 1

The magnitude of the average force exerted on the ball by the wall is 30 N.

Step-by-step explanation:

The magnitude of the average force exerted on the ball by the wall can be calculated using the impulse-momentum equation.

Impulse = change in momentum

Since the ball bounces back with the same speed, the change in momentum is given by: m * change in velocity = 0.10 kg * (0 - 3.0 m/s) = -0.30 kg·m/s

The force exerted on the ball by the wall is equal to the impulse divided by the time of contact: F = -0.30 kg·m/s / 0.01 s = -30 N

Therefore, the magnitude of the average force exerted on the ball by the wall is 30 N.

Answer 2

The magnitude of the average force exerted on the ball by the wall is calculated below.

The average force exerted by the ball on the wall is 3 N

Step-by-step explanation:

Given:

mass of the ball (m)=0.10 kg

speed (v) =3.0 m/s

time taken(t) =0.01 seconds

To calculate:

Average force(F) exerted by ball on the wall

We know;

F=(m×v)÷t

F=(0.10×3.0)÷0.01

F=3 N

Therefore the average force exerted by the ball on the wall is 3 N