Suppose a bouncy ball were thrown against a wall. The ball hit with 15 m/s and then bounced backwards with 15 m/s. The interaction time with the wall was 0.5 ms (5 x 10-3 s). If the ball's mass is 0.1 kg, what is the average force of the wall on the ball?

due to changing directions, the change in momentum is (mass(vf-vi))=.1kg*(30m/s)

force*time=changeinmomentum

6000N

To find the average force exerted by the wall on the ball, we can use Newton's second law of motion, which states that force is equal to the rate of change of momentum. The formula is:

Force = (m * Δv) / Δt

Where:
m = mass of the ball
Δv = change in velocity of the ball (final velocity - initial velocity)
Δt = time of interaction with the wall

Let's calculate the change in velocity first. The ball initially moves forward with 15 m/s and then bounces backward, so the change in velocity would be:

Δv = (-15 m/s) - (15 m/s)
Δv = -30 m/s

Now we can substitute the values into the formula:

Force = (0.1 kg * -30 m/s) / 0.005 s
Force = -600 N

The negative sign indicates that the force acts in the opposite direction to the motion of the ball. Therefore, the average force exerted by the wall on the ball is 600 Newtons.