Physics, please help..?

Question

a) A 3.0 kg ball has a velocity of 25 m/s downward just before it strikes the ground and bounces up with a velocity of 25 m/s upward. What is the change in momentum (impulse) of the ball?

b) If the ball is in contact with the ground for 0.20 seconds, what is the average force experienced by the ball on impact with the ground?

Thanks...

Answer 1

The average force experienced by the ball on impact with the ground is

779.4 N. Here's how this was found:

The UPWARDS momentum change M is 3.0 (25 - (-25)) kg m/s = 150 Ns.

To find the average force exerted by the ground during the contact time 0.20 s, we must not FORGET that there is a DOWNWARDS "impulse" D due to gravity during that time of 3.0 (9.80) (0.20) kg m/s = 5.88 Ns.

[Later edit: I'm afraid that 'Bekki B,' whose Physics answers are normally reliable, FORGOT about the fact that gravity continues to operate on the ball during its contact with the ground.]

Let the UPWARDS impulse on the ball from the floor be U; that upwards impulse has to both counter the magnitude of the downwards gravitational impulse D AND provide for the change in upwards momentum M; so

U = D + M = 155.88 Ns.

So the mean upward force applied to the ball is 155.88 Ns / 0.20 Ns

= 779.4 N.

Live long and prosper.

Answer 2

a) the change in momentum is the momentum after (+25 m/s times mass) minus the momentum before (-25 m/s times mass). Remember minus minus is plus.

b) Impulse force times time (for constant F). So force = impulse (answer above) divided by time.

Your welcome!