How Does Collision Timing Affect the Force Experienced?

[Peter G.]

My book is talking about how collisions can be very different, basically saying how altering the time of contact affects the force. It then gives an example:

"Consider a ball of 200g colliding with a hard floor and a trampoline. Before the collision, each ball travels downwards at 10 m/s and each bounces up with velocity 10 m/s. So the change in momentum, impulse, is the same for each:

0.2 x(-20) - 0.2 x 20 = -8 N s"​

I am a bit confused with the calculation they did:
What I did was:

Change in Momentum = mv - mu
(Considering up to be + velocity and down to be - velocity)

(0.2 x 10) - (0.2 x -10) = 4 N s

What are they doing different than I am?

Thanks,
PeterG

 

[cepheid]

I agree with your calculation, if the speed is 10 m/s both before and after.

EDIT: It also looks like they defined "down" to be the positive y-direction, which explains the sign discrepancy, at least.

 

[Peter G.]

Ok, thanks!