Question

Can someone check my work. I'm kind of stuck...

Answer 1

A 200 g ball is dropped from a height of 2.0 m, bounces on a hard floor, and rebounds to a height of 1.5 m. FIGURE P9.28 shows the impulse received from the floor. What maximum force does the floor exert on the ball?

Answer 2

I thought that since impulse= F x t

I could use this equation:

F x t = m Δv

Answer 3

Using some kinematics I could find out what the velocity of the ball is when it's coming down and its velocity when it goes up again.

Answer 4

which I did using \[v_f ^2 = v_i ^2 + 2ah\]

Answer 5

\[v_1 = 6.26 \frac{ m }{ s }\]
\[v_2 = 5.42 \frac{ m }{ s }\]

Answer 6

Plugging the velocities into the impulse equation and solving for F I get

1172 N ?

Answer 7

of course I converted the units to kg and s

Answer 8

change in velocity is 11.72 m/s right? since v1 is -6.26 m/s bc the ball is going down.

Answer 9

@agent0smith

Answer 10

yeah it tells me but symbolically.

Answer 11

hey the max force represents over 5 ms represents impulse right?

Answer 12

lol I thought I was going crazy or something

Answer 13

Fmax is just twice the average force (which is the one in F x t = m Δv)

Answer 14

oh ok yeah that is what I used and solved for F. It made sense to me but now my friend and I were comparing our results and they were different. He used "conservation of momentum" and the way he worked it out was confusing and didn't make sense. So that is why I want to know if my approach was right.

Answer 15

Is my approach right?

Answer 16

You can't use cons. of momentum (it is not conserved for the ball)

Answer 17

ohh yeah that makes sense

Answer 18

I got that the max force is 468.8 N

Answer 19

Wait, use the fact that the area under the Ft curve is impulse (m Δv)

Then, the graph is a triangle, so area of a triangle with a base of 5 ms and a height which happens to be Fmax.

Answer 20

oh ok, I still get the same results but that helps everything make more sense :)

Answer 21

Are you sure?

Answer 22

half

Answer 23

hehe I forgot to put 1/2

Answer 24

\[\frac{ 1 }{ 2 }*0.005*F_{\max} = 0.2*11.76\]

Answer 25

940 N