What is the average force exerted on a 0.54kg ball that falls and rebounds?

AI Thread Summary
The discussion revolves around calculating the average force exerted on a 0.54kg ball that falls from a height of 26m and rebounds to 16m, with a contact time of 1.8ms. Initial calculations for the velocity upon impact and rebound are confirmed as correct, yielding 22.6 m/s and 17.7 m/s, respectively. The average force is derived from the change in momentum over the contact time, with the correct approach accounting for the opposite directions of the velocities. The final calculation shows that the average force exerted on the ball is approximately 12088.9 N after correctly considering the signs of the velocities. This highlights the importance of accurately applying momentum principles in physics problems.
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Homework Statement


A 0.54kg ball falls from rest at a height of 26m and rebounds up 16m. If the contact between the ball and ground lasted 1.8ms, what average force was exerted on the ball?


Homework Equations


F=ma
v²=v0²+2a(x-x0)
v=v0+at
1000ms=1s (1.8ms is 0.0018s)

The Attempt at a Solution


(numbers are shortened for convenience but end results are with actual numbers)
Calculated for velocity:
v²=v0²+2a(x-x0)
v²=0+2(9.8)(26)
v=22.6m/s

Calculated acceleration as ball hits ground:
v=v0+at
0=22.6+a0.0018
a=12,541m/s²

Calculate for force:
F=ma
F=(0.54)(12,541)
F=6772.3N

This is the answer I would try but it is wrong and I am missing something, like the 16m bounce up.

I then tried to subtract the force calculated with the force exerted when the ball bounced back up, which had a force of:
v²=v0²+2a(x-x0)
v²=0+2(9.8)(16)
v=17.7m/s

v=v0+at
0=17.7+a0.0018
a=9838.2m/s²

F=ma
F=(0.54)(9838.2)
F=5312.6N

This is upward force so the resultant for is a downward (6772.3-5312.6)=1459.67N.
This is also wrong and might have some assumptions in it.

What is wrong with the calculations? Is there something that I didn't understand about the questions or the use of the equations?
 
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Through out the problem acceleration is g.
Your v1 = 22.6 m/s is correct. v2 = 17.7 m/s is correct.
Now the average force is equal to (final momentum - initial momentum)/time of contact.
Note that two momentum have opposite direction.
 
v1=22.6
v2=17.7

(final momentum - initial momentum)/time
(v2×m - v1×m)/time
(17.7×0.54 - 22.6×0.54)/0.0018
(9.56 - 12.2)/0.0018
-2.63/0.0018
-1459.67

Is this correct?
 
312213 said:
v1=22.6
v2=17.7

(final momentum - initial momentum)/time
(v2×m - v1×m)/time
(17.7×0.54 - 22.6×0.54)/0.0018
(9.56 - 12.2)/0.0018
-2.63/0.0018
-1459.67

Is this correct?
Remember that v1 and v2, and hence their respective momenta, are in opposite directions.
 
v1=-22.6
v2=17.7

(final momentum - initial momentum)/time
(v2×m - v1×m)/time
(17.7×0.54 - (-22.6)×0.54)/0.0018
(9.56 - (-12.2))/0.0018
21.76/0.0018
12088.9

Is this right?
 
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