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

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Homework Help Overview

The problem involves calculating the average force exerted on a 0.54kg ball that falls from a height of 26m and rebounds to a height of 16m, with a contact time of 1.8ms. The discussion centers around the application of kinematic equations and the concept of momentum in the context of forces during the ball's impact and rebound.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations for the ball's velocity just before impact and just after rebound, questioning the assumptions made regarding the direction of momentum. There are attempts to use momentum change over time to find the average force, with some participants exploring the signs of the velocities involved.

Discussion Status

Several calculations have been presented, with participants verifying each other's results and questioning the correctness of their approaches. There is an ongoing exploration of the relationship between initial and final momentum, and how to account for the direction of these values in the force calculation.

Contextual Notes

Participants note that the acceleration due to gravity is consistently applied, and there is a focus on the implications of the ball's upward and downward motion in the calculations. Some participants express uncertainty about the assumptions made regarding the forces involved during the impact and rebound phases.

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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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