How Do You Calculate the Compression of a Ball Hitting a Wall?

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To calculate the average force exerted on a rubber ball hitting a wall, the formula used is average force = change in momentum / time, resulting in 2mv/t. The change in kinetic energy (KE) from impact to rest is calculated as 0.5mv². When estimating the compression of the ball during the collision, the assumption of Hooke's law leads to a derived formula x = vt/2, but the expected answer is vt/4. The discrepancy may arise from not accounting for the non-steady force in typical Hooke's law scenarios. The discussion suggests using work-energy principles to reconcile the calculations.
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Homework Statement
i) Find the average force exerted on a rubber ball mass m hitting a wall at speed v. The ball spends a time t in contact with the wall before bouncing off at speed v.
ii) The change in KE of the ball from when it hits the wall to when it is instantaneously at rest
iii) Assuming a steady force is acting on the ball, estimate the amount by which the ball is compressed during the collision

The attempt at a solution
i) This is easy enough
Average force = change in momentum / time taken = 2mv/t
ii) Again, easy: 0.5mv²
iii) If we assume the ball obey's Hooke's law (very rough, I know)
F=kx
=>2mv/t=kx or equivalently 2mvx/t=kx²
If all the KE of the ball is converted to EPE internally: mv²=kx²
Sub in kx²=2mvx/t giving 2mvx/t=mv²
Cancel and rearrange for x, the compression, giving: x=vt/2

The answers give vt/4, is there something I'm doing wrong? I'm guessing they made similar assumptions to me to get a similar answer but I can't see what I've done wrong.

Cheers.
 
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RK455 said:
Homework Statement
i) Find the average force exerted on a rubber ball mass m hitting a wall at speed v. The ball spends a time t in contact with the wall before bouncing off at speed v.
ii) The change in KE of the ball from when it hits the wall to when it is instantaneously at rest
iii) Assuming a steady force is acting on the ball, estimate the amount by which the ball is compressed during the collision

The attempt at a solution
i) This is easy enough
Average force = change in momentum / time taken = 2mv/t
ii) Again, easy: 0.5mv²
iii) If we assume the ball obey's Hooke's law (very rough, I know)
F=kx
=>2mv/t=kx or equivalently 2mvx/t=kx²
If all the KE of the ball is converted to EPE internally: mv²=kx²
Sub in kx²=2mvx/t giving 2mvx/t=mv²
Cancel and rearrange for x, the compression, giving: x=vt/2

The answers give vt/4, is there something I'm doing wrong? I'm guessing they made similar assumptions to me to get a similar answer but I can't see what I've done wrong.

Cheers.

Note that the questioner said assume a steady force. This is certainly not the case in a typical Hookes Law situation, so perhaps you are supposed to do a simple F=ma etc solution.
Part (i) you calculated the average force - so the steady force needed for all this to happen.
Part (ii) you calculated the energy change.

I think it is now just work [Fx] = change in energy.
 
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