Basketball force/momentum problem

[quark001]

A basketball hits a wall at a speed of 20m.s-1 and leaves the wall at the same speed. At the moment when the ball's speed is momentarily zero, the ball is compressed by 0.3m. The mass of the ball is 0.5kg. Determine the force on the ball (exerted by the wall).

Getting change in momentum is easy, but how do you get from there to the force? No time interval is given.

 

[dx]

Use the fact that the ball is compressed by 0.3m at the moment when its speed reaches zero.

 

[haruspex]

It's not clear whether the question is asking for average force or the force when at maximum compression. Either way, I would assume the ball acts as a spring, so the compression phase will be SHM.

 

[quark001]

Just to clarify: the question is asking for the average force. In fact we are assuming that the force is constant for the time of contact.

 

[dx]

The change in kinetic energy from the moment of contact to the moment at which the ball is at rest is (1/2)mv²

This must be equal to the work done by the force. If the force is F, and the ball moves a distance 0.3m during this time, what is the work done?

 

[haruspex]

The change in kinetic energy from the moment of contact to the moment at which the ball is at rest is (1/2)mv²

This must be equal to the work done by the force. If the force is F, and the ball moves a distance 0.3m during this time, what is the work done?

It is possible, even likely, that that is the approach the question intends. However, it gives the wrong answer. Unless stated otherwise, the average of a quantity that varies over time should be taken as the average wrt time: ∫F.dt/∫dt. The average wrt distance, ∫F.dx/∫dx, will in general give a different answer.
E.g., suppose I go 10km at 10km/h, then 10km at 40km/h. My average speed is 16km/h, right? But if I average over distance I get (10km * (10km/h) + 10 km * (40km/h))/(10+10) = 25km/h.

 

[dx]

I was simply assuming that the force was a constant, like quark001 said, but yes you're right about average force.

 

[haruspex]

I was simply assuming that the force was a constant, like quark001 said,

Fair enough, but I do wish textbooks/teachers wouldn't set questions like this. It's misleading. The difference, fwiw, is that assuming constant force will give an acceleration of v²/2d, whereas an elastic bounce (SHM) gives v²/πd.

 

[quark001]

Thanks guys!