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Hello,

In the above question I could solve for average elastic force by taking velocity with respect to wall and finding change in momentum of the ball after that divided change jn momentum by time interval. Answer comes out to be option b.

But as it is written in the question that collision is elastic, therefore, Kinetic energy before the collision should be equal to kinetic energy after the collision and option d should also be correct. But the correct answers according to the book are b and c.

Please Explain why d is incorrect and c is correct.

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Kinetic energy before the collision should be equal to kinetic energy after the collision

In an elastic collision, the "total" kinetic energy and "total" momentum of the system will be conserved. You missed that point. The individual kinetic energy of the ball and wall could change. But the total kinetic energy of the system is conserved before and after collision. That is

$$T_{wall}+T_{ball}=T_{wall}'+T_{ball}'$$

where the primed quantities denote kinetic energy after collision and unprimed quantities denote kinetic energy after collision. This is what it is meant by kinetic energy of the system is conserved in elastic collision.

Unless

  • the massive wall is static (and the ball is not so fast enough to break the wall on collision), or
  • the ball and the wall have equal masses and equal velocities

then only we can say that the kinetic energy of each constituents of the system is conserved before and after collisions. But, in general, it is the total kinetic energy of the system (the sum of kinetic energies of all constituents making up the system) that is conserved.

Hence option (d) is incorrect, as the given problem does not satisfy any of the above mentioned criterions. So, on impact, the momentum of the ball increases and hence its kinetic energy also increases. This increase in kinetic energy of the ball will be equal to the decrease in the kinetic energy of the wall. Hence the kinetic energy of the system is conserved in the elastic collision.

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Alright this is what I got:

sorry for being so slow :(

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  • $\begingroup$ your good :) :) $\endgroup$ – Nicole. C Feb 10 '17 at 2:15
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I think d is incorrect because the kinetic energy increases due to the massive moving wall hitting against the ball.

Kinetic energy before the collision should be equal to kinetic energy after the collision

will only be true if the wall is static, not moving.

Just like hitting a ball with a bat.

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  • $\begingroup$ Please take a look at last line of the question, It says "elastic impact....". So d should be correct? $\endgroup$ – user411518 Feb 10 '17 at 0:42
  • $\begingroup$ Yes, that only means that the ball doesn' t lose KE hitting the wall, but it gains KE from the moving wall. $\endgroup$ – Nicole. C Feb 10 '17 at 0:48
  • $\begingroup$ How to calculate the change in kinetic enegry? $\endgroup$ – user411518 Feb 10 '17 at 0:51
  • $\begingroup$ wait lol let me think $\endgroup$ – Nicole. C Feb 10 '17 at 0:52
  • $\begingroup$ wait can you say again how you did (b), cuz there are many ways to solve this problem, i just want to know how you solved the first problem. I gt the answer already $\endgroup$ – Nicole. C Feb 10 '17 at 1:19

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