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According to the conservation of momentum and kinetic energy, in a perfectly elastic collision,

$m_{1i}v_{1i}+m_{2i}v_{2i}=m_{1f}v_{1f}+m_{2f}v_{2f}$

$ \frac{1}2 m_{1i}v_{1i}^2 + \frac{1}2 m_{2i}v_{2i}^2 = \frac{1}2 m_{1f}v_{1f}^2 + \frac{1}2 m_{2f} v_{2f}^2$

The momentum of the rubber ball: $$m_{1}v_{1i}=m_{1}v_{1f}+m_{2}v_{2f}$$

As a rule, in terms of kinetic energy

$$\frac{1}2m_{1}v_{1i}^2 = \frac{1}2m_{1}v_{1f}^2 + \frac{1}2m_{2}v_{2f}^2$$

As the ball bounces back and the door moves forwards, why is there more kinetic energy in the system? as in an elastic collision $(-v)^2=v^2$, so while in momentum: $$m_{1i}v_{1i}=m_{1f}(-v_{1i})+m_{2f}v_{2f}$$ $$2*m_{1i}v_{1i}=m_{2f}v_{2f}$$

in terms of the energy: $$\frac{1}2m_{1i}v_{1i}^2 = \frac{1}2m_{1f}v_{1i}^2 + \frac{1}2m_{2f}v_{2f}^2$$ $$0 = \frac{1}2m_{2f}v_{2f}^2$$

Am i missing a step? (This is collision between 2 dynamic objects rather than a dynamic and fixed object)

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Please note that v(final) of ball is not equal to v(initial) of ball. Therefore that assumption in ur answer was a mistake. To solve this question u need both the equations, the linear momentum conservation and energy conservation eq. Use the two equations to get the v(final) of the ball

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Sticky ball's kinetic energy is converted into thermal energy of (ball-door system) on its collision with the door. Law of conservation of energy doesn't get violated in the first place, in this way.

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  • $\begingroup$ Perfect collision, only measuring kinetic and potential energy, but u are correct. Also, I am asking about the rubber ball. $\endgroup$ – A man with a hat Mar 14 '18 at 9:51
  • $\begingroup$ What about rubber ball? Rubber ball's collision is much more elastic than that of the sticly ball i.e. rubber ball's kinetic energy doesn't get converted into as much thermal energy as that in the case of sticky ball. $\endgroup$ – Anonymous Mar 14 '18 at 13:23

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