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A speed skater with a mass of $m_1$ kg is moving at $v_{1i}$ m/s. He prepares to push the next speed skater of mass $m_2$ kg moving at $v_{2i}$ m/s. After the push, the velocity of the first skater with mass $m_1$ kg has dropped to $v_{1f}$ m/s. Determine the new velocity $v_{2f}$ of the skater with mass $m_2$ kg that was pushed. Assume there is no friction.

If I use the law of conservation of momentum, I get the an answer for the velocity as:

$$\text{total momentum before push} = \text{total momentum after push}$$ $$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$ $$v_{2f} = \frac{ m_1v_{1i} + m_2v_{2i} - m_1v_{1f} }{ m_2 }$$

However if I use the law of conservation of energy, I get a different answer for the velocity:

$$\text{total KE before push} = \text{total KE after push}$$ $$\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$$ $$m_1v_{1i}^2 + m_2v_{2i}^2 = m_1v_{1f}^2 + m_2v_{2f}^2$$ $$v_{2f}^2 = \frac{m_1v_{1i}^2 + m_2v_{2i}^2 - m_1v_{1f}^2}{m_2}$$ $$v_{2f} = \sqrt{ \frac{m_1v_{1i}^2 + m_2v_{2i}^2 - m_1v_{1f}^2}{m_2} }$$

In general, the two different approaches will give different results, however the final velocity should be the same regardless of approach.

Can someone help me figure out the reason that I'm getting inconsistent answers please?

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  • $\begingroup$ Please take a minute to read our guidelines for homework-like questions as well as check-my-work. We intend our questions to be potentially useful to a broader set of users than just the one asking, and we prefer conceptual questions over those just asking for a specific computation. $\endgroup$ – stafusa Dec 17 '18 at 1:15
  • $\begingroup$ Edited to be more general $\endgroup$ – NM_ Dec 17 '18 at 1:26
  • $\begingroup$ When the collision is elastic there should be a common solution to both approaches. Otherwise the given numbers may only work with the momentum approach, which means it was inelastic. $\endgroup$ – PhysicsDave Dec 17 '18 at 2:33
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    $\begingroup$ This does not seem to be a homework like question to me. The question is why conservation of KE and conservation of momentum give different answers. The OP worked out the answers both ways and was only asking for a conceptual understanding of why the results differ $\endgroup$ – Dale Dec 18 '18 at 3:45
  • $\begingroup$ See also en.wikipedia.org/wiki/Coefficient_of_restitution $\endgroup$ – J.G. Dec 19 '18 at 13:38
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The result should be the same regardless of approach. Can someone help me figure out the reason that I'm getting inconsistent answers please?

The reason that you are getting inconsistent answers is that your expression for the conservation of energy is incorrect. In a perfectly elastic collision KE is conserved, but in this case the skaters are not billiard balls colliding elastically. Some energy goes into thermal energy and there is a change in chemical potential energy also. A full conservation of energy approach needs to consider those also.

Unfortunately, there is no way to calculate such quantities from the information given. Therefore, all you can use is conservation of momentum. The conservation of momentum can, in fact, be used to figure out how inelastic the collision was.

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  • $\begingroup$ Isn't the collision "super-elastic" rather than "inelastic? The pusher does work on the pushed, increasing the total kinetic energy of the system. $\endgroup$ – DJohnM Dec 17 '18 at 7:46
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    $\begingroup$ I have never heard of the term “super elastic” before. But sure, sounds good! $\endgroup$ – Dale Dec 17 '18 at 12:26
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In collisions momentum is always conserved, but it does not have to be the case that kinetic energy is conserved. This is because Newton's third law always holds which gives rise to momentum conservation since there is no net force acting on the system. The internal forces cancel out, but those internal forces could be doing work that removes energy from the system.

There are four velocities here, $v_{1i}$, $v_{1f}$, $v_{2i}$, and $v_{2f}$. Once you specify three of these, the only way to have kinetic energy conserved is if you pick the three velocities to be just right$^*$. Without going through the derivation (I'll leave that to you), it must be that the relative velocities before the collision is opposite the relative velocity after the collision in order for kinetic energy to be conserved $$v_{1i}-v_{2i}=v_{2f}-v_{1f}$$

Using this relation you can show that your two answers are now equivalent (I will also leave this to you to show). If the above equation does not hold, then your two equations are not equivalent. See my answer here to see more analysis into collisions where energy is/isn't conserved for the case where $v_{2i}=0$


$^*$ With just the momentum conservation equation, you have one equation and four undermined variables. Therefore you need to specify three of the variables to get a full solution. However, if we include kinetic energy conservation, then we have two equations. If we specify three of the variables, we run the risk of over-determining the system of equations unless the three values are just right to keep the two equations consistent.

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