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I am a middle school science teacher. I'm a little rusty on my physics knowledge, and I came across a question that I feel like I should be able to solve, but it also feels like there is a piece of information missing. Keep in mind, I understand that in real life, this would not be a conservation of momentum problem. Please help me answer the question keeping in mind that friction is being ignored.

A football player runs at 8 m/s and plows into a 80 kg referee standing on the field, causing the referee to fly forward at 5 m/s. If this were a perfectly elastic collision, what would the mass of the football player be?

In order to solve this, do I need to also know the final speed of the football player, or is there some way it can be resolved without it.

The initial momentum is 8m (m being the mass of the football player). The final momentum of the referee is 400 kg*m/s and the final momentum of the football player is mv (v is the final velocity of the player).

Is there any way I can further simplify the solution of m=(8-v)/400 without knowing the final velocity?

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Because it is a perfectly elastic collision the kinetic energy and the momentum are conserved. So you have two equations for two unknowns which are the final velocity of the football player and his mass:

$$ m_f v_f^0+m_r v_r^0=m_f v_f^1+m_r v_r^1 $$ $$ \frac{m_f (v_f^0)^2}{2}+\frac{m_r (v_r^0)^2}{2}=\frac{m_f (v_f^1)^2}{2}+\frac{m_r (v_r^1)^2}{2} $$

and your data is $m_f$ is the mass of the football player (unknown), $m_r$ is the mass of the referee (known), $v_f^0=8 m/s$ - the initial speed of the football player, $v_f^1$ - the final speed of the football player (unknown), $v_r^0=0 m/s$ - the initial speed of the referee, $v_r^1=5$ - the final speed of the referee. You need to solve these equations for two unknowns.

The explicit solution to the equations $$ 8m=mv+400, ~32 m=m v^2/2+1000$$ is $$m=400/11~kg, ~v=-3 m/s.$$

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In a perfectly elastic collision, the final momentum of the system should be equal to the initial momentum of the system. It seems to be set up correctly, so I would say that you would need additional information for this.

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    $\begingroup$ Note that in perfectly elastic collisions, energy is also conserved, giving you a solvable system of equations. $\endgroup$ – Dave Coffman Mar 2 '15 at 23:00

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