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I have the following setup:

Mass $A$ is $4m$ kg and moves at 3 m/s on a smooth table. It collides directly with another mass $B$, which is $m$kg, which moves in the opposite direction to $A$ and had speed 6 m/s immediately before the collision. After the mass $A$ is said to have changed direction and move at 2 m/s.

So I want to find the coefficient of restitution.

When I applied the conservation of momentum, I found that $B$ had a speed of 14 m/s and also changed direction.

But the problem is when I calculate the coefficient of restitution I get $\frac{14 - - 2}{3 - - 6} = \frac{16}{9}$ and this is greater than 1!

I thought the coefficient of restitution was between $0$ and $1$?

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  • $\begingroup$ Since 3 speeds are given and the answer does not make sense, we can guess that the speed of B after the collision must be something other than 14 m/s. You may review your momentum equations and make sure all signs are correct. $\endgroup$ – V.F. Mar 29 '18 at 18:06
  • $\begingroup$ @V.F. The momentum equation is definitely correct, so it’s not that... $\endgroup$ – PhysicsMathsLove Mar 29 '18 at 18:28
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Working in units of $p_0 = m{\rm m/s}$, I get an initial state momentum:

$$ p_A + p_B = (4)(3) - (6)(1) = 12 - 6 = +6 $$

Given that $p'_A = (4)(-2) = -8$:

$$ p'_B = +14 $$.

The initial energy, in units of $p_0^2/(2m)$, is:

$$ E = \frac{p^2_A}{4} + \frac{p^2_B}{1} = 36 + 36 = 72 $$

Likewise the final state energy is:

$$ E' = \frac{p'^2_A}{4} + \frac{p'^2_B}{1} = 16 + 196 = 212$$

So the energy definitely went up.

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It appears that the conditions of this problem are faulty.

If we calculate a total kinetic energy before and after the collision, using speed 14m/s for B (found for the momentum equation), we'll find that the kinetic energy after the collision (assuming for simplicity m=1), 8+98=106, is greater than the kinetic energy before the collision, 18+18=36.

This explains by e>1.

Although it is not necessarily obvious at first glance, but we have to conclude that the speed of A after the collision cannot be -2m/s.

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    $\begingroup$ There are plenty of means by which a collision might liberate energy, so the conditions of the problem aren't necessary problematic. $\endgroup$ – J. Murray Mar 29 '18 at 22:49
  • $\begingroup$ Is the conservation of momentum still valid in this case though? $\endgroup$ – PhysicsMathsLove Mar 29 '18 at 22:54
  • $\begingroup$ @PhysicsMathsLove Was that question directed at me? If so, then the answer is yes. $\endgroup$ – J. Murray Mar 29 '18 at 23:24
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    $\begingroup$ @J.Murray Sure, but then e>1 is not an issue either. $\endgroup$ – V.F. Mar 30 '18 at 2:02
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    $\begingroup$ @V.F. Yes, I agree. I don’t see anything inherently wrong here. $\endgroup$ – J. Murray Mar 30 '18 at 2:04

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