If a particle with mass $m$ collides with a wall at right angles, and the collision is perfectly elastic. The particle hits the wall at $v\ ms^{-1}$. There is no friction or gravity. So the particle will rebound at $-v\ ms^{-1}$?

image description

What will the change in momentum be?

I did:

$$initial\ momentum = final\ momentum$$ $$mv = m(-v)$$ $$mv = -mv$$

But this doesn't seem right because it's like saying $1=-1$?

  • $\begingroup$ Both the total energy and total momentum are conserved. Hint: Treat the wall as a second very massive body in a 1D elastic two-body collision. $\endgroup$ – Qmechanic Jan 11 '13 at 13:24
  • $\begingroup$ On top of all these good answers, remember that momentum is a vector, not a scalar quantity. So change in momentum is also a vector. $\endgroup$ – Mike Dunlavey Jan 11 '13 at 14:00
  • $\begingroup$ @qmechanic, but the wall has no momentum, because no matter what mass it is, it's velocity is 0? $\endgroup$ – Jonathan. Jan 11 '13 at 15:12
  • $\begingroup$ @Jonathan.: It's not right to say that the the wall has no momentum after the collision (within the two-body idealization). Letting the mass ratio $M/m \to \infty$ go to infinity, the infinitely heavy wall (after the collision) will indeed have zero velocity (in the limit) but nevertheless carry the missing momentum $2mv$. $\endgroup$ – Qmechanic Jan 11 '13 at 15:24
  • $\begingroup$ @Qmechanic, I dont understand, momentum is $mv$, and $v = 0$, and anything times 0 is 0? $\endgroup$ – Jonathan. Jan 11 '13 at 23:37

The initial and final momentum are not the same because the ball is not an isolated system. The wall exerts a force on it. In principle the ball and the wall (and the planet it's connected to!) form an isolated system with a conserved momentum, but you'd have to take into account how much the wall moves after the collision.

The change of momentum is final momentum - initial momentum, and you have the correct values for the initial and final momentum.

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  • $\begingroup$ So the change in momentum is $-2mv$. I have the answer as positive $2mv$, is it possible to just use the magnitude? I would have thought that that implied to went into the wall? $\endgroup$ – Jonathan. Jan 11 '13 at 10:54
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    $\begingroup$ I wouldn't attach too much significance to the sign of the change. If your sign convention is that velocity towards the wall is positive, then momentum before is $+mv$ and after it's $-mv$, so the momentum has decreased by $2mv$. I would interpret this as meaning that the change is $-2mv$. $\endgroup$ – John Rennie Jan 11 '13 at 11:04

In presence of a force the momentum is not conserved, and the wall is a potential repulsive force. Instead, the momentum changes from a positive to a negative value, so the difference is positive.

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Your equation: $\text {initial momentum = final momentum}$, applies only to the total momentum. It does not apply to individual masses separately.

Here the initial momentum of the mass $m = \text {initial total momentum} = mv$ (since the wall is not moving)

The final total momentum is the sum of the momenta of the wall and the momentum of the mass $m$

The final total momentum is thus the initial total momentum = $mv = -mv + x$

Change in the momentum of $m = -mv-mv =-2mv$.

Change in the momentum of the wall $= + 2mv$.

Total change in the momentum of the system $= (-2mv + 2mv) = 0$ (by law of conservation of momentum). You may add the units to the quantities.

Comment: The diagram shows the velocity after collision as $-v \:\mathrm{ms^{-1}}$ with an arrow pointing to the left. That would be incorrect. $-mv$ with arrow pointing to the right or $mv$ with arrow pointing to the left would be correct.

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  • $\begingroup$ Hi user. Welcome to Physics.SE. Here, we use an unique TeX markup called MathJax, same as Math.SE. The markup is very much helpful in understanding equations, etc. Please have a look here for an introductory, or atleast have a look at our FAQ for an overview. For now, Manish has revised your post :-) $\endgroup$ – Waffle's Crazy Peanut Jan 16 '13 at 12:08

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