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Yesterday, while studying a simple question, a rather strange doubt hit my mind:

Consider a ball of mass $m$ moving with velocity $v$ that collides with a wall in a perfectly elastic collision and rebounds back with the same velocity. Taking the direction of initial motion of the ball as positive, the change in momentum of the ball is $-2mv$ and the wall is clearly at rest.

So, in order to obey the law of conservation of momentum, the wall must gain a momentum of $2mv$.

Now, this itself confused me a lot. I then applied the law of conservation of kinetic energy to this situation, and clearly the magnitudes of the initial and final velocities of the ball are equal, so kinetic energy of the ball is conserved and hence the final kinetic energy of the wall must be zero in order to conserve energy.

Now, this contradicts with the law of conservation of momentum, which says that the wall must gain momentum of $2mv$.

One more point: If we are talking about a wall then it is an important fact, which must be considered, that the wall is fixed on earth and so the mass of the wall is equivalent to that of earth, say $M$. If the law of conservation of momentum is applied now then we get velocity of the wall, say $V=2mv/M$ which is no doubt an extremely small value. But, the law of conservation of energy strictly says that the velocity of the wall must be exactly zero, not even the smallest value is allowed (theoretically speaking).

Please clarify this and correct me anywhere if I'm wrong. I have tried my best to make it clear what I want to ask :).

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    $\begingroup$ "clearly the magnitudes of the initial and final velocities of the ball are equal" Think again. $\endgroup$ – garyp Oct 7 '16 at 13:53
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    $\begingroup$ @garyp yeah, what a factually incorrect statement. That's like saying a $10m/s$ boost to an object already travelling at $10m/s$ would leave it travelling at exactly $20m/s$. What normal and sane person would ever make such a statement? (He said, attempting to portray obvious sarcasm but possibly failing miserably) $\endgroup$ – Jim Oct 7 '16 at 15:20
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    $\begingroup$ " I then applied the law of conservation of kinetic energy.." There is no such law. $\endgroup$ – Bill N Oct 7 '16 at 17:23
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    $\begingroup$ FWIW, this is basically the same as this other question. $\endgroup$ – Kyle Kanos Oct 7 '16 at 19:14
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    $\begingroup$ @PM2Ring I know, it's a travesty $\endgroup$ – Jim Oct 8 '16 at 16:07
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But, the law of conservation of energy strictly says that the velocity of the wall must be exactly zero

Actually, it doesn't. Energy is just conserved - nothing has to be zero for it to be conserved.

The mistake is that you assume the wall to be stationary also after collision. This is your own assumption, and as you clearly show with momentum conservation, that cannot be true.

You have already realized that if the whole Earth is included in the picture, the momentum conservation law makes sense in the way that the Earth is given a tiny, tiny, tiny speed after collision. Now redo the energy conservation considerations with this in mind - in other words, redo the energy calculations without assuming that the wall/Earth is stationary.

In fact, the acquired Earth speed is so tiny that it is negligible - that's why it is usually just assumed zero.


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    $\begingroup$ Additionally, if you're launching the ball from somewhere on Earth to begin with, the tiny, tiny, tiny amount of speed you're adding to the Earth is counteracted by the tiny, tiny, tiny amount of speed in the opposite direction from when you launched the ball. $\endgroup$ – Tin Man Oct 7 '16 at 20:55
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    $\begingroup$ @Amadeus9 Yes, but that's irrelevant to the calculations of the ball/wall collision. $\endgroup$ – user253751 Oct 8 '16 at 7:11
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When the ball hits the wall and bounces back, it transfers a momentum of about $2m_bv_b$ to the wall. The wall is fixed to the foundation of the building, so it similarly pushes off the building and transfers that momentum to the building. The building is firmly rooted into Earth and so, that momentum is transferred to Earth. In turn, Earth accepts the extra momentum and says "Hah! Such weak extra momentum. I shall ignore your pitiful attempts to affect me as you would ignore the weight of an extra bacteria on your hand". Technically, Earth's velocity increases such that

$$\sim2m_bv_b=M_E\Delta v_E$$ Whatever the mass and velocity of the ball may be, Earth's mass is $5.97\times10^{24}kg$, so the delta-v of Earth is basically nothing. Also remember there are probably 10 quadrillion other things of similar mass adding to and taking from Earth's momentum in all directions. It cancels out (aside from being negligible).

As for the energy, since kinetic energy is proportional to $v^2$ and momentum is only proportional to $v$, the change in Earth's total kinetic energy from this interaction is even less than the change in its total momentum (if we measure relative percentages, not absolute number values). This is simply because of how limited the change in Earth's velocity was by its colossal mass (sorry Earth, not even a glandular problem will excuse how fat you are. Earth is so fat it has its own gravitational pull). So strictly speaking, for it to be a perfectly elastic collision, kinetic energy of the system doesn't change, which means that the final velocity of the ball isn't exactly $-v$. Let's do some of the initial math and ignore things like angular momentum (that is, assume the ball hits normal to Earth's surface). Also, let's assume Earth's initial velocity is zero and call the initial velocity of the ball $v$ and the mass of the ball $m_b=1kg$.

$$-m_b\Delta v_b=M_Ev_{Ef}$$ $$m_bv^2=m_bv_{bf}^2+M_Ev_{Ef}^2$$

Insert some razzmatazz algebra here, then add some PVA glue, and it should look something like this:

$$v_{bf}^2+\frac{1}{5.97\times10^{24}}\Delta v_b^2=v^2$$

So there you have it. The final velocity of the ball can be found with this equation. Since we reasonably expect the change in velocity of the ball to be somewhere around $-2v$, that second term on the left side of this equation is always going to be around 24 orders of magnitude smaller than the first term. What does this mean? No calculator you own nor most computer-based calculators (outside of explicitly set up programs you might as well write yourself) will be precise enough to resolve the difference between the technically accurate final velocity of the ball and if you were to say it was just $v_f=-v$. It's also not hard to convince yourself that the change in Earth's velocity is about 24 orders of magnitude less than the change in the ball's velocity.

Okay, back to the point at hand. You wondered how momentum and energy were truly conserved. Well, the energy and momentum gets transferred through the wall and into Earth, which absorbs it. However, the change in velocity this causes in Earth is so small that it's practically impossible to measure. So what you end up seeing is the ball behaving as though it bounced off an immovable object. The truth is that the change in velocity isn't exactly $-2v$, but I promise you that everyone will be okay with it if you want to round to $-2v$.

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    $\begingroup$ So, I was wrong in thinking that final velocity of earth is exactly zero which is not. Thank you very much Jim, your explanation was exactly what I needed:) $\endgroup$ – user375072 Oct 8 '16 at 2:02
  • $\begingroup$ @user375072 that's what I'm here for $\endgroup$ – Jim Oct 8 '16 at 2:43
  • $\begingroup$ I'm a little fuzzy on your reference frame. (Okay; I'm not really, but...) If we fix the coordinate system to the Earth, then it is noninertial at one instant, but is inertial at all other times. What's the ball's final velocity in that slowly recoiling frame? (I predict it's $-v$.) $\endgroup$ – Eric Towers Oct 8 '16 at 5:57
  • $\begingroup$ @EricTowers yup, $-v$. However, my frame was Earth's initial frame. The collision with the ball boosted Earth out of that frame, but not me (I ain't 'fraid of no boost) $\endgroup$ – Jim Oct 8 '16 at 16:05
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    $\begingroup$ In turn, Earth accepts the extra momentum and says "Hah! Such weak extra momentum. I shall ignore your pitiful attempts to affect me as you would ignore the weight of an extra bacteria on your hand". Hah! Nice. $\endgroup$ – Lightness Races with Monica Oct 9 '16 at 11:31
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The wall is connected to the earth. The earth and the wall gain a momentum of 2mv which is so insignificant that it's effect on the earth and wall is basically non existent for all intents and purposes.

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I think you are all ignoring the fact that some proportion of the energy of the ball will be converted to heat on hitting the wall. This should strictly be accounted for in this equation. (Ex Physics Teacher)

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    $\begingroup$ We aren't forgetting that. It is a perfectly elastic collision, which by definition means all kinetic energy is conserved. If we allow for the conversion of kinetic energy into heat energy, then the collision becomes something other than perfectly elastic, which isn't in the spirit of the question $\endgroup$ – Jim Oct 8 '16 at 16:20

protected by Qmechanic Oct 8 '16 at 15:22

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