Why is conservation of momentum not valid here?

To explain my confusion, I would provide the following system:

The two masses $m$ and $M$, with $M\gg m$, are moving towards each other(as directed by the arrows) with a common constant speed $V_0$. There is no friction between any two surfaces and all collisions are perfectly elastic.

I take all velocities +ve towards right and I call the velocity of $m$ after collision $V$. As $M\gg m$, there will be negligible change in the velocity of $M$ after collision. Also, as the collision is elastic, the velocity with which the two masses approach each other must be equal to the velocity with which they get separated.

Therefore, \begin{align} -V_0-(V_0)&=-V_0-(V) \\V&=-3V_0 \end{align}

Now, this seems quite true to me.

But, when we apply conservation of momentum, \begin{align} mV_0-MV_0&=mV-MV_0 \\V&=V_0 \end{align}

So why is conservation of momentum not valid here?

• Possible duplicate of Conservation of momentum in an elastic collision Commented Sep 11, 2018 at 16:37
• @sammygerbil Sir, I do not think this is a duplicate of that question. My question asks why and how my working is not correct. The question you provided simply wanted an explanation of why a ball after hitting a rigid stationary wall rebounds with same speed. My question does not ask that. Though I know that this problem can be reduced to that problem if we look at things from the frame of reference of the big mass, but it can only be reduced to and the problems are different. Commented Sep 11, 2018 at 16:50
• @SeaDog Both questions ask why momentum is not conserved. Both questions make the assumption that because the mass of the larger object is so very much bigger than that of the smaller object then the change in velocity of the larger object (and therefore also its change of momentum) can be neglected. Commented Sep 11, 2018 at 16:57
• @sammygerbil There the wall was rigid and so, without any math and only clear conception, one can easily say that the change of momentum of the wall is zero Commented Sep 11, 2018 at 17:01
• @SeaDog That is exactly the same mistake which you have made. Commented Sep 11, 2018 at 17:03

1 Answer

As $M\gg m$, there will be negligible change in the velocity of $M$ after collision.

Yes, the change in the velocity of $M$ will be negligible, but what is conserved is not the velocity but the momentum and, since $M\gg m$, even a small change in the velocity of $M$ will translate in a relatively big change in its momentum.

• I agree with your first point. I would also say that your first point and @user190081 's comments fully answers my question. But I do not agree with your second point Commented Sep 11, 2018 at 16:24
• Coefficient of restitution $e=1$ for elastic collisions, whether or not masses are equal. So relative velocity of separation equals relative velocity of approach. This is a consequence of the conservation of kinetic energy and momentum. Commented Sep 11, 2018 at 16:31
• @sammygerbil Yes, rightly said. Commented Sep 11, 2018 at 16:33
• @SeaDog You, guys, are right. Thank you for the correction.
– V.F.
Commented Sep 11, 2018 at 17:42