Number of collisions with a block

[Abhishek11235]

The problem is given in David Morin's Classical mechanics.

Now, I jumped to solve part b of question. To find the number of bounces,we note that mass losses momentum of -2mV per bounce(This can be worked out from conservation of momentum and energy). Now initial momentum was MV. Then since per bounce loss is -2mV then after n bounces ,it comes to rest. Hence: Mv-2mnv=0 $\implies$ n=M/2m. However,when I checked solution,it deferred much from this one. There was factor of π/4 in solution. What is wrong with my reasoning?

 

[Hiero]

That book has such fun problems! Morin made me fall in love with physics. There are still a few 4 star problems I’ve yet to solve.

What you said about losing 2mv is only (approximately) true for the very first collision.

Somewhere in that book he derived the result that for elastic collisions, the relative velocity before and after the collision is the same; have you seen that? (If not then think about it from the center of mass frame; for non-relativistic speeds the relative velocity is (approximately) frame invariant.)

So with that in mind, your answer makes sense for the first collision, because the heavy block’s speed is approximately unchanged, hence the light ball must move forward at ≈2V.

But then, the light ball reflects off the wall and has speed -2V, and so then the relative velocity is ≈3V... So the next change in momentum is ≈-6mV!

Even with that correction though, I don’t think following this pattern will work because the approximation errors should add up (the number of collisions becomes infinite in the limit that the approximations are accurate, namely M>>m). In other words, we can’t keep saying the blocks velocity is ≈V_0... it has to decrease eventually!
(I don’t even think we can get a closed solution from this finite series.)

I’m not sure the best approach... But anyway, that is the flaw in your argument.

I don’t think any four star problem has a solution as brief as you suggested don’t be bothered if you can’t solve it.

 

[A.T.]

The problem is given in David Morin's Classical mechanics.

Now, I jumped to solve part b of question. To find the number of bounces,we note that mass losses momentum of -2mV per bounce(This can be worked out from conservation of momentum and energy). Now initial momentum was MV. Then since per bounce loss is -2mV then after n bounces ,it comes to rest. Hence: Mv-2mnv=0 $\implies$ n=M/2m. However,when I checked solution,it deferred much from this one. There was factor of π/4 in solution. What is wrong with my reasoning?View attachment 237359

This is similar, but about the total collisions (wall and block) and without time bound:

 

[neilparker62]

You could experiment on a spreadsheet - I tried with a large mass of 100kg, small mass 1kg and initial velocity 10 m/s :
$$
\begin{matrix}
Collision & Δp=2μΔv & v_m & v_M \\
0 & 0.000 & 0.000 & -10.000 \\
1 & 19.802 & 19.802 & -9.802 \\
3 & 58.622 & 38.820 & -9.216 \\
5 & 95.120 & 56.300 & -8.265 \\
7 & 127.851 & 71.551 & -6.986 \\
9 & 155.518 & 83.968 & -5.431 \\
11 & 177.027 & 93.059 & -3.661 \\
13 & 191.524 & 98.465 & -1.745 \\
15 & 198.437 & 99.971 & 0.239 \\
17 & 197.490 & 97.519 & 2.214 \\
19 & 188.722 & 91.204 & 4.101 \\
21 & 172.480 & 81.276 & 5.826 \\
23 & 149.407 & 68.131 & 7.320 \\
25 & 120.417 & 52.286 & 8.524 \\
27 & 86.658 & 34.372 & 9.391 \\
29 & 49.467 & 15.095 & 9.885 \\
31 & 10.317 & & \\
\end{matrix} $$

Hope I set up the formulae correctly - seems to be in keeping with the video per post #3 !

 

[haruspex]

I tried defining the velocities (positive towards the wall) as u_r for M and v_r for m after the r^th bounce from the wall. So initial condition is v₀=0.

I could see easily that (-1)^ru_r-v_r is interesting, so I also looked for constants α, λ such that u_r+λv_r=α(u_r-1+λv_r-1).
I.e., looking for the eigenvectors.
I ended up with M coming to rest when e^2irθ=i cot(θ) (or approximately) where cot²(θ)=M/m.

 

[A.T.]

This is similar, but about the total collisions (wall and block) and without time bound:

The solution:

 

[A.T.]

The solution:

Alternative solution: