Repeated Collision: Find Minimum Mass M

[Chestermiller]

Typo. He meant v₁=2Mv0/(M+m)

Thanks haruspex. That was a mistake in algebra, not a typo.

With this correction, the equation for δ becomes:
$$\delta=\frac{2Mv_0}{ω(M+m)}sin(ωt)$$
and
$$x_1(t)=\frac{Mv_0t}{(M+m)}+\frac{Mv_0}{ω(M+m)}sin(ωt)=\frac{Mv_0}{ω(M+m)}(ωt+sin(ωt))$$
So, if $x_M=\frac{v_0(M-m)t}{(M+m)}$, we get
$$\frac{v_0(M-m)t}{(M+m)}=\frac{Mv_0t}{(M+m)}+\frac{Mv_0}{ω(M+m)}sin(ωt)$$
From this, it follows that:
$$-mωt=Msin(ωt)$$
An additional relationship is required if M barely catches up with m for a second time. This is that the velocities must match.
$$\frac{v_0(M-m)}{(M+m)}=\frac{Mv_0}{(M+m)}+\frac{Mv_0}{(M+m)}cos(ωt)$$
From this, it follows that
$$-m=Mcos(ωt)$$
The angle we're looking for is in the third quadrant where cosine and sine are both negative. So, $sin(ωt)=-\sqrt{1-(\frac{m}{M})^2}$
This is sufficient to solve for the value of M/m that makes good on both these equations.

Chet

 

[peripatein]

Actually, I didn't want to bring this up, but you're still wrong as far as I am concerned. Recheck your algebra. I have checked my calculations numerous times and v1 = Mv0/(m+M) not 2Mv0/(m+M).

 

[peripatein]

Adding that latter condition, I finally obtained:
tg(x)=x
where x=wt
That gives t~4.49/w

 

[Chestermiller]

Actually, I didn't want to bring this up, but you're still wrong as far as I am concerned. Recheck your algebra. I have checked my calculations numerous times and v1 = Mv0/(m+M) not 2Mv0/(m+M).

Uh Oh. I check it numerous times too, and got 2Mv0/(m+M). Here is the sequence of steps:

I started with the following:
$$v_M=v_0\frac{M-m}{M+m}$$
$$v_1=\frac{M(v_0-v_M)}{m}$$

Substituting the first equation into the second equation, I get:
$$v_1=\frac{Mv_0}{m}\left(1-\frac{(M-m)}{(M+m)}\right)=\frac{Mv_0}{m}\frac{(M+m)-(M-m)}{(M+m)}=\frac{Mv_0}{m}\frac{2m}{(M+m)}=\frac{2Mv_0}{M+m}$$
I can't see where I made a mistake.

 

[Chestermiller]

From the previous equations I presented, I get:

$$ωt=\sqrt{\left(\frac{M}{m}\right)^2-1}$$
so,
$$\cos\left(\sqrt{\left(\frac{M}{m}\right)^2-1}\right)=-\frac{m}{M}$$
This needs to be solved for m/M.

 

[peripatein]

I have checked mine several times as well and obtained the following:
v_M = (Mv₀ - 2mv₁)/M
v₁ = Mv₀/(M+m)
Hence,
v_M=v₀(M-m)/(M+m)

 

[haruspex]

Uh Oh. I check it numerous times too, and got 2Mv0/(m+M). Here is the sequence of steps:

I started with the following:
$$v_M=v_0\frac{M-m}{M+m}$$
$$v_1=\frac{M(v_0-v_M)}{m}$$

Substituting the first equation into the second equation, I get:
$$v_1=\frac{Mv_0}{m}\left(1-\frac{(M-m)}{(M+m)}\right)=\frac{Mv_0}{m}\frac{(M+m)-(M-m)}{(M+m)}=\frac{Mv_0}{m}\frac{2m}{(M+m)}=\frac{2Mv_0}{M+m}$$
I can't see where I made a mistake.

I get the same. peripatein, please post your calculation.

what do you mean by "sketch that"?

Sketching is not the same as solving graphically (which requires an accurate drawing).
I was trying to get you to see that the straight line representing the subsequent motion of M had to be tangential to the curve for the motion of m. This is the same as Chester's observation that the velocities must match as well as the positions, and therefore produces the same equation.

 

[peripatein]

I am going to leave it the way it is, so no need to repost. In any case, it doesn't matter much. The physics is the essence.
But thank you very much for your help; I am sincerely grateful! :)

 

[Chestermiller]

I have checked mine several times as well and obtained the following:
v_M = (Mv₀ - 2mv₁)/M
v₁ = Mv₀/(M+m)
Hence,
v_M=v₀(M-m)/(M+m)

The first of these equations is not consistent with the momentum balance. Where did the 2 come from? See post #6 by fightfish (which is correct).

 

[Chestermiller]

Adding that latter condition, I finally obtained:
tg(x)=x
where x=wt
That gives t~4.49/w

This result for x is approximately correct. So, m/M=-cos(4.49) is the solution you have been looking for. That completes the solution.