Elastic collision between ball and block

[utkarshakash]

Homework Statement

A block with large mass M slides with speed V0 on a frictionless table towards a wall. It collides elastically with a ball with small mass m, which is initially at rest at a distance L from the wall. The ball slides towards the wall, bounces elastically, and then proceeds to bounce back and forth between the block and the wall.

a) How close does the block come to the wall ?
b) How many times does the ball bounces off the block, by the time the block makes its closest approach to the wall?
Assume that M >> m, and give your answers to leading order in m/M.

Homework Equations

http://www.luiseduardo.com.br/mechanics/conservationlaws/conservationlawsproblems_arquivos/image023.jpg

The Attempt at a Solution

Let the velocity of block and ball be v1 and v2 after collision. Applying conservation of momentum and law of restitution,

v_1 = \dfrac{M-m}{M+m} v_0 \\<br /> v_2 = \dfrac{2M}{M+m} v_0

Using approximations
v_1 = v_0 \\ v_2 = 2v_0

At distance of closest approach the velocity of bigger block should become zero. But I can't think how to find that distance as the velocity of bigger block is not constant and changes after every collision.

 

[dauto]

You're going to have to be a bit cleverer that that aren't you. Start by making a better choice of notation. Don't call the speeds after the first collision v₁ and v₂. Call them V₁ and v₁. In general Call the speeds after the ith collision V_i and v_i. Your job is, given V_i and v_i, find V_i+1 and v_i+1. Off course, you also need to keep track of L_i - the distance from the location of the ith collision and the wall. Also, don't be so fast at making hasty approximations.

Come back after you've done that.

 

[utkarshakash]

You're going to have to be a bit cleverer that that aren't you. Start by making a better choice of notation. Don't call the speeds after the first collision v₁ and v₂. Call them V₁ and v₁. In general Call the speeds after the ith collision V_i and v_i. Your job is, given V_i and v_i, find V_i+1 and v_i+1. Off course, you also need to keep track of L_i - the distance from the location of the ith collision and the wall. Also, don't be so fast at making hasty approximations.

Come back after you've done that.

V_{i+1} = \dfrac{M-m}{M+m} V_i - \dfrac{2m}{M+m} v_i

Let L_i be the distance from the point of ith collision to wall. By the time i+1 th collision occurs, let the block travel a distance Δx.
∴L_i+1 = L_i - Δx

In terms of L_i
L_{i+1} = L_i \left( 1- \dfrac{2V_i}{V_i+v_i} \right)

Am I correct so far?

 

[utkarshakash]

Please help me.

 

[SammyS]

V_{i+1} = \dfrac{M-m}{M+m} V_i - \dfrac{2m}{M+m} v_i

Let L_i be the distance from the point of ith collision to wall. By the time i+1 th collision occurs, let the block travel a distance Δx.
∴L_i+1 = L_i - Δx

In terms of L_i
L_{i+1} = L_i \left( 1- \dfrac{2V_i}{V_i+v_i} \right)

Am I correct so far?

What is your result for $\ v_{1+1} \ ?$

 

[utkarshakash]

What is your result for $\ v_{1+1} \ ?$

V_{1+1} = \dfrac{M-m}{M+m} v_0 - \dfrac{2m}{M+m} 2v_0

 

[SammyS]

V_{1+1} = \dfrac{M-m}{M+m} v_0 - \dfrac{2m}{M+m} 2v_0

I asked about $\ v_{1+1} \ ,$

not about $\ V_{1+1} \ .$

Or are you not taking dauto's suggestion for notation ?

 

[utkarshakash]

I asked about $\ v_{1+1} \ ,$

not about $\ V_{1+1} \ .$

Or are you not taking dauto's suggestion for notation ?

Sorry about that.

v_{i+1} = \left( \dfrac{2M}{M+m} \right) V_i + \left( \dfrac{M-m}{M+m} \right) v_i

If I put i=1

v_2 = \left( \dfrac{2M}{M+m} \right) V_1 + \left( \dfrac{M-m}{M+m} \right) v_1

 

[dauto]

Now you can start making approximations. You should try to simplify those expressions. may be define the parameter $\mu = m/M$ and expand those expressions in power series of that parameter keeping only the independent term and the linear term. Any quadratic or higher term may be dropped. That will help clear things up quite a bit while still preserving the essence of the problem. You should be able to get the correct order of magnitude for the solution but probably not much more than that, but it seems that's all the problem text asked you to do.

 

[utkarshakash]

Now you can start making approximations. You should try to simplify those expressions. may be define the parameter $\mu = m/M$ and expand those expressions in power series of that parameter keeping only the independent term and the linear term. Any quadratic or higher term may be dropped. That will help clear things up quite a bit while still preserving the essence of the problem. You should be able to get the correct order of magnitude for the solution but probably not much more than that, but it seems that's all the problem text asked you to do.

V_{i+1} = (1-2 \mu) V_i - 2 \mu (1- \mu) v_i \\<br /> v_{i+1} = 2(1- \mu ) V_i + (1-2 \mu) v_i

But I really don't see how does this help me. The problem asks me to find the closest distance.

 

[dauto]

One step at a time. The expession for the distance depends on the velocities, so you should solve the velocities first because their expressions do not depend on the distances. There is still one quadratic term in your expressions that may be dropped. Try writing the pair of equations in matrix form by defining the column vectors X_i = (V_i v_i)^T and express X_i+1 as a matrix product

X_i+1 = M X_i,

where the 2 by 2 matrix may be written as M = A + μB.

A and B are independent of μ.

Once you've done that calculate M², M³, and M⁴, and look for patterns that will allow you to find the general expression for Mⁿ.

Remember to drop any quadratic terms that arise as you do your calculation.

Once you have M_n you can easily find X_n = Mⁿ X₀.

That should get you busy for a while.

 

[dauto]

Alternatively you might find the expression for the momenta change δP_i = M(V_i+1-V_i) and δp_i = μM(v_i+1-v_i) and approximate

δV = dV/dn and δv = dv/dn,

and solve the differential equations.

 

[dauto]

OK, I solved the problem using both methods described above, The second method turned out to be easier and surprisingly accurate (I compared my answers with a numerical solution produced with a spread-sheet using M = 100000 m). You might want to go that way.

 

[utkarshakash]

One step at a time. The expession for the distance depends on the velocities, so you should solve the velocities first because their expressions do not depend on the distances. There is still one quadratic term in your expressions that may be dropped. Try writing the pair of equations in matrix form by defining the column vectors X_i = (V_i v_i)^T and express X_i+1 as a matrix product

X_i+1 = M X_i,

where the 2 by 2 matrix may be written as M = A + μB.

A and B are independent of μ.

Once you've done that calculate M², M³, and M⁴, and look for patterns that will allow you to find the general expression for Mⁿ.

Remember to drop any quadratic terms that arise as you do your calculation.

Once you have M_n you can easily find X_n = Mⁿ X₀.

That should get you busy for a while.

Transforming the equation in matrix notation, I get

\left( \begin{array}{c} V_{i+1} \\ v_{i+1} \end{array} \right) = \left( \begin{array}{cc} 1-2\mu &amp; -2\mu \\ 2(1-\mu) &amp; 1-3\mu \end{array} \right) \left( \begin{array}{c} V_{i} \\ v_{i} \end{array} \right)

M^2 = \left( \begin{array}{cc} 1-8\mu &amp; -4\mu \\ 2(2-6\mu) &amp; 1-8\mu \end{array} \right) \\ <br /> M^3 = \left( \begin{array}{cc} 1-18\mu &amp; -6\mu \\ 2(3-19\mu) &amp; 1-18\mu \end{array} \right)

But I can't see a definite pattern except that the matrix is symmetric and M₁₂ = 2ⁿ (-μ) .

[EDIT] Oops, I'm sorry. The matrix is not symmetric. Only the principal diagonal elements are equal.

 

[utkarshakash]

OK, I solved the problem using both methods described above, The second method turned out to be easier and surprisingly accurate (I compared my answers with a numerical solution produced with a spread-sheet using M = 100000 m). You might want to go that way.

Your second method is not clear to me. Can you please elaborate?

 

[dauto]

Transforming the equation in matrix notation, I get

\left( \begin{array}{c} V_{i+1} \\ v_{i+1} \end{array} \right) = \left( \begin{array}{cc} 1-2\mu &amp; -2\mu \\ 2(1-\mu) &amp; 1-3\mu \end{array} \right) \left( \begin{array}{c} V_{i} \\ v_{i} \end{array} \right)

M^2 = \left( \begin{array}{cc} 1-8\mu &amp; -4\mu \\ 2(2-6\mu) &amp; 1-8\mu \end{array} \right) \\ <br /> M^3 = \left( \begin{array}{cc} 1-18\mu &amp; -6\mu \\ 2(3-19\mu) &amp; 1-18\mu \end{array} \right)

But I can't see a definite pattern except that the matrix is symmetric and M₁₂ = 2ⁿ (-μ) .

[EDIT] Oops, I'm sorry. The matrix is not symmetric. Only the principal diagonal elements are equal.

There is a mistake in your matrix. The right bottom term is not correct, I don't think.

 

[dauto]

Your second method is not clear to me. Can you please elaborate?

You already have the expressions for V_i+1 and v_i+1. Find the momentum change δP_i = M δ V_i = M (V_i+1 - V_i) and similarly for δp_i.

Its possible to express
$$\delta P = \frac{\delta P}{1} = \frac{\delta P}{\delta n} \approx \frac{d P}{d n}$$
where n counts the collision number.

That way you find a pair of differential equations for dP/dn and dp/dn. This differential equations turn out to be easily solvable if you keep in mind That P >> p >> μP.

 

[utkarshakash]

There is a mistake in your matrix. The right bottom term is not correct, I don't think.

That's a typo. It should be 1-2μ

 

[dauto]

That's a typo. It should be 1-2μ

OK.

Don't you see the pattern? Only the bottom left term is difficult (and you won't need the to get a zeroth order solution). all the other ones either grow linearly or quadratically.

 

[utkarshakash]

You already have the expressions for V_i+1 and v_i+1. Find the momentum change δP_i = M δ V_i = M (V_i+1 - V_i) and similarly for δp_i.

Its possible to express
$$\delta P = \frac{\delta P}{1} = \frac{\delta P}{\delta n} \approx \frac{d P}{d n}$$
where n counts the collision number.

That way you find a pair of differential equations for dP/dn and dp/dn. This differential equations turn out to be easily solvable if you keep in mind That P >> p >> μP.

I get these two equations.

\dfrac{dP}{dn} = -2(\mu P + p) \\<br /> \dfrac{dp}{dn} = 2\mu (P-p)

If I follow your approximations, the first equation changes to
\dfrac{dP}{dn} = -2p

and second becomes

\dfrac{dp}{dn} = 2\mu P

If I divide both equations I get something like this

\mu P dP + pdp = 0

Integrating both sides

\mu \dfrac{P^2}{2} + \dfrac{p^2}{2} + c=0

How to find the constant of integration?

 

[dauto]

Find the constant of integration by plugging in the initial values for p and and P. You're almost there.

 

[utkarshakash]

Find the constant of integration by plugging in the initial values for p and and P. You're almost there.

I get
c= - \dfrac{\mu M^2 v_0 ^2}{2}

Is this correct? If yes, what should be my next step?

 

[dauto]

Your next step would be solving that last equation for p and plugging it back in the expression for dP/dn. But there is a simpler method. Differentiate the expression for dP/dn with respect to n to find an expression for d²P/dn² and simplify it by using the expression for dp/dn. You should get to a very familiar differential equation.

 

[utkarshakash]

Your next step would be solving that last equation for p and plugging it back in the expression for dP/dn. But there is a simpler method. Differentiate the expression for dP/dn with respect to n to find an expression for d²P/dn² and simplify it by using the expression for dp/dn. You should get to a very familiar differential equation.

\dfrac{d^2 P}{dn^2} = -4 \mu P

The solution for this differential equation is
P=P_0 \cos (2 \sqrt{\mu} n)

where P_0 = Mv_o

Am I on the right track?

[EDIT] I arrived at the correct answer for second part. But what about first? How to get that?

 

[dauto]

What second part? Your solution is correct. To find n now you just have to remember that after the last collision the particle m will be either at rest or almost. All the energy is transferred back to the particle M which will have speed V_n = - V₀. Plug that in your solution and find n.

 

[SammyS]

What second part? Your solution is correct. To find n now you just have to remember that after the last collision the particle m will be either at rest or almost. All the energy is transferred back to the particle M which will have speed V_n = - V₀. Plug that in your solution and find n.

If OP wants to find n for the block (mass M) when it's closest to the wall, then V must be zero.

When V < 0, the block is moving away from the wall.

 

[dauto]

If OP wants to find n for the block (mass M) when it's closest to the wall, then V must be zero.

When V < 0, the block is moving away from the wall.

I thought he wanted the total number of collisions.

 

[SammyS]

I thought he wanted the total number of collisions.

The OP states:

"a) How close does the block come to the wall ?
b) How many times does the ball bounces off the block, by the time the block makes its closest approach to the wall?"​

Without working through all of this, I'm pretty sure that the value of n asked for will be 1/2 of what you were trying to get.

 

[dauto]

The OP states:

"a) How close does the block come to the wall ?
b) How many times does the ball bounces off the block, by the time the block makes its closest approach to the wall?"​

Without working through all of this, I'm pretty sure that the value of n asked for will be 1/2 of what you were trying to get.

OK. Yes the difference is just a factor of 2 because the second half of the motion looks just like the first but in reverse.

 

[utkarshakash]

What second part? Your solution is correct. To find n now you just have to remember that after the last collision the particle m will be either at rest or almost. All the energy is transferred back to the particle M which will have speed V_n = - V₀. Plug that in your solution and find n.

By second part I meant b) which asks me to find number of collisions. Plugging P=0 in the equation gives me n= \dfrac{\pi}{4} \sqrt{\dfrac{M}{m}} which is correct. Now I have to find the distance of closest approach. Do I need to derive some other equations? If yes, please give me some hints.

 

[dauto]

Try converting your expression for L_n (Post #3) into a differential equation dL/dn.

 

[utkarshakash]

Try converting your expression for L_n (Post #3) into a differential equation dL/dn.

OK. I could transform it into something like this:

\dfrac{dL}{dn} = \left( \dfrac{-2 \mu P }{p} \right) L

I followed a similar approach to find p(n) and got

p = 2mv_0 \sin (2 \sqrt{\mu}n)

If I plug the values for P and p in the DE, I get

\dfrac{dL}{L} = \dfrac{-\cos(2\sqrt{\mu}n)}{\sin(2\sqrt{\mu}n)} dn

Integrating both sides,

\log L = \dfrac{-1}{2 \sqrt{\mu}} \log \left( \sin 2 \sqrt{\mu} n \right) + c

For n=0, L=L. But then the logarithmic term tends to infinity. What's wrong here?

 

[dauto]

I'm getting

(1/L) dL/dn = -2P/(P+p/μ).

Try integrating that instead.

 

[utkarshakash]

I'm getting

(1/L) dL/dn = -2P/(P+p/μ).

Try integrating that instead.

But if you further approximate it, you will arrive at the same equation as mine.

 

[utkarshakash]

I'm getting

(1/L) dL/dn = -2P/(P+p/μ).

Try integrating that instead.

\dfrac{dL}{L} = \dfrac{-2 \cos \lambda n}{\cos \lambda n + 2 \sin \lambda n} dn \\<br /> \lambda = 2 \sqrt{\mu}

Assuming sinλn = t, the RHS transforms to

\dfrac{-2}{\lambda (\sqrt{1-t^2}+2t)} dt

I'm stuck here. Please help.

 

[dauto]

\dfrac{dL}{L} = \dfrac{-2 \cos \lambda n}{\cos \lambda n + 2 \sin \lambda n} dn \\<br /> \lambda = 2 \sqrt{\mu}

I'm getting
\dfrac{dL}{L} = \dfrac{-2 \cos \lambda n}{\cos \lambda n + \mu^{-1/2} \sin \lambda n} dn

 

[utkarshakash]

I'm getting
\dfrac{dL}{L} = \dfrac{-2 \cos \lambda n}{\cos \lambda n + \mu^{-1/2} \sin \lambda n} dn

I'm posting my solution step-by-step.

\dfrac{dL}{L} = \dfrac{-2 Mv_0 \cos \lambda n}{Mv_0 \cos \lambda n + (2 mv_0/ \mu) \sin \lambda n} dn

If I substitute the value of μ, I get my equation. What's wrong with it?

 

[dauto]

I don't think that's correct either. I'll have to go now. I'll have more time later today

 

[utkarshakash]

I don't think that's correct either. I'll have to go now. I'll have more time later today

OK. Please take your time.

 

[dauto]

OK. Here's what I got

L_i+1=L_i(1−2V_i/(V_ii+v))

(1/L) dL/dn = -2V/(V+v)

(1/L) dL/dn = -2P/(P+p/μ)

(1/L) dL/dn = -2cos(λn)/[cos(λn)+μ^1/2sin(λn)/μ]

(1/L) dL/dn = -2cos(λn)/[cos(λn)+μ^-1/2sin(λn)],

where I used

P=P₀cos(λn)

and

p=μ^1/2P₀sin(λn)

 

[utkarshakash]

OK. Here's what I got

L_i+1=L_i(1−2V_i/(V_ii+v))

(1/L) dL/dn = -2V/(V+v)

(1/L) dL/dn = -2P/(P+p/μ)

(1/L) dL/dn = -2cos(λn)/[cos(λn)+μ^1/2sin(λn)/μ]

(1/L) dL/dn = -2cos(λn)/[cos(λn)+μ^-1/2sin(λn)],

where I used

P=P₀cos(λn)

and

p=μ^1/2P₀sin(λn)

Looks like there was a mistake in my equation for p(n). OK, how do you integrate it then? Substitutions aren't working here. What else should I try?

 

[dauto]

Here's the trick

\frac{cos\theta}{cos\theta+A sin\theta} = \frac{cos\theta}{cos\theta+A sin\theta} +\xi -\xi,
where \xi is to be chosen later to our own convenience.
\frac{cos\theta d\theta}{cos\theta+A sin\theta} = \frac{[(1+\xi)cos\theta+\xi A sin\theta]d\theta}{cos\theta+A sin\theta} -\xi d\theta,
We are planning on making the change of variables
u=cos\theta+A sin\theta, and du=(-sin\theta+A cos\theta)d\theta,
So we want to chose \xi in such a way that the numerator [(1+\xi)cos\theta+\xi A sin\theta]d\theta is proportional to du
[(1+\xi)cos\theta+\xi A sin\theta]d\theta = \eta du = \eta (-sin\theta+A cos\theta)d\theta which gives us two equations
(1+\xi) = \eta A and
\xi A = - \eta The set is easily solved to \xi = -\frac{1}{1+A^2} and \eta=\frac{A}{1+A^2}. Now that we know what \xi and \eta are, we have
\frac{cos\theta d\theta}{cos\theta+A sin\theta} = \frac{[(1+\xi)cos\theta+\xi A sin\theta]d\theta}{cos\theta+A sin\theta} -\xi d\theta = \frac{\eta du}{u} - \xi d\theta.
Integration now is a trivial matter.

 

[utkarshakash]

Here's the trick

\frac{cos\theta}{cos\theta+A sin\theta} = \frac{cos\theta}{cos\theta+A sin\theta} +\xi -\xi,
where \xi is to be chosen later to our own convenience.
\frac{cos\theta d\theta}{cos\theta+A sin\theta} = \frac{[(1+\xi)cos\theta+\xi A sin\theta]d\theta}{cos\theta+A sin\theta} -\xi d\theta,
We are planning on making the change of variables
u=cos\theta+A sin\theta, and du=(-sin\theta+A cos\theta)d\theta,
So we want to chose \xi in such a way that the numerator [(1+\xi)cos\theta+\xi A sin\theta]d\theta is proportional to du
[(1+\xi)cos\theta+\xi A sin\theta]d\theta = \eta du = \eta (-sin\theta+A cos\theta)d\theta which gives us two equations
(1+\xi) = \eta A and
\xi A = - \eta The set is easily solved to \xi = -\frac{1}{1+A^2} and \eta=\frac{A}{1+A^2}. Now that we know what \xi and \eta are, we have
\frac{cos\theta d\theta}{cos\theta+A sin\theta} = \frac{[(1+\xi)cos\theta+\xi A sin\theta]d\theta}{cos\theta+A sin\theta} -\xi d\theta = \frac{\eta du}{u} - \xi d\theta.
Integration now is a trivial matter.

Integrating the equation I get

\log L = \left( \dfrac{A}{1+A^2} \log (\cos \theta + A \sin \theta) + \dfrac{\theta}{1+A^2} \right) (-2) + \log L_0

Substituting n=\frac{\pi}{4 \sqrt{\mu}}

\log L = \dfrac{\sqrt{\mu}}{1+\mu} \log \mu - \dfrac{\mu \pi}{1+\mu} + \log L_0

But this is not even close to the correct answer. Are you getting the same result?

 

[dauto]

Integrating the equation I get

\log L = \left( \dfrac{A}{1+A^2} \log (\cos \theta + A \sin \theta) + \dfrac{\theta}{1+A^2} \right) (-2) + \log L_0

Substituting n=\frac{\pi}{4 \sqrt{\mu}}

\log L = \dfrac{\sqrt{\mu}}{1+\mu} \log \mu - \dfrac{\mu \pi}{1+\mu} + \log L_0

But this is not even close to the correct answer. Are you getting the same result?

I got

log (L/L_0) =-\left[\frac{log [\mu^{1/2}cos\theta +sin\theta] + \mu^{1/2}\theta}{1+\mu} \right]_{\theta=0}^{\pi/2}

What do you think the correct answer should be?

 

[utkarshakash]

I got

log (L/L_0) =-\left[\frac{log [\mu^{1/2}cos\theta +sin\theta] + \mu^{1/2}\theta}{1+\mu} \right]_{\theta=0}^{\pi/2}

What do you think the correct answer should be?

The correct answer is L₀ √μ. Your equation seems to give a different answer.

 

[dauto]

The correct answer is L₀ √μ. Your equation seems to give a different answer.

Look closely. There are a couple of terms that are negligible. One of the terms in the numerator may be dropped. And the term μ^1/2 in the numerator is also negligible. It simplifies as

log (L/L_0) =-\left[log [\mu^{1/2}cos\theta +sin\theta] \right]_{\theta=0}^{\pi/2}

Do you see it now?

 

[TSny]

Here is another approach to part (a). In post #3 you have $$L_{i+1} = \frac{v_i-V_i}{v_i+V_i}L_i$$
So, in particular,

$$L_2 = \frac{v_1-V_1}{v_1+V_1}L_1$$ and $$L_3= \frac{v_2-V_2}{v_2+V_2}L_2 = \frac{v_2-V_2}{v_2+V_2} \frac{v_1-V_1}{v_1+V_1}L_1$$

For an elastic collision, the relative speed of approach before the collision equals the relative speed of separation after the collision.

For example, at the first collision

$V_0 =v1-V_1$, which means $L_2 = \frac{V_0}{v_1+V_1}L_1$

For the second collision

$v_1+V_1=v2-V_2$ which can be used to show $L_3= \frac{V_0}{v_2+V_2}L_1$.

Generalize this to write $L_N$ in terms of $L_1, v_{N-1}, V_{N-1}$ and $V_0$.

See if you can apply this to the case where $\small N -1$ is the index of the collision for which $\small M$ essentially comes to rest.

 

[haruspex]

This is an intriguing question. I wasn't sure what approximation would be reasonable, so I set out to solve it exactly to start with - looks like I did what TSny suggests. I took u_n, v_n to be the speeds, to the right, of M, m respectively, just before each time they collide (so the v_n are negative). I got $\left[\stackrel{u_n}{v_n}\right] = \left(\frac{M-m}{M+m}\right)^nPD^nP^{-1}\left[^u_0\right]$, where u is the initial speed of M, and PDP^-1 is the diagonalisation of $A = \left[\begin{array}{cc} 1 & b \\ a & 1 \end{array} \right]$, $a = \frac{2m}{M-m}$, $b = -\frac{2M}{M-m}$. For u_n = 0, this gave
$(1+\sqrt{ab})^n = -(1-\sqrt{ab})^n$. (Note that b is negative.) I.e. $n = \frac{i \pi}{\ln(\frac{1+i\mu}{1-i\mu})}$, where $\mu = \frac{2\sqrt{Mm}}{M-m}$. For M>>m this reduces to $n = \frac{\pi}{4}\sqrt{\frac Mm}$.
I'm sure this was all a lot harder than picking the right approximation up front, but it was interesting.
Applying the same approach to the distance, I got this dreadful formula, where x_n is the remaining distance at the nth collision:
$x_{n+1} = x_n\frac{(1+i\rho)^n(1-i\alpha)-(1-i\rho)^n(1+i\alpha)}{(1+i\rho)^n(1+i\alpha)-(1-i\rho)^n(1-i\alpha)}$, where $\rho = 2\frac{\sqrt{Mm}}{M-m}, \alpha = \sqrt{\frac mM}$.
For M>>m, $\rho ≈ 2\alpha << 1$, reducing it to $\frac{\tan(2n\alpha)-\alpha}{\tan(2n\alpha)+\alpha}$, further approximating to $\frac{2n-1}{2n+1}$. Thus x_n ≈ x₁/(2n-1). Does that give the right answer?

 

[utkarshakash]

Look closely. There are a couple of terms that are negligible. One of the terms in the numerator may be dropped. And the term μ^1/2 in the numerator is also negligible. It simplifies as

log (L/L_0) =-\left[log [\mu^{1/2}cos\theta +sin\theta] \right]_{\theta=0}^{\pi/2}

Do you see it now?

Yes. I got the correct answer. Thanks for helping !

 

[utkarshakash]

This is an intriguing question. I wasn't sure what approximation would be reasonable, so I set out to solve it exactly to start with - looks like I did what TSny suggests. I took u_n, v_n to be the speeds, to the right, of M, m respectively, just before each time they collide (so the v_n are negative). I got $\left[\stackrel{u_n}{v_n}\right] = \left(\frac{M-m}{M+m}\right)^nPD^nP^{-1}\left[^u_0\right]$, where u is the initial speed of M, and PDP^-1 is the diagonalisation of $A = \left[\begin{array}{cc} 1 & b \\ a & 1 \end{array} \right]$, $a = \frac{2m}{M-m}$, $b = -\frac{2M}{M-m}$. For u_n = 0, this gave
$(1+\sqrt{ab})^n = -(1-\sqrt{ab})^n$. (Note that b is negative.) I.e. $n = \frac{i \pi}{\ln(\frac{1+i\mu}{1-i\mu})}$, where $\mu = \frac{2\sqrt{Mm}}{M-m}$. For M>>m this reduces to $n = \frac{\pi}{4}\sqrt{\frac Mm}$.
I'm sure this was all a lot harder than picking the right approximation up front, but it was interesting.
Applying the same approach to the distance, I got this dreadful formula, where x_n is the remaining distance at the nth collision:
$x_{n+1} = x_n\frac{(1+i\rho)^n(1-i\alpha)-(1-i\rho)^n(1+i\alpha)}{(1+i\rho)^n(1+i\alpha)-(1-i\rho)^n(1-i\alpha)}$, where $\rho = 2\frac{\sqrt{Mm}}{M-m}, \alpha = \sqrt{\frac mM}$.
For M>>m, $\rho ≈ 2\alpha << 1$, reducing it to $\frac{\tan(2n\alpha)-\alpha}{\tan(2n\alpha)+\alpha}$, further approximating to $\frac{2n-1}{2n+1}$. Thus x_n ≈ x₁/(2n-1). Does that give the right answer?

This was indeed a difficult problem, much difficult than what I assumed. Lots of approximations are required. Your method seems complicated to me. But, anyways, thanks for posting your solution. Knowing different methods to solve a problem is certainly beneficial.