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.