Elastic collision between ball and block

  • #1
utkarshakash
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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 [Broken]

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,

[itex]v_1 = \dfrac{M-m}{M+m} v_0 \\
v_2 = \dfrac{2M}{M+m} v_0 [/itex]

Using approximations
[itex]v_1 = v_0 \\ v_2 = 2v_0 [/itex]

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.
 
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Answers and Replies

  • #2
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You're gonna 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 Vi and vi. Your job is, given Vi and vi, find Vi+1 and vi+1. Off course, you also need to keep track of Li - 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.
 
  • #3
utkarshakash
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You're gonna 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 Vi and vi. Your job is, given Vi and vi, find Vi+1 and vi+1. Off course, you also need to keep track of Li - 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.
[itex]V_{i+1} = \dfrac{M-m}{M+m} V_i - \dfrac{2m}{M+m} v_i [/itex]

Let Li 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.
∴Li+1 = Li - Δx

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

Am I correct so far?
 
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  • #4
utkarshakash
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Please help me.
 
  • #5
SammyS
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[itex]V_{i+1} = \dfrac{M-m}{M+m} V_i - \dfrac{2m}{M+m} v_i [/itex]

Let Li 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.
∴Li+1 = Li - Δx

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

Am I correct so far?
What is your result for ##\ v_{1+1} \ ?##
 
  • #6
utkarshakash
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What is your result for ##\ v_{1+1} \ ?##
[itex]V_{1+1} = \dfrac{M-m}{M+m} v_0 - \dfrac{2m}{M+m} 2v_0[/itex]
 
  • #7
SammyS
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[itex]V_{1+1} = \dfrac{M-m}{M+m} v_0 - \dfrac{2m}{M+m} 2v_0[/itex]
I asked about ## \ v_{1+1} \ , ##

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

Or are you not taking dauto's suggestion for notation ?
 
  • #8
utkarshakash
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I asked about ## \ v_{1+1} \ , ##

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

Or are you not taking dauto's suggestion for notation ?
Sorry about that.

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

If I put i=1

[itex]v_2 = \left( \dfrac{2M}{M+m} \right) V_1 + \left( \dfrac{M-m}{M+m} \right) v_1 [/itex]
 
  • #9
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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.
 
  • #10
utkarshakash
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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.
[itex]V_{i+1} = (1-2 \mu) V_i - 2 \mu (1- \mu) v_i \\
v_{i+1} = 2(1- \mu ) V_i + (1-2 \mu) v_i [/itex]

But I really don't see how does this help me. The problem asks me to find the closest distance.
 
  • #11
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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 Xi = (Vi v_i)T and express Xi+1 as a matrix product

Xi+1 = M Xi,

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 Mn you can easily find Xn = Mn X0.

That should get you busy for a while.
 
  • #12
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Alternatively you might find the expression for the momenta change δPi = M(Vi+1-Vi) and δpi = μM(vi+1-vi) and approximate

δV = dV/dn and δv = dv/dn,

and solve the differential equations.
 
  • #13
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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.
 
  • #14
utkarshakash
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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 Xi = (Vi v_i)T and express Xi+1 as a matrix product

Xi+1 = M Xi,

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 Mn you can easily find Xn = Mn X0.

That should get you busy for a while.
Transforming the equation in matrix notation, I get

[itex]\left( \begin{array}{c} V_{i+1} \\ v_{i+1} \end{array} \right) = \left( \begin{array}{cc} 1-2\mu & -2\mu \\ 2(1-\mu) & 1-3\mu \end{array} \right) \left( \begin{array}{c} V_{i} \\ v_{i} \end{array} \right) [/itex]

[itex]M^2 = \left( \begin{array}{cc} 1-8\mu & -4\mu \\ 2(2-6\mu) & 1-8\mu \end{array} \right) \\
M^3 = \left( \begin{array}{cc} 1-18\mu & -6\mu \\ 2(3-19\mu) & 1-18\mu \end{array} \right) [/itex]

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.
 
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  • #15
utkarshakash
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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?
 
  • #16
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Transforming the equation in matrix notation, I get

[itex]\left( \begin{array}{c} V_{i+1} \\ v_{i+1} \end{array} \right) = \left( \begin{array}{cc} 1-2\mu & -2\mu \\ 2(1-\mu) & 1-3\mu \end{array} \right) \left( \begin{array}{c} V_{i} \\ v_{i} \end{array} \right) [/itex]

[itex]M^2 = \left( \begin{array}{cc} 1-8\mu & -4\mu \\ 2(2-6\mu) & 1-8\mu \end{array} \right) \\
M^3 = \left( \begin{array}{cc} 1-18\mu & -6\mu \\ 2(3-19\mu) & 1-18\mu \end{array} \right) [/itex]

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.
 
  • #17
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Your second method is not clear to me. Can you please elaborate?
You already have the expressions for Vi+1 and vi+1. Find the momentum change δPi = M δ Vi = M (Vi+1 - Vi) and similarly for δpi.

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.
 
  • #18
utkarshakash
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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
 
  • #19
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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.
 
  • #20
utkarshakash
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You already have the expressions for Vi+1 and vi+1. Find the momentum change δPi = M δ Vi = M (Vi+1 - Vi) and similarly for δpi.

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.

[itex] \dfrac{dP}{dn} = -2(\mu P + p) \\
\dfrac{dp}{dn} = 2\mu (P-p) [/itex]

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

and second becomes

[itex]\dfrac{dp}{dn} = 2\mu P [/itex]

If I divide both equations I get something like this

[itex]\mu P dP + pdp = 0 [/itex]

Integrating both sides

[itex]\mu \dfrac{P^2}{2} + \dfrac{p^2}{2} + c=0[/itex]

How to find the constant of integration?
 
  • #21
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Find the constant of integration by plugging in the initial values for p and and P. You're almost there.
 
  • #22
utkarshakash
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Find the constant of integration by plugging in the initial values for p and and P. You're almost there.
I get
[itex] c= - \dfrac{\mu M^2 v_0 ^2}{2} [/itex]

Is this correct? If yes, what should be my next step?
 
  • #23
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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.
 
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  • #24
utkarshakash
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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.
[itex]\dfrac{d^2 P}{dn^2} = -4 \mu P [/itex]

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

where [itex]P_0 = Mv_o [/itex]

Am I on the right track?

[EDIT] I arrived at the correct answer for second part. But what about first? How to get that?
 
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  • #25
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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 Vn = - V0. Plug that in your solution and find n.
 

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