# Homework Help: Velocity of spheres after collision

1. May 13, 2013

### Saitama

1. The problem statement, all variables and given/known data
Three spheres, each of mass m, can slide freely on a frictionless, horizontal surface. Spheres A and B are attached to an inextensible inelastic cord of length / and are at rest in the position shown when sphere B is struck directly by sphere C which is moving to the right with a velocity $v_o$. Knowing that the cord is taut when sphere B is struck by sphere C and assuming perfectly elastic impact between B and C, determine the velocity of each sphere immediately after impact.

2. Relevant equations

3. The attempt at a solution
Let the velocity of A, B and C after collision be $v_A$, $v_B$ and $v_C$ respectively.

Conserving linear momentum,
$mv_0=mv_A+mv_B+mv_C$

As the string is inextensible, the components of velocity of A and B along the string should be equal i.e $v_A\sin\theta=v_B\sin\theta \Rightarrow v_A=v_B$ but this is wrong because as per the answer key the velocities of A and B are different.

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• ###### after collision.png
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2. May 13, 2013

### consciousness

Hmm interesting question I think that the problem lies in the directions you have decided for the velocities of A and B. How can the velocity of A have a component perpendicular to the string if the tension can only provide impulse to it? Also for B the tension will give an impulse and the velocity vector wont be parallel to Vo. You have to use vector equations for conservation of linear momentum.

3. May 13, 2013

### haruspex

It's a string, not a rod.

4. May 13, 2013

### Saitama

Let the velocity of B in horizontal and vertical direction be $v_{Bx}$ and $v_{By}$ and similarly for A, $v_{Ax}$ and $v_{Ay}$

Conserving momentum in the horizontal direction,
$mv_o=mv_{Ax}+mv_{Bx}+mv_C$

Conserving momentum in the vertical direction,
$0=mv_{Ay}+mv_{By} \Rightarrow v_{By}=-v_{Ay}$

I still need more equations to solve the problem. I can only think of conserving energy.
$$\frac{1}{2}mv_o^2=\frac{1}{2}mv_C^2+\frac{1}{2}m(v_{Ax}^2+v_{Ay}^2)+ \frac {1}{2}m(v_{Bx}^2+v_{By}^2)$$
$$\Rightarrow v_o^2=v_C^2+v_{Ax}^2+2v_{Ay}^2+v_{Bx}^2$$

Is it right to assume that C moves only in horizontal direction even after the collision?
As the collision is elastic, $v_o=v_{Bx}-v_{C}$.

Am I doing this right? I still need more equations.

5. May 13, 2013

### consciousness

One equation is to equate the velocities along the string. As it is an elastic collision don't use conservation of energy just use coefficient of restitution as energy conservation gives complicated equations and the same result as coefficient of restitution.

EDIT:

And yes C will move in horizontal direction only as no vertical impulse acts on it. We have 5 unknowns and 5 equations but I don't think we will get the answer as conserving energy isn't a "unique equation" and using coefficient of restitution is the same thing.

Last edited: May 13, 2013
6. May 13, 2013

### Saitama

I think I cannot do that. Check haruspex's post.

7. May 13, 2013

### consciousness

Well I thought that since the string remains tight after the collision the components along it would be equal. As for how it remains tight I don't have any mathematical reason but I cannot imagine practically how the the string would become loose after the collision.

8. May 14, 2013

### haruspex

Ive come around to your view. Initially, at least, it will remain taut. And since it is taut before the collision, you can treat the collision as work conserving.
No, that ignores the role of A. Consider KE of whole system.

9. May 14, 2013

### Saitama

Does this mean I can conserve energy in this problem?

And can I equate the components of velocity of A and B along the string?

10. May 14, 2013

### haruspex

Yes, I think you can justify both of those.

11. May 14, 2013

### Saitama

I wrote the equation for conservation of energy in post #4.

(I have assumed that x component of velocity both A and B is in the right direction and y-component in vertically upward direction)
Equating the components of velocity along the string,
$$v_{By}\cos \theta+v_{Bx}\sin \theta=v_{Ax}\sin \theta+v_{Ay}\cos \theta$$
$$(v_{Bx}-v_{Ax})\sin \theta=2v_{Ay}\cos \theta$$

From the figure, $\sin \theta=1/2$ and $\cos \theta=\sqrt{3}/{2}$
$$\Rightarrow v_{Bx}-v_{Ax}=2\sqrt{3}v_{Ay}$$

That's going to be really dirty. :yuck:

Can you give me a few ideas how should I begin to solve this? I have got enough equations I think but substituting expressions in the energy equation gives so many squared terms. :uhh:

Last edited: May 14, 2013
12. May 14, 2013

### haruspex

When you substitute in the energy equation, you should find a common factor (a velocity) that cancels out, reducing it to a linear form.

13. May 14, 2013

### Saitama

I had $v_o=v_{Ax}+v_{Bx}+v_C \Rightarrow v_C=v_o-(v_{Ax}+v_{Bx})$
Also $v_{Bx}-v_{Ax}/2\sqrt{3}=v_{Ay}$

Substituting them in the energy equation,
$$v_o^2=v_o^2+v_{Ax}^2+v_{Bx}^2-2v_o(v_{Ax}+v_{Bx})+\frac{v_{Bx}^2+v_{Ax}^2-2v_{Ax}v_{Bx}}{6}+v_{Bx}^2$$
$$\Rightarrow 12v_o(v_{Ax}+v_{Bx})=7v_{Ax}^2+19v_{Bx}^2-2v_{Ax}v_{Bx}$$
I don't see how can I simplify that.

14. May 14, 2013

### ehild

One equation is still missing: conservation of angular momentum.

ehild

15. May 14, 2013

### Saitama

About what point should I do this?

16. May 14, 2013

### ehild

I would do it around the point above C, in the horizontal line between A and B. But A is good, too.

Last edited: May 14, 2013
17. May 14, 2013

### Saitama

I am assuming the point X shown in attachment.

$$0=\frac{mv_{By}\ell}{2}+\frac{mv_{Ax}\sqrt{3}\ell}{2}$$
$$\Rightarrow v_{By}=-\sqrt{3}v_{Ax}$$

Correct?

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Last edited: May 14, 2013
18. May 14, 2013

### ehild

correct

19. May 14, 2013

### Saitama

I further solved it but I am not able to reach the right answer.

I had $v_{Bx}-v_{Ax}x=2\sqrt{3}v_{Ay}$. As $v_{By}=-\sqrt{3}v_{Ax}=-v_{Ay}$

$$v_{Bx}-v_{Ax}=6v_{Ax} \Rightarrow v_{Bx}=7v_{Ax}$$

From the energy equation:
$$12v_o(v_{Ax}+v_{Bx})=7v_{Ax}^2+13v_{Bx}^2-2v_{Ax}v_{Bx}$$
$$\Rightarrow 12v_o(8v_{Ax})=(7+13\times 49-14)v_{Ax}^2$$
$$\Rightarrow v_{Ax}=\frac{16v_o}{105}$$

$$v_A=\sqrt{v_{Ax}^2+v_{Ay}^2}=2v_{Ax}=\frac{32v_o}{105}$$

But this is wrong.

20. May 14, 2013

### ehild

You might have an error in the energy equation. Check.

ehild

21. May 14, 2013

### Saitama

I did find a mistake in the energy equation [strike]but I still do not end up with the right answer[/strike] .

I will make the equations again.
I had $v_C=v_o-(v_{Ax}+v_{Bx})=v_o-8v_{Ax}$
$\Rightarrow v_C^2=v_o^2+64v_{Ax}^2-16v_ov_{Ax}$

From the energy equation,
$$v_o^2=v_C^2+v_{Ax}^2+2v_{Ay}^2+v_{Bx}^2$$
$$\Rightarrow v_o^2=v_o^2+64v_{Ax}^2-16v_ov_{Ax}+v_{Ax}^2+2(\sqrt{3}v_{Ax})^2+49v_{Ax}^2$$

Solving this, I get $v_{Ax}=2v_o/15$ and therefore $v_{A}=2v_{Ax}=4v_o/15$ which is correct as per the answer key.

$$v_C=v_o-8v_{Ax}=v_o-\frac{16}{15}v_o=\frac{-v_o}{15}$$

I have $v_{Bx}=7v_{Ax}=14v_o/15$ and $v_{By}=-\sqrt{3}v_{Ax}=-2\sqrt{3}v_o/15$
Hence,
$$v_B=\sqrt{v_{Bx}^2+v_{By}^2}=\frac{v_o}{15}\sqrt{196+12}=\frac{\sqrt{208}v_o}{15}$$

Thank you ehild and haruspex!

Last edited: May 14, 2013
22. Nov 8, 2017

### Arnav jamale

What should I do if collision was head on inelastic

23. Nov 8, 2017

### Arnav jamale

24. Nov 8, 2017

### haruspex

I assume you mean completely inelastic. What can you think of? What equation would you have instead?

25. Nov 8, 2017

### Arnav jamale

Conservation of linear momentum but we can't conserve kinetic energy