# Collision of a particle with a thin rod

1. Jan 14, 2015

### Satvik Pandey

1. The problem statement, all variables and given/known data
A thin rod AB of mass $m_{1}$ and of length $l$ is placed on a smooth horizontal ground. A point sized particle of mass $m_{2}$ having a velocity $v$ making an angle $\theta$with the line perpendicular to $AB$ collides with the rod at point $A$.

Assuming the collision to be elastic find $\theta$ such that the point sized particle again collides with the rod at point $B$.

2. Relevant equations

3. The attempt at a solution

T
he image was provided with the question.

When two things collide equal and opposite impulses acts along the direction perpendicular to the contact surface. So in this situation if the particle collides with rod at point A then the impulse which acts on the rod passes through the CoM of the rod. So no impulsive torque acts on the rod about its CoM. So the rod will not rotate. And without rotating, it is not possible for the particle to collide again at point B.
So, how this can be solved?

2. Jan 14, 2015

### phinds

You might want to rethink this. Do you understand force vectors?

3. Jan 14, 2015

### ehild

No, why do you think so? The particle collides with the side of the rod near the end, not with the cross-section.

4. Jan 15, 2015

### Satvik Pandey

OK. But it is given that we have to assume that the particle is point size and the rod is thin. So if that particle collide at the $edge$ we have to assume that edge to be point (because rod is very thin). So how common normal direction can be drawn perpendicular to the contact surface. So I have assume that it collides little bit to the right of the edge.

I think velocity of the CoM of the rod will be in vertical direction because the impulse acts in vertical direction only.
As the collision is elastic so $e=1$.
So we can equate the Velocity of approach and velocity of separation in common normal direction.

$V_{app}=v cos\theta$ and $V_{sep}=V+\frac{l\omega}{2}+v_{0} cos\alpha$ ($\alpha$ is the angle between $v_{0}$ and the common normal direction.

As no force in common tangent direction so $v sin\theta=v_{0} sin\alpha$

Impulse acting on the particle is equal ti its change in velocity in the common normal direction. right?

So $J=\triangle V$

So $J=-m_{1}(v_{0}cos\alpha+vcos\theta)$ (Considering downward direction as -ve).

The same impulse acts on the rod.

${ \tau }_{ imp }=\frac{l}{2}J$ (${ \tau }_{ imp }$ is the impulsive torque)

This impulsive torque is equal to the change in the angular momentum of the rod. right?

So $\frac{l}{2}J=L_{f}$

or $\frac{l}{2}J=I\omega$

Is the impulse due to collision is equal to the change in the momentum of the CoM of the rod?

If yes then $J=MV$

Am I right till here?

5. Jan 15, 2015

### phinds

Really?

6. Jan 15, 2015

### Satvik Pandey

If that particle collide at the edge we have to assume that edge to be point (because rod is very thin). So how common normal direction can be drawn perpendicular to the contact surface. So I have assume that it collides little bit to the right of the edge. Is it fine to do so?

During collision equal and opposite impulse acts in the common normal direction. So impulse acts in the vertical direction.

7. Jan 15, 2015

### phinds

You are treating the point of collision as though it was not part of the rod. No, it is not fine to do so.

8. Jan 15, 2015

### Satvik Pandey

I don't get you.In the diagram I have shown that the particle collides with the rod so the point of collision lies on the rod.

9. Jan 15, 2015

### phinds

Yeah, I expressed that badly. What I'm trying to point out to you is that you have to treat the rod as a macro object. You hit the rod at an angle and then you are totally ignoring the angle by somehow deciding that it doesn't matter and that you can just throw away the horizontal component of the strike force. That's why I asked you in post #2 if you understand force vectors.

10. Jan 15, 2015

### Satvik Pandey

So are trying to say that I am neglecting the horizontal component of impulse. I have a confusion. During collision equal and opposite impulse acts in common normal direction.right? Isn't this true for macro object?

11. Jan 15, 2015

### ehild

Assuming the collision instantaneous, the tangential force (friction) between the rod and particle can be ignored.
The rod can be considered very thin, practically one-dimensional, like a line segment, and the particle arrives at the line segment very near of the end, and rebounds from it. You can take the force of interaction perpendicular to the line representing the rod. The tangential component of momentum of the particle does not change during the collision. The normal components of the momenta do change, both that of the particle and the rod while the total momentum of the system is conserved.
The angular momentum of the whole system is also conserved. The particle has some angular momentum with respect to the CoM of the rod. It will be different after the collision, and the rod gains some angular momentum.

Last edited: Jan 15, 2015
12. Jan 15, 2015

### Satvik Pandey

Thank you ehild for the explanation.Are we neglecting tangential force (friction) between the rod and particle because the surfaces are friction less or due to instantaneous collision? If the surfaces were not friction less then we have to consider horizontal impulse due to friction. right?

What do you think about my calculations in #post4.

Last edited: Jan 15, 2015
13. Jan 15, 2015

### phinds

Sorry. I screwed up. Ehild has given a good response.

14. Jan 15, 2015

### ehild

Edit: Wrong statement: In case of instantaneous collision, you can ignore friction even then when it exists.
Check your calculations, I can not follow them. Explain the notations. You seem to ignore the masses.

Last edited: Jan 16, 2015
15. Jan 16, 2015

### ehild

The velocity of separation is meaningless here. You can not treat the road as a point where the collision occurs.
You have two write out conservation of energy taking both translational and rotational energies into account.

Do not forget the mass.

It looks correct , but I would use conservation of momentum and angular momentum instead of the impulse-change of momentum relations. And you miss the equation for conservation of energy.

16. Jan 16, 2015

### haruspex

It's not relevant to this question, but that statement is not correct.
In an oblique impact, the normal impulse being J, the tangential impulse is up to $J \mu_s$. What happens past that is an interesting question. Does $\mu_k$ come into play?

17. Jan 16, 2015

### ehild

You are right, as always.:) I think the tangential force would be impulsive, too, and would change the tangential velocities.
Fortunately, the problem says that the collision is elastic, so there is no friction.

18. Jan 16, 2015

### Satvik Pandey

I have a doubt regarding the validity of the coefficient of restitution. I have equated the velocity of approach and separation of the particle and the point on which the particle collided with the rod. Can this concept be applied for extended bodies which might involve a combination of both rotation and translation ?

Sorry. That was a typo.:s

By conservation of energy I got

$\frac { 1 }{ 2 } m{ v }^{ 2 }=\frac { 1 }{ 2 } M{ V }^{ 2 }+\frac { 1 }{ 2 } I{ \omega }^{ 2 }+\frac { 1 }{ 2 } m{ v }_{ 0 }^{ 2 }$

By conserving angular momentum about the CoM of the rod I got

$\frac { mvlcos(\theta ) }{ 2 } =-\frac { m{ v }_{ 0 }lcos\alpha }{ 2 } +I\omega$

Considering 'clock' as -ve.

Here $m$ and $M$ are the masses of the particle and rod respectively.

What is the condition necessary for the particle to collide again at point B? I know that point B and particle must be at same place at a given time. But I don't know how to frame this constraint equation.

19. Jan 16, 2015

### Satvik Pandey

It's OK.:);)

20. Jan 16, 2015

### TSny

Yes, that concept applies to this problem. It can take the place of the equation for conservation of energy.

OK

The signs can be a headache. Are you taking $\omega$ to be a negative quantity if it is clockwise? Anyway, double check that you have the signs correct. Also, keep in mind that the particle need not "bounce back". It might retain an "upward" component of velocity after the collision.

It might be good to consider things from the reference frame of the CM of the rod after the collision.