Problem about an elastic collision between a rod and a ball at an angle

[ubergewehr273]

Homework Statement

Homework Equations

Conservation of linear momentum
Conservation of angular momentum
$\frac{-v_{relative after collision}} {v_{relative before collision}}=e$ where e is coefficient of restitution

The Attempt at a Solution

Components of velocity given as $v_{x}=5v_{0}sin\theta$ & $v_{y}=5v_{0}cos\theta$ ($\theta=37^o$)
The $v_{y}$ component doesn't contribute to collision and hence applying linear momentum conservation along x axis,
$5mv_{0}sin\theta=mv_{1} + mv_{2}$ here $v_{1}$ is velocity of ball after collision and $v_{2}$ is velocity of COM of rod after collision.
Using equation 3,
$v_{2}-v_{1}=5v_{0}sin\theta$
Adding the above 2 equations, I get $v_{1}=0$, therefore the horizontal component of velocity of ball gets nullified which means only $v_{y}$ component remains and hence final velocity of ball should be $5v_{0}cos37=4v_{0}$. Where am I going wrong ?

P.S. In my approach I never used conservation of angular momentum. Is it really required ?

 

[BvU]

Yes it is.

Where am I going wrong ?

Imagine the rod is not pivoting around P but not fixed at all there: would you obtain the same result ? Isn't that strange ?

 

[ubergewehr273]

Yes it is.
Imagine the rod is not pivoting around P but not fixed at all there: would you obtain the same result ? Isn't that strange ?

Of course, I'd obtain the same result. Then how does rotation about P change something about the ball. That is my doubt.

 

[BvU]

So according to you, it does not matter for the result that the rod is fixed to only rotate around P and can not move freely ?
(actually, this is re-phrasing my second question)

 

[ubergewehr273]

Well by intuition, it must make a difference but I am unable to frame the given scenario mathematically.

 

[BvU]

It does make a difference, yes.
Why not try and see if angular momentum conservation gives you the same answer or not ?

 

[ubergewehr273]

Well I tried but didn't end up with a proper answer.
$5mv_{0}sin37=mv_{1}+mv_{2}$
$5mv_{0}(\frac{l}{2})sin37=mv_{1}\frac{l} {2} + \frac{ml^2}{3}\omega$
$\omega\frac{l} {2}=v_{2}$
Solving the above, I get $v_{2}=0$ & $v_{1}=5v_{0}sin37$.

 

[BvU]

I like this conservation of angular momentum.
But you haven't abandoned conservation of linear momentum yet. Probably because you realize you need a second equation.
What did you do so far with the other given fact ?

 

[ubergewehr273]

Actually, I don't see the point of incorporating the restitution equation. Because I have already considered the fact while writing angular momentum conservation equation (i.e, the term $mv_{1}\frac{l}{2}$ in the RHS)

 

[BvU]

Neither do I. But I don't see that you have made use of it either ($mv_1 {l\over 2}$ ?).

That coefficient isn't mentioned in the problem statement.
The coefficient of restitution follows from the answer to the problem, not the other way around.

Let me try to hint rather bluntly: if the pivot pin isn't strong enough, what will happen during the collision ?

 

[ubergewehr273]

Let me try to hint rather bluntly: if the pivot pin isn't strong enough, what will happen during the collision ?

I think I'm getting some insights. I suppose I have to consider the hinge reaction force applied by the pivot during collision. Is that right ?

 

[haruspex]

The coefficient of restitution follows from the answer to the problem, not the other way around.

No, we are told it is perfectly elastic, so the coefficient is 1. But it does not help because the reaction force from the hinge invalidates use of that equation anyway.

 

[haruspex]

I think I'm getting some insights. I suppose I have to consider the hinge reaction force applied by the pivot during collision. Is that right ?

Or find a way to apply a second conservation law (i.e. in addition to the energy law) that does not involve that reaction.

 

[BvU]

No, we are told it is perfectly elastic, so the coefficient is 1. But it does not help because the reaction force from the hinge invalidates use of that equation anyway.

OP writes $\frac{-v_{\rm relative \ after \ collision}} {v_{\rm relative \ before \ collision}}=e$. Are you saying $e=1$ ?

 

[haruspex]

OP writes $\frac{-v_{\rm relative \ after \ collision}} {v_{\rm relative \ before \ collision}}=e$. Are you saying $e=1$ ?

I suppose it depends how you define the coefficient. If you define it by that equation then it is not 1, but I view it as a property of the bodies in collision, in which case e=1 because it is perfectly elastic, but that equation is not applicable because there is an external impulse from the hinge...(edit) ... and because some of the energy goes into rotation.

 

[BvU]

I suppose it depends how you define the coefficient. If you define it by that equation then it is not 1, but I view it as a property of the bodies in collision, in which case e=1 because it is perfectly elastic, but that equation is not applicable because there is an external impulse from the hinge...(edit) ... and because some of the energy goes into rotation.

I agree (not a fan of restitution coefficients).
The other thing: We wanted the OP to find that out by himself... hence the ever firmer questions/hint

 

[ubergewehr273]

But it does not help because the reaction force from the hinge invalidates use of that equation anyway.

How does the hinge reaction invalidate the equation?

 

[BvU]

It is a force, so a $d\vec p\over dt$ which breaks momentum conservation

 

[BvU]

Let me try to hint rather bluntly: if the pivot pin isn't strong enough, what will happen during the collision ?

It breaks. That means a force is exercised on it and - according to Newton 3 - it exercises a force on the rod.

A tricky one is: which way will it break ? You can imagine it breaks off to the right if the ball hits the rod close enough to the pivot point (because the rod will want to rotate clockwise) and it breaks to the left if the ball hits at the bottom somewhere (because the rod will want to rotate counterclockwise). That's a nice exercise for when we've finished this one: where must the ball hit so that linear momentum IS conserved ? ​

I suppose I have to consider the hinge reaction force applied by the pivot during collision. Is that right ?

If you choose your axis of rotation wisely, the force on the pin exerts zero torque and you can do an angular momentum balance.

 

[ubergewehr273]

If you choose your axis of rotation wisely, the force on the pin exerts zero torque and you can do an angular momentum balance.

Ok, conserving angular momentum about hinge so that there is zero torque by the hinge reaction,
$3mv_{0}\frac{l}{2}=mv_{1}\frac{l}{2}+\frac{ml^2}{3}\omega$

 

[BvU]

Sounds as if you aren't all that convinced yet ...
So angular momentum is conserved (*) . What else ?

(*) Depending on what you mean with $v_1$, I agree.