# Pendulum collision problem

1. The problem statement, all variables and given/known data

A stick of mass m and length l is pivoted at one end. It is held horizontal
and then released. It swings down, and when it is vertical the free end elastically collides.
Assume that the ball is initially held at rest and then released a split second before the
stick strikes it

a. Which conservation laws apply here?
b. If the stick loses half of its angular velocity during the collision, what is the mass of
the ball, M?
c. Determine the speed V of the ball right after the collision

2. Relevant equations

Moment of inertia of the stick:

$$\frac{1}{3}mL^2$$

$$L=I\omega$$

$$\omega=\frac{v}{r}$$

$$K_{rotational}=\frac{1}{2}I\omega^2$$

3. The attempt at a solution

For (a), I said that conservation of energy, and conservation of linear/angular momentum also apply

For (b) I used conservation of energy just before the collision:

$$mg\frac{L}{2}=\frac{1}{2}I\omega^2$$

$$mg\frac{L}{2}=\frac{1}{2}(\frac{1}{3}mL^2)\omega^2$$

This gives me an omega of $$\omega=\sqrt{\frac{3g}{L}}$$

I then used conservation of energy before and after collision:

$$\frac{1}{2}I{\omega^2}=\frac{1}{2}I\frac{\omega^2}{4}+\frac{1}{2}M{V_{b }}^2$$

I reduce this down to: $$M{V_{b}}^2=\frac{3}{4}mLg$$

Then I used conservation of linear momentum (which is the part I’m not entirely sure about). I assumed the momentum of the mass of the stick before the collision was equal to the sum of the momentum of the CM and ball after the collision:

$$mv_0=mv_1 + MV_b$$
$$V_{cm}=\frac{L}{2}\omega=\frac{L}{2}\sqrt{\frac{3g}{L}}=\sqrt{\frac{3Lg }{4}}$$

==> $$\frac{mL\omega}{4}=MV_b$$

Isolating v and putting it into the $$M{V_{b}}^2$$ expression I get $$M=\frac{1}{4}m$$

Then the velocity of the ball comes out to be $$\sqrt{3gL}$$

Is this correct? For example, I did not use angular momentum, when my gut is telling me that I should have. But assuming that conservation of linear momentum holds, I don’t need it right?

Many thanks for your help guys.

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