Write this out and try it for yourself in your notebook!!! You won’t fully absorb how to do it if you simply skim the solution.

**Question:** A bullet of mass m and speed v passes completely through a pendulum bob of mass M as shown. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length L and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

First off all, let’s draw out what this question is actually describing in the before/after situations.

The question is asking what the *minimum* velocity required to make the mass go around in a circle once. That means that it should just *barely* get to the top, so we can assume mass M has no velocity at the top. Let’s consider the conservation of momentum for this situation to see how fast the block M moves after the bullet goes through it.

∑P_{before} = ∑P_{after}

mv_{1} = Mv_{2} + m(v_{1}/2)

Now, solve for v_{2}.

mv_{1} – (1/2)mv_{1} = Mv_{2}

(1/2)mv_{1} = Mv_{2}

(1/2)(m/M)v_{1} = v_{2}

Great, now we know how fast the block will move based on the speed that the bullet hits it. Now we want to find out how fast the block needs to move in order to have enough energy to get into position. Consider what the block is doing:

If we call the bottom position our y=0 point, then the top is just height h=2L. At the bottom, there’s no potential energy, and later the block just *barely* passes the top (so v at the top is zero as well).

This means that the conservation of energy gives

(1/2)Mv_{2}^{2} = Mg(2L)

Simplifying a bit…

(1/2)v_{2}^{2} =2gL

v_{2}^{2} =4gL

v_{2}=√4gL

Now you can throw in the result from earlier to sub out v_{2} in terms of v_{1}, and then solve for v_{1}

(1/2)(m/M)v_{1}= √4gL

(1/2)(m/M)v_{1}= 2√gL

v_{1}= 4(M/m)√gL

And you’re done!!!

Now, consider this: If the bullet hit the pendulum in an **inelastic** collision, would the velocity needed be greater than, less than, or equal to the previous velocity we found?