Question

A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. Use conservation of energy and momentum to find the minimum value of v such that the bob will barely swing through a complete vertical circle.

Answer 1

Answer:

$v = \frac{2M}{m}(\sqrt{4gL})$

Explanation:

After bullet passing through the pendulum we know that the pendulum will raise to its maximum height

So the minimum speed required for this is in such a way that it will reach at highest point with zero speed

so here we can use energy conservation

$\frac{1}{2}Mv_1^2 = Mg(2L)$

so we will have

$v_1 = \sqrt{4gL}$

now we got the speed of the pendulum so we can use now momentum conservation of bullet + pendulum system

here we have

$mv + 0 = Mv_1 + m\frac{v}{2}$

$mv = M(\sqrt{4gL}) + m\frac{v}{2}$

so we have

$m\frac{v}{2} = M(\sqrt{4gL})$

so the speed is given as

$v = \frac{2M}{m}(\sqrt{4gL})$