A bullet of mass m and speed v passes completely through
a pendulum bob of mass M as shown in Figure P6.50. The
bullet emerges with a speed of v/2. The pendulum bob is
suspended by a stiff rod of length  and negligible mass.
What is the minimum value of v such that the bob will
barely swing through a complete vertical circle?

Question

A bullet of mass m and speed v passes completely through
a pendulum bob of mass M as shown in Figure P6.50. The
bullet emerges with a speed of v/2. The pendulum bob is
suspended by a stiff rod of length  and negligible mass.
What is the minimum value of v such that the bob will
barely swing through a complete vertical circle?

rock_mit182 · Answer

attachment

rock_mit182 · Answer

$$m*v _{0b} + M * 0 = (m+M)V _{f} $$

anonymous · Answer

Sorry, I can't.

rock_mit182 · Answer

I'm stuck since then (momentum equation)

rock_mit182 · Answer

$$V _{f}= \frac{ 1 }{ 2 }v$$

anonymous · Answer

conserve momentum

then note that the energy needed to make that circle implies a change in potential energy of the bob from the lowest point to the highest point on the circle, m g (2L), where L is the  is the length of the wire holding the bob.

rock_mit182 · Answer

mgh =1/2mv*v 

solving for v:
$$V =\sqrt{2gh } =\sqrt{4gl}$$

rock_mit182 · Answer

@douglaswinslowcooper right ?

anonymous · Answer

That's the speed the bob will need. Since it got v/2 from the bullet, double this to indicate the speed needed by the bullet.

rock_mit182 · Answer

what you mean, we double v/2 cause is the min momentum in order to get a swing, and the L lenght is deduced from the height that reached the bob ?