Answer to Question #1183 in Mechanics | Relativity for Jon Allen

Question

A bullet of mass 0.00114 kg and speed v passes completely through a pendulum bob of mass
69.6 kg.
The bullet emerges with a speed v/2.
The pendulum bob is suspended by a stiff rod
of length 0.449 m and negligible mass.
The acceleration of gravity is 9.8 m/s2 .


What is the minimum value of v such that the pendulum bob will barely swing through
a complete vertical circle?
Answer in units of m/s.

Answer 1

We should use the Law of the Moment and Energy Conservation :
mv = mv/2 + MV;
MV²/2 = 2MgL.
Here m - is the mass of the bullet, M - the mass of the pendulum, v - speed of the bullet, V - the speed of the pendulum after interaction, L - is the length of the stiff.
V = 2√(gL)
mv/2 = 2M√(gL)
v = 4M/m √(gL) = 4*69.6/0.00114*√(9.8*0.449) = 512,271.964 m/s