Question
A bullet of mass m is shot at speed v toward a pendulum bob of mass M. The bullet penetrates through the block and emerges on the other side traveling at speed v/2.
1. What is the speed v_block of the block immediately after the bullet emerges from the block (In terms of M, m and v)?
2. To what maximum height h does the block rise (In terms of M, m and v)?
1. What is the speed v_block of the block immediately after the bullet emerges from the block (In terms of M, m and v)?
2. To what maximum height h does the block rise (In terms of M, m and v)?
Answers
m-mass of bullet, v- initial velocity of bullet, let v'be the velocity of block.
conservation of momentum,
m*v+ M*0= m*(v/2)+ M*v'
solve for v'.
conservation of momentum,
m*v+ M*0= m*(v/2)+ M*v'
solve for v'.
velocity imparted to block will cause the the block to rise to a height h at this instant the kinetic energy is converted into potential energy completely so,
(1/2)*M*(v')^2 = M*g*h
use the value of v' from above solution and solve for h.
(1/2)*M*(v')^2 = M*g*h
use the value of v' from above solution and solve for h.