A 5.00-g bullet is fired into a 500-g block of wood suspended as a ballistic pendulum.
The combined mass swings up to a height of 10.00 cm. What was the magnitude of the
momentum of the combined mass immediately after the collision?
The Ans is 0.707 kg m/sec
I do not know how to go get this answer
PHYSICS - drwls, Wednesday, October 31, 2012 at 3:27am
Energy is conserved during the upward swing. Therefore
(M+m)V = (M+m)gH
V = gH = 0.98 m/s
is the block velocity just after the bullet (mass m) is embedded inside the block (mass M). The momentum at that time is
(M+m)V = (0.505 kg)*V = 0.495 kg m/s
I do not agree with the 0.707 answer.