Asked by Bobbie
A 2.0kg bag is held by a string to the ceiling. A 10g bullet travelling at 200m/s strikes the stationary bag. The height of the bag after the collision is 10cm, assuming there is no friction, determine the speed, in m/s, of the bag after the collision.
Answers
Answered by
Fredr
mb = mass of bullet
vb = velocity of bullet
ma = mass of bag
vab = velocity of bag and bullet
from conservation of momentum:
mb*vb = (ma + mb)*vab
then from conservation of energy:
(ma + mb)*vab^2/2 = (ma + mb)*g*h
vb = velocity of bullet
ma = mass of bag
vab = velocity of bag and bullet
from conservation of momentum:
mb*vb = (ma + mb)*vab
then from conservation of energy:
(ma + mb)*vab^2/2 = (ma + mb)*g*h