Two bodies, one of mass 4kg and the other of mass 6 kg, approach each other head-on with speed of 2m/s and 8m/s, respectively. Calculate the velocity of each body after collision if the collision is perfectly elastic.

Answer 1

Given: M1 = 4kg, V1 = -2m/s.
M2 = 6kg, V2 = 8m/s.
V3 = velocity of M1 after collision.
V4 = velocity of M2 after collision.

Momentum before = Momentum after
M1*V1 + M2*V2 = M1*V3 + M2*V4
4*(-2) + 6*8 = 4*V3 + 6*V4
Eq1: 4V3 + 6V4 = 40.

V3 = ((M1-M2)V1 + 2M2*V2)/(M1+M2)
V3 = ((4-6)(-2) + 12*8)/(4+6) = (4 + 96)/10 = 10 m/s.
In Eq1, replace V3 with 10 and solve for V4:
4*10 + 6V4 = 40
V4 = 0.

Answer 2

equestrian worker accidentally drop a brick from a high scaffold what is the velocity of a break after 4.0s