a toy car of mass 0.01 kg, gets into an elastic collision with a toy train mass 0.35 kg. if the train is at rest and the car is moving at 2 m/s initially. how fast is the train moving after the collision?

Answer 1

Conservation of momentum:
Mc*V1 + Mt*V2 = Mc*V3 + Mt*V4.
0.01*2 + 0.36*0 = 0.01V3 + 0.36V4.
Eq1: 0.01V3 + 0.36V4 =0.02. V3 = 2-36V4

Conservation of KE:
0.5Mc*V1^2 = 0.5Mc*V3^2 + 0.5Mt*V4^2.
0.5*0.01*2^2 = 0.005V3^2 + 0.18V4^2.
Eq2: 0.005V3^2 + 0.18V4^2 = 0.02.

Eq1 = 0.02; Eq2 = 0.02. Therefore, Eq1 = Eq2:
0.01V3 + 0.36V4 = 0.005V3^2 + 0.18V4^2.
V3 + 36V4 = 0.5V3^2 + 18V4^2
0.5V3^2-V3 + 18V4^2-36V4 = 0
0.5V3(V3-2) + 18V4(V4-2) = 0

Replace V3 with 2-36V4:
0.5(2-36V4)(-36V4) + 18V4(V4-2) = 0
(1-18V4) - 0.5(V4-2) = 0
1-18V4 - 0.5V4 + 1 = 0
18.5V4 = 2
V4 = 0.108 m/s. = Velocity of the train.