Answer:
3.7 m/s
Explanation:
M = 444 kg
U = 5 m/s
m = 344 kg
u = - 5 m/s
Let the velocity of train is V and the car s v after the collision.
As the collision is elastic
By use of conservation of momentum
MU + mu = MV + mv
444 x 5 - 344 x 5 = 444 V + 344 v
500 = 444 V + 344 v
125 = 111 V + 86 v .... (1)
By using the formula of coefficient of restitution ( e = 1 for elastic collision)
-5 - 5 = V - v
V - v = - 10
v = V + 10
Substitute the value of v in equation (1)
125 = 111 V + 86 (V + 10)
125 = 197 V + 860
197 V = - 735
V = - 3.7 m/s
Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.