Answer
Given,
Mass of truck, M = 1400 Kg
speed of car, u = 14 m/s
mass of car, m = 641 Kg
speed of car before collision, u' = 0 m/s
Using conservation of momentum
M u + m u' = M v + m v'
1400 x 14 + 0 = 1400 v + 641 v' .....(i)
For elastic collision
Relative velocity of approach = relative velocity of separation
14 = v' - v
v' = 14 + v
Now putting value in equation (i)
19600 = 1400 v + 641 (14+v)
v = 5.21 m/s
So,
v' = 14 + 5.21 = 19.21 m/s
Speed of truck after collision = 5.21 m/s
Speed of car after collision = 19.21 m/s.