A truck with a mass of 1400 kg and moving with a speed of 14.0 m/s rear-ends a 641-kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision.

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.

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it