Question

A truck with a mass of 1350 kg and moving with a speed of 12.0 m/s rear-ends a 821-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.

vcar = ___________________ m/s

vtruck = ____________________ m/s

Answer 1

Using conservation of momentum, we have

p_initial = p_final

(1350 kg) (12 m/s) + 0 = (1350 kg) v_truck + (821 kg) v_car

(16200 kg.m/s) = (1350 kg) v_truck + (821 kg) v_car                                         { eq.1 }

We know that, relative velocity of approach = relative velocity of separation

For an elastic collision, we have

(12 m/s) = v_car - v_truck

v_car = v_truck + (12 m/s)                                               { eq.2 }

inserting the value of v_car in eq.1 & we get -

(16200 kg.m/s) = (1350 kg) v_truck + (821 kg) [v_truck + (12 m/s)]

(16200 kg.m/s) = (1350 kg) v_truck + (821 kg) v_truck + (9852 kg.m/s)

(6348 kg.m/s) = (2171 kg) v_truck

v_truck = 2.92 m/s

Using eq.2 ;   v_car = v_truck + (12 m/s)   

v_car = [(2.92 m/s) + (12 m/s)]

v_car = 14.9 m/s