Question

A carof mass 1000 kg travelling at 12 m/s dashed into the rear of a bus of mass 2000 kg moving in the same direction at 8 m/s. The car bounces backward after the colision with a speed of 9 m/s. What is the velocity of the bus after the collision?

Answer 1

According to law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
i.e, momentum of system before collision = momentum after collision.
1000(12)+2000(8) = 1000(-8) + 2000(v2)

-8, because car is moving backwards.

12000+16000=-8000 + 2000v2
38000+8000 = 2000v2
46000 = 2000v2
23 = v2

So, bus will move forward with velocity =23m/s

Answer 2

Answer:

23.5m/s

Explanation:

According to law of conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2

i.e, momentum of system before collision = momentum after collision.

1000(12)+2000(8) = 1000(-9) + 2000(v2)

-9, because car is moving backwards.

12000+16000= -9000 + 2000v2

38000+9000 = 2000v2

47000 = 2000v2

23.5 = v2

So, bus will move forward with velocity =23.5m/s