Question

A 6450-kg railroad car travels alone on a level frictionless track with a constant speed of 20.0 m/s . A 3850-kg load, initially at rest, is dropped onto the car.

Part A

What will be the car's new speed?

Express your answer to three significant figures and include the appropriate units.

Answer 1

This is a momentum question involving the equation p = mv and the conservation of momentum between the two states:

p (before) = p (after)

Conservation of momentum, here's how you find it...

P1 = m1u2
P2 = m2u2

m1 = 6450 kg
m2 = 3850 kg

u1 = 20 m/s
u2 = 0 m/s

P1 = (6450 kg) * (20 m/s) = 129,000 kg-m/s
P2 = (3850 kg) * (0.0 m/s) = 0 kg-m/s

Equation

P1 + P2 = (m1 + m2) * v
(129,000 kg-m/s) = [ (6450 kg) + (3850 kg) ] * v

New velocity = 12 .52m/s