Question

A 3.00 kg box sliding west at 2.00 m/s makes an inelastic collision with a second box sliding 1.50 m/s east. Afterwards, they both come to a stop. What was the mass of the second box?​

Answer 1

Let west be the negative direction, and east be positive. Then the total momentum of the system is conserved such that

(3.00 kg) (-2.00 m/s) + m (1.50 m/s) = 0

Solve for m :

m = (3.00 kg) (2.00 m/s) / (1.50 m/s) = 4.00 kg