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?​

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

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.