Two carts on a frictionless track are launched toward each other at the same spe

ID: 2257649 • Letter: T

Question

Two carts on a frictionless track are launched toward each other at the same speed. Cart !, on the left, has a mass of 1.00 kg and cart B, on the right, has a mass of 0.25 kg. After the collision, it is observed that cart A is still moving to the right, albeit at a slower speed of 1 m/s. How fast is cart B moving after the collision?

This question was already posted and someone worked out the answer, but I don't understand why kinetic energy is being used or how this person set up their equations. If you could just exaplin it a little better that would be great and appreciated!

Initial momentum = maua + mbub = 1*u + 0.25*(-u) = 0.75*u

Final momentum = mava + mbvb = 1*1 + 0.25*vb = 1 + 0.25*vb

Initial KE = 1/2*maua2 + 1/2*mbub2 = 1/2*1*u2 + 1/2*0.25*(-u)2 = 0.625*u2

Final KE = 1/2*mava2 + 1/2*mbvb2 = 1/2*1*12 + 1/2*0.25*vb2 = 0.5 + 0.125*vb2

Conservation of momentum and energy yields,

0.75*u = 1 + 0.25*vb and 0.625*u2 = 0.5 + 0.125*vb2

Solving both these equations, we get,

vb = -1 m/s

Explanation / Answer