Question

A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the first car moves in the same direction as before with a speed v/3.

(a) Find the final speed of the second car.

(b) Is this collision elastic or inelastic?

Answer 1

(a) mv1 + mv2 = mv1' + mv2'
mv + (.5m)(0) = m(v/3) + (.5m)v'
mv = m(v/3) + .5mv'
m(2v/3) = .5mv'
v' = m(2v/3) / .5m = 4v/3
v' = 4v/3

(b) in order to be elastic, KE must be conserved, so
.5mv^2 + .5mv^2 = .5mv'^2 + .5mv'^2
.5mv^2 + 0 = .5m(v/3)^2 + .5(.5m)(4v/3)^2
.5mv^2 = .5mv^2 / 9 + 4mv^2 / 9
.5mv^2 = 4.5mv^2 / 9 = .5mv^2
That means it is elastic

If you want a simpler way,
va -vb = -(v'a -v'b)
v - 0 = -((v/3) - (4v/3))
v = -(-v)
v=v this means it is elastic