Block m1 of mass 2m and velocity v0 is traveling to the right (+x) and makes an elastic head-on collision with block m2 of mass m and velocity - 2v0 (i. e., traveling to the left). What is the velocity v1′ of block m1 after the collision?

1) In any collision the momentum is conserved

(2*m) * (vo) + (m) * (-2*vo) = (2*m) (v1') + (m) (v2')

candel all the m factors (because they appear in all the terms on both sides of the equation)

2 (vo) - 2 (vo) = 2 (v1') + (v2') = > 2 (v1') + v (2') = 0 = > (v2') = - 2 (v1')

2) Elastic collision = > conservation of energy

=> [1/2] (2*m) (vo) ^2 + [1/2] (m) * (2*vo) ^2 = [1/2] (2*m) (v1') ^2 + [1/2] (m) (v2') ^2

cancel all the 1/2 and m factors = >

2 (vo) ^2 + 4 (vo) ^2 = 2 (v1') ^2 + (v2') ^2 = >

4 (vo) ^2 = 2 (v1') ^2 + (v2') ^2

now replace (v2') = - 2 (v1')

=> 4 (vo) ^2 = 2 (v1') ^2 + [-2 (v1') ]^2 = 2 (v1') ^2 + 4 (v1') ^2 = 6 (v1') ^2 = >

(v1') ^2 = [4/6] (vo) ^2 = >

(v1') ^2 = [2/3] (vo) ^2 = >

(v1') = [√ (2/3) ] * (vo)

Answer: (v1') = [√ (2/3) ] * (vo)