summer014
Answered

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?

Verified Answer :

Edufirst

1) In any collision the momentum is conserved

(2m)(vo) + (m)(-2vo) = (2m)(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] (2m) (vo)^2 + [1/2](m)(2vo)^2 = [1/2](2m)(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)




 

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