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Question

Question

science physics

Answer 1

Answer #1

let the speed after collision of 2kg be v1 and of 4kg be v2

by the conservation of momentum we have

m1u1 + m2u2 = m1v1 + m2v2

2*2 + 4*-4 = 2v1 + 4v2

v1 + 2v2 = -6 -----(1)

velocity of approach = velocity of recess

v2 - v1 = 2 - (-4)

v2 - v1 = 6 ------(2)

adding both equation we have

v2 = 0

v1 = -6 m/s

Answer 2

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