Question

A 1.3 KG blocks flies along a frictionless surface at 1.0 M/S.a2 block sliding at a faster 5.0 M/S collides with the first from behind and sticks to it. The final velocity of the combine blocks is 2.0 M/S. What was the mass of the second block?

Answer 1

the initial momentum of the system of block m1 and block m2 is

Pi= m1v1 + m2v2

the final momentum of the combine blocks is

Pf= (m1+m2)V

according to the law of convervation of momentum

Pi = Pf

m1v1 + m2v2 = (m1+m2)V

1.3 × 1 + 5m2 = 1.3 × 2 + 2m2

m2= 1.3/3 kg