Question

a block of mass 120kg moves with a speed pf 6m/s on frictionaless horizontal surface towards another block of mass 180g kept at rest .they collide and the first block stops .find the speed of the other block after collision

Answer 1

let the speed of other block be X

Since, momentum before collision = momentum after collision

=120/1000*6 + 180/1000*0 = 120/1000*0 + 180/1000 X

= 48/100 + 0 = 0 + 18/100 X

= 18/100 X = 48/100

= X = 48/100*100/18

So, X = 9/24

Answer 2

It is based on law of conservation of momentum
so,
m1u1 +m2u2 = m1v1 + m2u2
120×6 + 180×0 = 120×0 + 180×?
720 = 180×?
?= 720/180
?=4
Therefore the speed of second body after collision is 4m/s