A 1.1 kg block slides along a frictionless surface at 1.5m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.1 m/s .
Conservation of momentum
m1v1 m2v2 = (m1 m2)*vf = m1vf m2vf
m2v2- m2vf = m1vf-m1v1
m2 =m1 (vf-v1)/(v2-vf)
m1 = 1.1 kg (2.1m/s -1.5m/s )/(4.3m/s 2.1 m/s) = 1.1 kg *0.6/2.2=0.3 kg
This post is last updated on hrtanswers.com at Date : 1st of September – 2022