Question

A 2—kg ball B travelling at 22 m.s–¹ overtakes a 4—kg ball travelling in the same direction at 10 m.s—¹. Find the velocity of each ball after collision

Answer 1

Answer

Initial momentum is given

$P_i=mu+m'u'\\=2\times22+4\times10\\=84kg-m/sec$

So velocity after collision

For first particle B

$v=\frac{P_i+m'(u'-u) }{m+m'}\\=\frac{84+4(10-22) }{2+4}\\=6m/sec$

Now A

$v'=\frac{P_i+m(u-u') }{m+m'}\\=\frac{84+2(22-10) }{2+4}\\=18m/sec$