Question

A man weighing 60 kg runs along the rail with a velocity of 18 km/hr and jumps into a car of mass 1 quintal standing on the rail . calculate the velocity

Answer 1

Let m1=60, u1=5, m2=100, u2=0 
We need to find v(common)
So, m1u1+m2u2=(m1+m2)v
   i.e 300+0=160v

v=300/160
  = 15/8

Answer 2

Answer:

1.875m/s

Explanation:

18km/h=5m/s

for this q we have to change the formula of conservation of linear momentum a bit

m1u1+m2u2=m1v1+m2v2

m1u1+m2u2=(m1+m2)v

because the velocity v1=v2

60*5+100*0=(60+100)v

300=160*v

300/160=v

therefore v=1.875m/s