Question

A man weighing 60 kg runs along the rails with a velocity of 18 km/h and jumps into a car of mass 1 quintal standing on the rails. Calculate the velocity with which the car will start travelling along the rails. (1 quintal=100kg).

Answer 1

Mass of man (m1) = 60kg            
Initial velocity of man (u1) = 18 km/h=18x5/18=5m/s
Final velocity of man (v1) = (v2) = v 
Mass of car (m2)= 100kg
Initial velocity of car (u2) = 0 (as it is in rest)
Final velocity of car (v2) = v
According to law of conservation of momentum,
m1u1 +m2u2 = m1v1 + m2v2
60x5 + 100x0 = 60 x v + 100 x v
300+0 = 160v 
v = 300/160
v = 1.875 m/s

Answer 2

Explanation:

Mass of man (m1) = 60kg

Initial velocity of man (u1) = 18 km/h=18x5/18=5m/s

Final velocity of man (v1) = (v2) = v

Final velocity of man (v1) = (v2) = v Mass of car (m2)= 100kg

Initial velocity of car (u2) = 0 (as it is in rest)

Final velocity of car (v2) = v

According to law of

conservation of momentum,

conservation of momentum,m1u1 +m2u2 = m1v1 + m2v2

60x5 + 100x0 = 60 x v + 100 x v

60x5 + 100x0 = 60 x v + 100 x v300+0 = 160v

60x5 + 100x0 = 60 x v + 100 x v300+0 = 160v v = 300/160

60x5 + 100x0 = 60 x v + 100 x v300+0 = 160v v = 300/160v = 1.875 m/s

Himanshu Himanshu Himanshu Himanshu