Ramu saw in a NEWS channel that a goods train of mass 200 metric ton moving with a velocity of 72 km `h^(-1)` collides with a passenger train of mass 300 metric ton with a velocity of 54 km `h^(-1)` coming in the opposite direction on the same track and both the trains move together after collision. He calculates their common velocity. Find what could be his answer.

Explanation

Passenger and goods trains are moving on the same track in opposite directions. Let us consider the direction of motion of passenger train as positive (left to right in the figure) and that of goods train as negative (right to left in the figure).
Mass of passenger train,
`m_(P) = 300 " metric ton = 300"xx 103 kg`
Velocity of passenger train before collision,
`u_(P) = 54` kmph
`=54 xx (5)/(18) m s^(-1) = 15 m s^(-1)`
`therefore` Momentum of passenger train before collision,
`P_(ip) = 300 xx 10^(3) xx 15 = 45 xx 10^(5) " kg m s"^(-1)`
Mass of goods train,
`m_(g) = 200 " metric ton = 200"xx 10^(3)kg`
Velocity of goods train before collision,
`u_(g) = -72 km`
`h^(-1) = - 72 xx (5)/(18) m s^(-1) = -20 ms^(-1)`
`therefore` Momentum of goods train before collision,
`P_(ig) = 200 xx 10^(3) xx (-20) = -40 xx 10^(5) " kg m s"^(-1)`
Since total momentum is positive, both the trains after collision, move in the same direction as that of the passenger train.
Let 'v' be the magnitude of the common velocity with which both trains move together after collision.
Then, after collision, the momentum of passenger train is, `P_(fp) = m_(p)v` and that of the goods train is, `P_(fg) = m_(g)v`.
`therefore` Total momentum after collision `P_(f("total"))`
`=P_(fp) + P_(fg) = m_(p)v + m_(g)v = (m_(p) + m_(g))v`
`=(300xx10^(3) + 200 xx 10^(3)) v = 500 xx 10^(3) xx v`
According to principle of conservation of momentum,
`P_(i("total")) = P_(f("total"))`
`implies 5 xx 105 = 500 xx 103 xx v`
`implies v = 1 " m s"^(-1)= 3.6 " km h"^(-1)`