Question

a single 1,300 jk cargo car is rolling along a train track at 2.0 m/s when 400 kg of coal is dropped vertically into it. what is it’s velocity right afterward? assume a closed system.

Answer 1

1.53 m/s

Step-by-step explanation:

Given:

Mass of the car (M) = 1300 kg

Mass of the coal (m) = 400 kg

Initial velocity of the car (U) = 2 m/s

Initial velocity of the coal (u) = 0 m/s (Since it is dropped)

When the coal is dropped into the car, then they move with same final velocity.

Let the final velocity be 'v' m/s.

For a closed system, the law of conservation of momentum holds true.

So, initial momentum is equal to final momentum of the car-coal system.

Initial momentum of the car =
`MU=1300* 2=2600\ Ns`

Initial momentum of the coal =
`mu=0\ Ns`

Total initial momentum is the sum of the above two momentums.

So, total initial momentum = 2600 + 0 = 2600 Ns

Now, final momentum is given as the product of combined mass and final velocity. So,

Final momentum of the system =
`(M+m)v=(1300+400)v=1700v`

Now, from law of conservation of momentum,

Initial momentum = Final momentum

`2600=1700v\\\\v=(2600)/(1700)\\\\v=1.53\ m/s`

Therefore, the final velocity of either of the two masses is same is equal to 1.53 m/s.