Question

Consider a system to be one train car moving toward another train car at rest. When the trains collide, the two cars stick together. What is the total momentum of the system after the collision?

Before Collison
Car 1 Car 2
m1 =600kg m2 =400kg
v1 =4 m/s v2 =0 m/s
Choices:
800kg × m/s
1,600kg × m/s
2,400kg × m/s
4,000kg × m/s
Someone please help! The lady who tought me did a terrible job of explaining how to do this!

Answer 1

Answer : 2400 kgm/s.

Explanation :

It is given that,

Mass of train 1,
`m_1=600\ kg`

Mass of train 2,
`m_2=400\ kg`

When the trains collide, the two cars stick together. So, it is a perfectly inelastic collision. The momentum of the system will remain conserved.

Momentum of train car 1,
`p_1=m_1v_1=2400\ Kgm/s`

Momentum of train car 2,
`p_2=m_2v_2=0`

So, total momentum of the system after the collision will be same as before the collision.

`p_f=p_1+p_2`

`p_f=2400\ kgm/s`

Hence, after the collision total momentum of the system would be 2400 kgm/s.

Answer 2

As per the question, the system consists of two cars.

The the masses of two cars are denoted as-
`m_(1) \ and\ m_(2) \ respectively`

The two cars undergoes collision with each other. Here collision must be inelastic in nature.

Let the velocities of each car is denoted as-
`v_(1) \ and\ v_(2) \ respectively`

It is given that masses of two cars and initial velocities-

`m_(1) =600 kg`
`v_(1) =4m/s`

`m_(2) = 400kg`
`v_(2) =0 m/s`

First we have to calculate the total momentum of the system before collision.

The momentum of a body is the product of its mass with velocity.

Momentum of car 1:

`p_(1) =m_(1) v_(1)`

`=600kg*4m/s`

`=2400\ kg\ m\ s^-1`

The momentum of car 2:

`p_(2) =m_(2) v_(2)`

`=400\ kg*0\ m/s`

`=\ 0\ kg\ m\ s^-1`

Now the total momentum of the system is :

`p_(1) +p_(2)`

`=[2400+0]\ kg\ m\ s^-1`

`=2400\ kg\ m\ s^-1`

Now we are asked to calculate the total momentum of the system after collision.

The basic condition in case of any type of collision is that it is in accordance with law of conservation of linear momentum.

Hence, the momentum will be conserved here.

It means the total momentum before collision must be equal to the total momentum after collision.

Hence, the correct answer to the question will be 2400 kg m/s.