Collision Velocity Calculation for Box Cars

[stargirl17]

Homework Statement

A 24,000 kg box car traveling north at 2 m/s collides with an 18,000 kg box car. Upon collision the two box cars lock together. What is the velocity of the 18,000 box car before the collision if after the collision the two box cars are traveling south at 1 m/s?

Homework Equations

MaVai + MbVBi = MaMaf + MbVbf

The Attempt at a Solution

 

[thebigstar25]

initially for object a (with mass ma) you have an initial velocity of 2 m/s , after collision they are locked together (so they have the same final velocity) with velocity 1m/s..

you have got everything , make proper choice for the direction (since the velocity is a vector quantity, e.g you can choose the north to be your positive y-axes and the south is the negative) ..

I think it is clear from there .. :)

 

[ManyNames]

Homework Statement

A 24,000 kg box car traveling north at 2 m/s collides with an 18,000 kg box car. Upon collision the two box cars lock together. What is the velocity of the 18,000 box car before the collision if after the collision the two box cars are traveling south at 1 m/s?

Homework Equations

MaVai + MbVBi = MaMaf + MbVbf

The Attempt at a Solution

I won't solve your problem, but i will demonstrate a similar situation, and from there you can use some of that brain matter ;)

we have a van of 1000kg which lock into a collision with another van at 20 m/s into a second van of 1200kg which is moving at 5m/s. We use the equation p=Mv to calculate the resultant force.

Before collision, we have 1000 x 20(+1200 x 0) = 1000 x v(1200 x 5)

=20,000 = 1000 x v +6000

so 6000 - 20,000 is 14,000 = 1000 x v

so the impact of velocity is 14m/s through conversion of units.

 

[lightgrav]

what is v_A,f ... and what is v_B,f ... ?

 

[ManyNames]

I assume she has meant initial and final velocity. It's all written quite terribly.