Solving for Velocity After Two Objects Collide

[EthanVandals]

Homework Statement

A 5kg mass moving at 20m/s collides with a stationary 10kg mass. If the 5kg mass comes to rest, how fast will the 10kg mass be moving?

M1 = 5kg
M2 = 10kg
V1 = 20m/s

Homework Equations

KE = 1/2mv²

The Attempt at a Solution

I am fairly confident in this, and that's what scares me. Can it really be so simple as a transference of energy? The 5kg mass has KE of approximately 1,000J. When it hits the other mass, the kinetic energy of the 10kg block would also be 1,000J. Then to find the velocity, I would take those 1,000J, divide them by 10, and multiply by 2, giving me 200, then I'd take the square root of that, giving me a new velocity of 14.14m/s...Is it possible I got this correct?

 

[kuruman]

Can it really be so simple as a transference of energy?

No. What is always conserved in a collision? Hint: It is not energy.

 

[EthanVandals]

No. What is always conserved in a collision? Hint: It is not energy.

My first guess would be momentum, but I am not quite sure how that would come into play here...Unless you're saying that it would also be moving at 20m/s...

 

[kuruman]

My first guess would be momentum, ...

No need to guess. Momentum is always conserved in a collision. Energy may or may not be conserved.

Unless you're saying that it would also be moving at 20m/s...

What do you think is "it" that should also be moving at 20 m/s?

Momentum conservation simply means that if you add all the momenta before the collision to get p_before and then add all the momenta after the collision to get p_after, then p_before = p_after.

Apply this simple idea here.

 

[EthanVandals]

No need to guess. Momentum is always conserved in a collision. Energy may or may not be conserved.

What do you think is "it" that should also be moving at 20 m/s?

Momentum conservation simply means that if you add all the momenta before the collision to get p_before and then add all the momenta after the collision to get p_after, then p_before = p_after.

Apply this simple idea here.

Then the second box (the 10kg one) based on the before and after equaling each other should also move at 20m/s when it is hit by the smaller mass?

 

[gneill]

Then the second box (the 10kg one) based on the before and after equaling each other should also move at 20m/s when it is hit by the smaller mass?

I don't see why you're guessing. Why not compute the value of the momentum and check?