Elastic Collision - two carts

[Millacol88]

Homework Statement

Two carts equipped with spring bumpers on an air track have an elastic collision. The 253-g cart has an initial velocity of 1.80 m/s [N]. The 232-g cart is initially stationary. What is the velocity of each cart after the collision?

Homework Equations

I know because cart 2 is initially stationary:
m1v1 = m1v1' + m2v2'
and 1/2m1v1^2 = 1/2 m1v1'^2 + 1/2m2v2'^2

The Attempt at a Solution

I've been trying to solve this by rearranging the equations and subbing them into each other, but I keep ending up with really complicated equations that I can't seem to solve. Is there a simpler way to look at a problem like this? I'm fine with most collisions that aren't elastic, but the ones that are I can't do. Thanks.

 

[PotentialE]

With regards to the spring bumpers, was there any given information in the problem about the spring constants?

 

[Millacol88]

Nothing. I posted the question verbatim, and there was no diagram either.

 

[gneill]

I've been trying to solve this by rearranging the equations and subbing them into each other, but I keep ending up with really complicated equations that I can't seem to solve. Is there a simpler way to look at a problem like this? I'm fine with most collisions that aren't elastic, but the ones that are I can't do. Thanks.

The squares of the velocities in the KE formula can definitely complicate things algebraically. Another relationship that you can use in place of it (and is derivable from the conservation of momentum and KE) is that the relative velocities of the two objects after collision is equal to the negative of the relative velocities before the collision. Using variables, suppose that the initial velocities of the objects are v1 and v2, and the final velocities are u1 and u2. Then

(u2 - u1) = -(v2 - v1)

That should make your life *much* easier!

 

[asteriatic]

I understand that elastic collisions involve conservation of both momentum and kinetic energy. In this case, since both carts are on an air track, there are no external forces acting on them and therefore we can assume conservation of momentum. This means that the total momentum of the system before the collision is equal to the total momentum after the collision.

Using the equation you provided, we can set up the following equation:
m1v1 = m1v1' + m2v2'

Since the initial velocity of cart 2 is 0, we can simplify this to:
m1v1 = m1v1' + 0

We also know that the total kinetic energy before the collision is equal to the total kinetic energy after the collision, which can be represented by the equation:
1/2m1v1^2 = 1/2 m1v1'^2 + 1/2m2v2'^2

Now we can solve for v1' and v2' by rearranging these equations:
v1' = (m1v1 - m1v1') / m1
v2' = (2m1v1 - m1v1') / m2

Substituting the values given in the problem, we get:
v1' = (0.253*1.80 - 0.253*v1') / 0.253 = 1.80 - v1'
v2' = (2*0.253*1.80 - 0.253*v1') / 0.232 = 3.60 - 1.09*v1'

Solving for v1' and v2', we get:
v1' = 0.900 m/s
v2' = 1.10 m/s [N]

Therefore, after the elastic collision, the 253-g cart will have a velocity of 0.900 m/s and the 232-g cart will have a velocity of 1.10 m/s [N].