Collision with Center of Mass

[postfan]

Homework Statement

A cart of mass m moving at 12 m/s to the right collides elastically with a cart of mass 4.0 kg that is originally
at rest. After the collision, the cart of mass m moves to the left with a velocity of 6.0 m/s. Assuming an elastic
collision in one dimension only, what is the velocity of the center of mass (vcm) of the two carts before the collision?
(A) v_cm = 2.0 m/s
(B) v_cm = 3.0 m/s
(C) v_cm = 6.0 m/s
(D) v_cm = 9.0 m/s
(E) v_cm = 18 m/s

Homework Equations

The Attempt at a Solution

I just gave an arbitrary distance of 60 meters initially between the carts with the moving cart being at x=0 at t=0. During t=0 the center of mass is (60-0)/2=30 and at t=1 the center of mass is (60+12)/2=36, implying that C is correct however this is B, what am I doing wrong?

 

[showzen]

During t=0 the center of mass is (60-0)/2=30 and at t=1 the center of mass is (60+12)/2=36

For the velocity of the center of mass, you need to sum the $m_i \dot{\vec{r}}_i$ and divide by M.

 

[postfan]

OK since the collision is elastic both momentum and kinetic energy is conserved so I came up with 2 equations :
1) m1v1=m1v2+m2v3
2).5m1v1^2=.5m1v2^2+.5m2v3^2
where v1=12, v2=6, and m2=4 with everything else being unknown.

I got m=15/8 ( didn't bother with v3 since before collision the second block is at rest).
I summed up all the momentums, took it over the masses and got 180/47 which doesn't correspond to any of the answer choices. What am I doing wrong?

 

[showzen]

$v_1$ and $v_2$ are vectors going in opposite directions.

 

[postfan]

OK, making sure that my vectors were in the right direction so v2=-6 I got m1=4/3. using your equation (I assumed you meant velocity instead of radius), I got an answer of 3. Assuming that's right, one more question how do you derive that formula for center of mass velocity?

 

[showzen]

You take the center of mass position equation:

$x_{cm}=\frac{m_1x_1+m_2x_2+~...~+m_nx_n}{m_1+m_2+~...~+m_n}$

And replace position by velocity:

$v_{cm}=\frac{m_1v_1+m_2v_2+...+m_nv_n}{m_1+m_2+...+m_n}$

Basically just taking the time derivative.

 

[postfan]

Oh OK, was my answer right though?

 

[lightgrav]

Oh OK, was my answer right though?

the easy way to check:
velocity after, relative to com_velocity, is negative 1 times velocity before, relative to com_velocity ... (-1, for perfectly elastic; 0 for perfectly inelastic):
-6 (m/s) minus 3 (m/s) = -1 ( 12 minus 3 ) (m/s) ... -9 m/s = - 9 m/s

 

[postfan]

What?

 

[lightgrav]

Do you actually not recognize the values from your own problem?
Any velocity is always relative to some other velocity. relative means to subtract it ("minus" above)
The velocity of a cow, relative to an InterState trucker driving to California, is 70 mph Eastward.
Collisions are most simple if viewed from the center-of-mass; in that frame, total momentum is zero.
Before collision: the massive car approaches slower, from the other direction
After bouncing: both cars charge direction, and recoil away (outward); the massive car is slower.