# Collision with Center of Mass

1. Jan 19, 2015

### postfan

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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?

2. Jan 19, 2015

### showzen

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

3. Jan 19, 2015

### 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?

4. Jan 19, 2015

### showzen

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

5. Jan 19, 2015

### 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?

6. Jan 19, 2015

### 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.

7. Jan 19, 2015

### postfan

Oh OK, was my answer right though?

8. Jan 19, 2015

### lightgrav

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

9. Jan 19, 2015

### postfan

What?

10. Jan 19, 2015

### 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.