# Girl on a Wagon

1. Jan 15, 2009

### abbie

A 50.0-kg girl stands on a 10.0-kg wagon holding two 15.0-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 6.5 m/s relative to herself . Assuming that the wagon was at rest initially, what is the speed of the girl relative to the ground after she throws both weights at the same time?

conservation of momentum

I can't seem to get this and I don't know why.. I used conservation of momentum and got a perfectly reasonable answer. I even used center of mass and got the same answer...

0 = -(15*2)*6.5 + (50+10)*V
V = 3.25
I tried both positive and negative and both are apparently wrong

with Center of Mass I get:
10 seconds after throw - weights travel 65 m, so-
(-65*30 + 60*X)/90 ----> 1950 = 60X
using the distance I get V = 3.25

So obviously I'm doing something wrong, and I don't care how stupid I feel finding out what it is cause this question is really bugging me..

Thanks

2. Jan 15, 2009

### LowlyPion

The speed of the masses is in the frame of reference of the girl/wagon. And they separate at 6.5m/s.

Now the mass thrown is half the mass of the girl/wagon. In the frame of reference of the observer (that is grading the question) that means that the masses |Vm| are going twice the speed of the wagon/girl |Vwg| but in opposite directions of course.

If in the frame of the wagon that speed of separation is 6.5 ...
... and 2*|Vwg| = |Vm| then ...

|Vm| = 2/3*(6.5)

And |Vwg| = 1/3*6.5

This preserves your momentum relationship and of course the center of mass of the system remaining in the same position