Girl Throws Weights: Calculate Speed Relative to Ground

[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

 

[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

 

[thomate1]

for your question. It looks like you are on the right track with using conservation of momentum to solve this problem. The equation you used is correct, but it looks like there may be a mistake in your calculation. When solving for the velocity of the girl after the weights are thrown, you should be dividing by the total mass of the system (50 kg + 10 kg = 60 kg), not just the mass of the girl (50 kg). This should give you a final velocity of 3.25 m/s, which matches your answer using the center of mass method.

One thing to keep in mind when using conservation of momentum is that the initial and final momenta must be calculated relative to the same reference frame. In this case, the initial momentum is calculated relative to the girl, while the final momentum is calculated relative to the ground. This is why we divide by the total mass of the system to get the final velocity relative to the ground.

I hope this helps clear up any confusion and you can feel more confident in your solution to this problem. Keep up the good work in your scientific studies!