# Trolley on flat land

1. Oct 4, 2007

### kateman

1. The problem statement, all variables and given/known data
While coasting along a street at .5m/s, Mieko, a 15kg girl in a 5kg trolly, sees a vicious dog in front of her. She has with her only a 3kg bag of sugar which she is bringing home from the store, and throws it with forward velocity of 4m/s relative to her original motion. How fast is she moving after she throws the bag of sugar? whit is her final direction of motion?

2. Relevant equations
p = mv

3. The attempt at a solution
right so i did p=mv of the trolley (worked out to be 11.5) and then of the bag being thrown (worked out to be 12)

the next step confused me. if i divide 11.5 by 12, i get the right answer of .9582 backwards. but i cannot see how that is mathematically correct.

now if i use the m1v1i + m2v2i = m1v1f + m2v2f formula, i only get a value half of the correct answer!
can someone please explain this to me?

2. Oct 5, 2007

### learningphysics

Are you sure that .9582 is the correct answer? I'm not getting that.

You can consider Mieko and the trolley together as one system of 20kg. so m1 is 20kg. m2 is 3kg.

use:

$$m1*v_{trolleyinitial} + m2*v_{baginitial} = m1*v_{trolleyfinal} + m2*v_{bagfinal}$$

You're right that initial total momentum is 11.5. The velocity of the bag is 4m/s relative to the initial motion, so the final velocity of the bag is 4.5m/s.

3. Oct 5, 2007

### kateman

well the correct answer is .1m/s backwards
i just realised what i had done. i misread .9 and though it was .09 so that it could be rounded up to .1

and then i didnt think that the bag would also have the two velocities (of bag and trolley added together)

haha i have to stop doing physics late at night :P

thanks learningphysics, its always helpful to have someone else to see what i have overlooked :)