# Easy momentum/velocity problem

1. Aug 4, 2008

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

A 55kg person is standing at the left end of a 240kg cart moving to the left at 7.6 m/s. She runs to the right end and continues horizontally off the cart. What should be her speed relative to the cart in order to leave the cart with no horizontal velocity component relative to the ground?

2. Relevant equations

$$m_{c} =$$mass of cart

$$m_{p} =$$mass of person

$$v_{c} =$$initial velocity of cart

$$v'_{c} =$$final velocity of cart

$$v_{p} =$$initial velocity of person

$$v'_{p} =$$final velocity of cart

initial momentum = final momentum
$$m_{i} v_{i} = m_{f} v_{f}$$

3. The attempt at a solution

The answer given in the book is 9.3 m/s.

My reasoning must involve a fundamental flaw, because I can't convince myself that is the right answer. To get the answer of 9.3 m/s they must have started from the following equation and solved for $$v'_{c} = -v'_{p}$$
$$m_{c} v_{c} = m_{c} v'_{c} + m_{p} v'_{p}=(m_{c} - m_{p})v'_{c}$$

I tried the following:
$$v_{c} (m_{c} + m_{p})=m_{p} v'_{p} + m_{c} v'_{c}$$
but that would just be a guess also since neither of the above to equations look right to me, and I couldnt suggest another equation to use.

Any Explanations/reasoning behind the method would be appreciated

2. Aug 4, 2008

### Staff: Mentor

Nothing wrong with that. What's $$v'_{p}$$? (It's given.)

Hint: First worry about speed relative to the ground. Then translate to speed relative to the cart.

3. Aug 4, 2008

i would say $$v'_{p} = -v'_{c}$$
so
$$295 \times 7.6 = 2242 = v'_{c}(m_{c}-m_{p}) = 185 \times v'_{c}$$
so
$$v'_{c} = \frac{295 \times 7.6}{185} = 12.12 m/s$$

which is wrong according to the book. perhaps i cant say $$v'_{p} = -v'_{c}$$?

4. Aug 4, 2008

### Staff: Mentor

No you can't. (Not sure where you got the idea that you could.) $$v'_{p}$$ is the woman's velocity with respect to the ground. It's given:

5. Aug 5, 2008