I'm trying to solve a homework problem as review for an exam I have tomorrow and I was wondering if someone could help explain it to me. It is as follows:

You are at Lowe’s shopping for bricks with your best friend. You fill your cart and put the remaining bricks in your friend's cart. As it turns out, your cart has twice the mass of your friend’s. In the parking lot you and your friend do a physics experiment. Ignoring friction (and assuming the parking lot police are not watching),

a. if you each start from rest and push your respective carts with equal force for 3 seconds, what is the ratio of the momentum of your car friend's cart?
b. what is the ratio of the work you do on does on her/his cart during this race

Approach for a

Let $p_1$ = my cart, $p_2$ = friend cart, where p is a vector denoting momentum

$\Delta p = p_f - p_i$ , $p_i$ = 0 since it's at rest

$$p_1 = 2m_1v_1,\quad p_2 = m_1v_2$$



$$\frac{p_2}{p_1} =\frac{v_2}{2v_1}$$

The answer is a 1:1 ratio but I'm not sure where to go from here to prove that answer


closed as off-topic by John Rennie, Emilio Pisanty, Brandon Enright, jinawee, Kyle Kanos Feb 28 '14 at 17:09

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You've probably learned about a quantity called "impulse" - try using that to solve the problem.

Let $J$ be impulse, defined for constant forces as $J = F t$ where $F$ is the force applied and $t$ is the time for which the force is applied. Since $F=ma$, we can substitute this to get:

$$\begin{aligned} J &= Ft \\ J &= mat \\ J &= m(at) \\ J &= m \Delta v \\ J &= \Delta (mv) \\ J &= \Delta p \end{aligned}$$

That is, $F t$ is equal to the change in momentum of the cart. Since both carts have the same force applied to them for the same amount of time, the momentum change of both carts must be equal.

For part (b), consider conservation of energy. The force does a certain work on the cart, which causes the cart to gain kinetic energy. To find the work done on each cart, simply find the increase in kinetic energy of each one. Try this part by yourself, and comment if you need further assistance.

  • $\begingroup$ So if I was to show my work for the first one I would just do your steps in reverse right? Also for the second one would it be a 4:1 ratio? $\endgroup$ – sreya Feb 28 '14 at 2:56
  • $\begingroup$ Sorry, it should be a 2:1 ration no, for part B? $\endgroup$ – sreya Feb 28 '14 at 3:02
  • $\begingroup$ You wouldn't necessarily need to do the steps in reverse; you could just demonstrate that $Ft = \Delta p$ using the same argument I presented, and then note that both $F$ and $t$ are the same in both situations. Yes, (b) is a 2:1 ratio. $\endgroup$ – Shivam Sarodia Feb 28 '14 at 14:35

For (a), force equals the rate with which momentum changes

$$\frac{d\vec p}{dt} = \vec F$$

Since the force on each cart is equal, constant, and applied for the same amount of time, the change in momentum for each cart is...?

For (b), keep in mind that the less massive cart will have greater acceleration during the time the force is applied.


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