2
$\begingroup$

I need some help.

I am not quite sure how to know what sign criteria to use in the conservation of momentum equation. For instance, this exercise really confused me:

problem

Here, when I see the solution to the problem to find the velocity of the cart A, they do the following: $$v-v_A=v'=3\frac ms$$so that $v_A = v-3$. Up until now I agree with all signs. However, then they say that by momentum conservation, $$0=mv+m_Av_A$$ My question is, why is this not $$0=mv-m_Av_A$$since the boy and the cart have opposite signed directions for their velocities? Why are the signs taken into account in the first equation and not in the second one?

I am really confused, because in other exercises that are mixed with Kinematics, they do take the signs into account.

All help greatly appreciated.

$\endgroup$
2
$\begingroup$

The convention is that adding up all of the momenta in a closed system must give you a constant. So if you begin without any momentum, you have

$$ 0 = \vec p_\text{child} + \vec p_A $$

The arrows are a hint that this is actually three coupled equations, one in each of the coordinate directions, which are commonly named after the three letters at the end of the alphabet:

\begin{align} 0 &= p_{\text{child},x} + p_{A,x} \\ 0 &= p_{\text{child},y} + p_{A,y} \\ 0 &= p_{\text{child},z} + p_{A,z} \\ \end{align}

Now in your problem, two of these equations are trivially satisfied, because you're pretending like there is only motion along one dimension, so you have

$$ 0 = p_{\text{child},x} + p_{A,x} $$

This gives you the result that, in order to conserve momentum, the cart and the boy must end up moving in opposite directions: one along the positive $x$-axis, and one along the negative $x$-axis.

Your version with a minus sign works in this case, but will fail you in more complicated problems in the future where it is not correct to assume that the final momenta will be oppositely directed.

For an example where the sign convention matters even though you're only working in one dimension, imagine exactly the same problem but analyzed by someone who is traveling along a parallel track at some speed $|\vec v_i|$. This observer thinks the initial momentum is not zero, and would describe the "jumping-off" event with the relation

\begin{align} (m_\text{child} + m_A)v_i &= m_\text{child}v_\text{child} + m_A v_A \end{align}

In this formulation, where the momenta on both sides are added, the signs of $v_\text{child}$ and $v_A$ are results that tell you whether the child or the cart are outrunning or being outrun by the observer. However if you change the sign of the term $m_A v_A$ manually, your results will be incorrect except in cases where the boy and the cart end up moving in opposite directions.

$\endgroup$
  • $\begingroup$ Okay, I think I understood. Then, since we are only working in the x direction, we are just using the modulus, which is always positive, right? $\endgroup$ – Bee May 16 '17 at 20:37
  • $\begingroup$ @Bee: Not sure I understand your comment. Because you only consider the $x$-direction (problem is essentially one-dimensional), you can ignore the $y$ and $z$ directions. I would avoid putting things like modulus by hand as this can easily lead to errors in more complicated situations. Start by defining a "coordinate system" (here, e.g. x axis pointing right), then write the equations for the conservation of momentum as vector equations (you should only have + signs). Do the calculation and if you end up with a negative/positive velocity it means that the velocity is pointing left/right. $\endgroup$ – user1583209 May 16 '17 at 20:49
  • $\begingroup$ @Bee No, that's an oversimplification too. I've added a counterexample. $\endgroup$ – rob May 16 '17 at 21:08
  • $\begingroup$ Thank you to both of you!! I understand what you mean with the observer example, but what I don't understand about what @user1583209 said is that, aren't vectors supposed to always carry their sign? If so, how can I only have positive signs? $\endgroup$ – Bee May 16 '17 at 21:16
  • 1
    $\begingroup$ As you saw in Farcher's answer, @Bee: it's a choice to have "the positive direction" go the same way for all the vectors. This is usually the better choice. $\endgroup$ – rob May 17 '17 at 6:56
2
$\begingroup$

My question is, why is this not $0=mv-m_Av_A$ since the boy and the cart have opposite signed directions for their velocities?
Why are the signs taken into account in the first equation and not in the second one?

The answer is that it could be $0=mv-m_Av_A$ but you would have had be consistent when setting up your equations.

Let $\hat i$ represent the unit vector in the left direction.

The relative velocity equation is:

the velocity of the boy relative to the ground, $\vec v$, is equal to the velocity of the boy relative to cart $A$, $\vec v'$ plus the velocity of cart $A$ relative to the ground, $\vec v_{\rm A}$.

You chose to make $\vec v = v \hat i, \vec v' = v' \hat i$ and $\vec v_{\rm A} = v_{\rm A} \hat i$ where $v, v'$ and $v_{\rm A}$ are componets of the velocity vectors.

$\vec v = \vec v' + \vec v_{\rm A} \Rightarrow v \hat i= v' \hat i+ v_{\rm A} \hat i \Rightarrow v = v' + v_{\rm A} \Rightarrow v-v_{\rm A} = v'$

Now apply conservation of linear momentum.

$\vec 0 = m \vec v + m_{\rm A} \vec v_{\rm A} \Rightarrow 0 \hat i = m v \hat i + m_{\rm A} v_{\rm A} \hat i \Rightarrow 0 = m v + m_{\rm A} v_{\rm A} $

Note that it was your choice to make $\vec v_{\rm A} = v_{\rm A} \hat i$ and you must therefore not change your mind when applying your choice to the linear momentum equation.

Not surprisingly when you do the sums you will find that $v_{\rm A}$ is negative.

I have been long winded in my explanation so far because I want to show you that you could have written your conservation of momentum equation as $0 = m v - m_{\rm A} v_{\rm A} $.

To do that you would have had to choose that $\vec v_{\rm A} = - v_{\rm A} \hat i$ and there is nothing wrong with doing that because if you follow this through correctly you will get the correct positive value for $v_{\rm A}$.
The consistent application of your choice that $\vec v_{\rm A} = - v_{\rm A} \hat i$ comes in using the relative velocity equation as follows.

$\vec v = \vec v' + \vec v_{\rm A} \Rightarrow v \hat i= v' \hat i+ (-v_{\rm A}) \hat i \Rightarrow v = v' - v_{\rm A} \Rightarrow v+v_{\rm A} = v'$

Solving for $v_{\rm A}$ give you the correct positive value.

In such problems it is probably better/easier/more likely to get the correct answer? to use $\vec v_{\rm A} = v_{\rm A} \hat i$ rather than $\vec v_{\rm A} = - v_{\rm A} \hat i$ although I hope I have shown that it is not essential.

After all when you do mechanics problems where you have a falling mass being pulled back by a string in tension I would guess that you tend to use $mg-T$ where the tension $T$ comes out to be a positive quantity rather than $mg+T$ which results in $T$ being a negative quantity?

$\endgroup$
  • $\begingroup$ Thank you so much!! I think my mistake was with the concept of relative motion, since we did not see it in class and I just went with my intuition not really understanding what I was plugging into the equation! Your explanation is perfect. $\endgroup$ – Bee May 17 '17 at 6:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.