Imagine you have two carts gliding on a frictionless rail, sliding to the right at $1.0\, m/s$. The spring between them has a spring constant of $110\, N/m$ and is compressed $4.4\, cm$ . The lighter cart is $0.1\, kg$ while the heavier cart is $0.3\, kg$. The carts slide past a flame that burns through the string holding them together. Prove that the lighter cart moving to the left after hitting the flame while the heavier cart is moving to the right. Assume that the $0.1\, kg$ cart is traveling behind the $0.3\, kg$ cart.

Here's what I've done so far:

I've set up two systems of equations showing the conservation of energy and momentum:

$$ (m_1 + m_2)v=m_{1}v_{1} + m_{2}v_{2}\\ (m_1 + m_2)v_{0}^2/2+kx^2/2=m_{1}v_{1}^2/2 + m_{2}v_{2}^2/2 $$

Let $m1$ be $0.1\, kg$ and $m2$ be $0.3\, kg$. Plugging this into the conservation of energy equation gives me the following equation:


I then take this and I substitute it into the v2 of my conservation of energy equation:

$$ 0.1/2 v_{1}^2/2+ 0.3/2 v_{2}^2/2$$

This becomes a quadratic equation and I get the zeroes for v1 as $-2.263$ and $0.2602$. I don't think these are correct and I'm not sure where I messed up. On the other hand how will I know which zero to accept and which one to reject. I mean obviously question tells me but assume it didn't and I was asked: "which one's moving left, which one's moving right, and what velocity?" How will I know then?

  • $\begingroup$ The question of which zero to accept boils down to a matter of "which cart is in front". If the lighter cart (with mass $m_1$) is initially traveling in front of the heavy cart, then we should have $v_1 > v_0$. If the lighter cart is traveling behind the heavy cart, then we should have $v_1 < v_0$. The equations that you wrote down do not encode any information about which cart is in front, so it makes sense that there are two solutions. $\endgroup$ – Jasha Nov 3 '17 at 3:49
  • $\begingroup$ I understand. I set it up so that the lighter one is traveling behind the heavier one. $\endgroup$ – cambelot Nov 3 '17 at 4:01
  • 2
    $\begingroup$ @Jasha. There is no before and after. They are side to side, left and right. $\endgroup$ – William Elliot Nov 3 '17 at 4:35
  • $\begingroup$ actually, the question boils down to a lot more than that. It boils down to how things are set up exactly in the beginning, and also what "after hitting the flame" means. If I understand correctly, we have both a spring and a string, the spring compressed (and the string keeping it that way until it is burned). So immediately after the burn, they are both moving right! What happens after, if the spring is attached to them, is oscillation superimposed on top of the motion to the right, so proving that one is moving either way is meaningless unless you specify at what moment. $\endgroup$ – Nick Pavlov Nov 3 '17 at 11:26
  • $\begingroup$ @WilliamElliot, there is before and after. What the OP does with the conservation laws finds also what would have led up to the initial situation with the spring compressed, if there had been no string at all: the two objects moving towards each other, or the lighter one catching up with the heavier one, then they collide and the spring starts compressing, until their velocities become equal. It is is exactly this "hypothetical past" that one of the roots can be viewed to represent. $\endgroup$ – Nick Pavlov Nov 3 '17 at 12:55

I haven't gone through the numbers to see if your answers are correct or not. This answer is mainly to address your confusion about the two roots. It's possible to interpret both of them by asking a related question:

Imagine you have two carts moving towards each other on a frictionless rail. Attached to a bumper of one of the carts is a spring with a stiffness of $110\, N/m$. The two carts collide, compressing the spring. At the moment when the spring is most compressed, it is compressed by $4.4\, cm$, and both cars are sliding to the right at the same speed of $1.0\, m/s$. The lighter cart is $0.1\, kg$ while the heavier cart is $0.3\, kg$. What were the carts' velocities before and after the collision?

If you try to solve this problem, you'll find that the equations you write down are exactly the same as the ones you have above! However, in this case, there's a more natural interpretation: one of the solutions you get for $\{v_1, v_2 \}$ gives you the speed of the carts before the collision, and the other gives you the speed after. To distinguish between these two, you just need to figure out which solution involves the carts approaching each other, and which solution involves them moving away from each other

Moreover, the "after" of my problem is exactly the same as yours. Imagine that you turned on the lights in the room the instant after the string was broken (and ignored the trailing bits of string and the flame, etc.) From the perspective of the motion of the carts, there is no difference between the motion after the string breaks in your problem, and the motion after the spring is maximally compressed in my problem. Thus, the solution for $v_1$ and $v_2$ that corresponds to the "after" of my collision problem is the meaningful one for your problem.


In order to explain why you get two roots, and how we know which is the "right" one, let me consider the system of two carts with the spring attached to them at both ends (I have reason to believe that this is not what the question intended, but it is equivalent to what you are doing with the conservation laws). Let $M = m_1 + m_2$ be the total mass of the carts and $t = 0$ be the moment the string gets burned and so the spring starts to uncompress. Going straight to Newton's laws, we then have a pair of differential equations: $$ \begin{align} m_1\ddot x_1 &= k(x_2 - x_1)\\ m_2\ddot x_2 &= k(x_1 - x_2) \end{align} $$ This can be easily decoupled by rewriting in terms of $X = (m_1/M)x_1 + (m_2/M)x_2$ (this is the position of center of mass) and $Y = x_2-x_1$ (the distance between the carts). The first one obeys $\ddot X = 0$, which means the center of mass keeps going right at constant velocity (equal to the initial $v$) and expresses the same thing as conservation of momentum. The second one obeys $\ddot Y + \omega^2 Y = 0$ where $\omega$ is a quantity that can be expressed in terms of the masses and the spring constant (it is the angular frequency of oscillations of the system of carts connected by the spring). This is the equation of simple harmonic motion, and if we put in it the initial conditions $Y(0) = -4.4$ and $\dot Y(0) = 0$ (because $\dot Y = v_2(t) - v_1(t)$ and at $t=0$, $v_1 = v_2$) we can get $$ Y(t) = x_2(t) - x_1(t) = -4.4\cos\omega t $$ Combining this together with $X(t) = vt$ we can obtain explicit forms for $x_1(t)$ and $x_2(t)$, and also for the velocities.

Now I can draw a parallel to your approach, and show how it is different. What you have done with the conservation of energy amounts to asking, "what are the velocities at the moment when the energy stored in the spring is released, that is, at the moment when the spring is at its normal length?" In the oscillating system, in one cycle there are two such moments - when the spring reaches zero elongation while getting longer (which happens at $t_1 = T/4 = \pi/2\omega$, and while getting shorter (at $t_3 = 3T/4 = 3\pi/2\omega$, after having stretched to a maximum positive elongation at $t_2 = T/2 = \pi/\omega$). The two roots you have found correspond to the velocities of the objects at $t_1$ and $t_3$, respectively: both of these satisfy the exact same conservation law equations. Once the first of these happens, I presume what was intended is that the spring, which is not actually attached, "drops" since nothing is pushing on it from the sides any more, and after that each cart, without any forces acting on it, just keeps going with the velocity it had at that moment. So it really is about which happens first, but your approach doesn't contain any explicit information about time, so it is not immediately obvious which is the root you want.

The conservation laws approach is similar to thinking of the process as a collision. But you have to realize that in a collision, there are two sets of solutions satisfying the equations - the values before the collision, and the values after (in your problem, the initial moment is actually in between those - it is the moment of maximum deformation). If you want to be certain to distinguish them, you could write the momentum conservation law as a pair of equations like this $$ \begin{align} m_1 v_1 - m_1 v &= -J\\ m_2 v_2 - m_2 v &= J \end{align} $$ (or if, in general the initial velocities are different, replace $v$ with $u_1$ and $u_2$ accordingly). The quantity $J$ in these equations is the impulse of the force by which the colliding objects interact (it is just an integral of the force over time), so the equal magnitude and opposite signs of the right-hand-sides are an expression of Newton's third law. If you add them up, you get conservation of momentum. But, if you get two roots, one of them will correspond to positive $J$ (meaning the $m_2$ object was pushed/pulled to the right), while the other to negative $J$ ($m_2$ pushed/pulled to the left). We know which one we need here, since while uncompressing, the spring pushes outward.


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