As simple as this question might seem, I failed to intuitively answer it.

Let's assume there is a $10,000$ $kg$ truck moving at $1$ $m/s$, so its momentum and KE are: $p=10,000$ $kg.m/s$ and $KE=5,000$ $J$.

Now if we want to stop this truck, we can simply send another truck with the same mass and velocity in the opposite direction to collide with the first one, and both of them will stop because of the conservation of momentum.

But what if we want to stop the truck by doing work in the opposite direction of motion ?

Let's assume there is a rope with an end tied to the back of the the truck, and the other end is tied to a $400$ $kg$ motorcycle moving at $5$ $m/s$, then its $p=2,000$ $kg.m/s$ and $KE=5,000$ $J$.

Now we have a truck moving in one direction with the same kinetic energy as the motorcycle which is moving in the opposite direction, but the truck has more momentum. So will the truck stop by the energy (or work) of the motorcycle ?

If yes, then how is the momentum conserved, and if no, then where does the energy of the motorcycle go?

Ignore any friction forces.

  • $\begingroup$ There is insufficient momentum in the motorcycle to stop the truck, so it will keep moving, albeit slower. What does that mean for the remaining energy? $\endgroup$
    – CuriousOne
    Oct 17, 2014 at 23:13
  • $\begingroup$ Just an observation: But what if we want to stop the truck by doing work in the opposite direction of motion? well, that's exactly what happened in the first scenario too! $\endgroup$ Nov 22, 2014 at 14:22

5 Answers 5


The other answers are great. I decided to plot it, however, because it's nice visualizing these things. Since your biggest doubt is about kinetic energy, be sure to pay attention to the last graph.

SYSTEM. Motorcycle going to the left, truck going to the right, bound by an elastic rope ten meters long ($k=100\frac{\mathrm N}{\mathrm m}$). Masses and speeds are as per OP. Note: the rope is not a spring. A spring pushes when compressed. The rope simply gets loose (i.e. it has a deadband).


position plot

Points of note:

  • 0.0 s: bike and truck start together.
  • 1.7 s: rope stretches, beginning to pull vehicles together.
  • 7.9 s: rope lets go of vehicles, sending bike the other way.
  • 9.5 s: bike passes truck
  • 11 s: rope stretches again, starting to send bike the initial direction.

This will repeat forever, as there's no energy dissipation in this model.

The rope's actions are more clearly visible in the speed plot.


speed plot

As the truck is much more massive than the bike, it barely feels the rope's actions. The bike, however, is being tossed all over.


momentum plot

As everyone said, total momentum is always conserved.


energy plot

Some things are really interesting here, and they show how tricky it is to rely on energy instead of momentum for accounting of such situations.

  • The bike's KE varies much more than the truck's. This is the core of this question: while the variation of momentum of one is always the same as the other's, the same can't be said about energy. Energy is always being transferred to other places, for instance:
  • Energy is shared between vehicles and rope. But its sum is always the same, as long as you don't forget the rope.
  • One could wrongly say "the total energy is constant only because the scenario is perfectly elastic." No, the total energy would be constant even if it was an inelastic direct collision. In that case, the purple graph would be potential energy of the vehicles' deformations, perhaps later becoming heat lost to surroundings, but either way: total energy would still conserve, only not as KE.
  • If this was an elastic direct collision instead of a rope scenario, the graph would still be the same, stopping at 9s. In that case, the purple graph would be potential energy of the vehicles' deformations, which would be zero at the end. Thus, not only total energy would remain constant, but total KE would too (which is what the special case elastic collision is all about).

Note: in an inelastic direct collision scenario, the graph would stop at ~4.75s, where vehicle speeds are equal (it's a bit after the bike's zero energy point; it would have ~100J of KE).


Being $x_1$ the bike's position, $x_2$ the truck's and $L$ the rope's length, Hooke's Law gives: $$|F|=k\left(\left|x_2-x_1\right|-L\right)$$

Note the deadband has appeared already. Signs must be handled with care.

Being $m_t$ the mass of the truck and $m_m$ the motorcycle's, we have:

$$\ddot x_1 = \frac{k}{m_m}\left(x_2-x_1\right)$$ $$\ddot x_2 = \frac{k}{m_t}\left(x_1-x_2\right)$$

Since I don't remember if I ever knew how to solve a system of ODE's analitically, I fired up Xcos (Scilab's former Scicos) and modeled it there. In case anyone's curious, here's how the diagram ended up. I can send the .zcos file if anyone wants it.


Regarding scenario 1:

we can simply send another truck with the same mass and velocity in the opposite direction to collide with the first one, and both of them will stop because of the conservation of momentum.

If one assumes a totally inelastic collision then, by conservation of momentum, it is true that both trucks (objects) stop.

This requires that the initial kinetic energy of the objects is entirely converted to another form. The total change in kinetic energy equals the total work that is done during the collision process.

So, to be sure, work is done during the collision process since the trucks (objects) are, shall we say, permanently deformed.

I assume you know this already and have taken this into account.

Regarding scenario 2:

If we assume that the rope can stretch, dissipating energy in the process, then we have the following results:

(1) eventually, the truck and motorcycle have the same velocity which is equal to

$$v_f = \frac{(10,000 - 2,000)\mathrm{\frac{kg\cdot m}{s}}}{10,400 \mathrm{kg}} = 0.769\mathrm{\frac{m}{s}}$$

(2) the final kinetic energy is

$$KE_f = \frac{1}{2}10,400 \; \mathrm{kg}\cdot \left(0.769\mathrm{\frac{m}{s}}\right)^2 = 3,077 \mathrm{J}$$

so the work done (energy dissipated) by the rope is

$$W = \Delta KE = 10,000 - 3,077 = 6,923 J $$


A momentum-based analysis is the way to go for the motorcycle-rope-truck scenario.

In your kinetic energy argument, you are assuming that kinetic energies add like vectors. This is not the case.

If you want to properly apply a kinetic-energy-work argument, you need to think about the force $F$ that the rope exerts on the truck and the distance $d$ over which this force acts. Only by doing this will your momentum and kinetic energy methods agree on the answer. (Note that this ignores any energy-storing capability of the rope.)


The truck keeps going. Assuming the rope does not break, then the kinetic energy ends up in elastic potential energy in the rope. The rope will stretch. If you ask what happens if you assume a rope that cannot stretch, bzzt. No such thing. The rope will either stretch or break.

  • $\begingroup$ Someone downvoted this? It might not be the best answer, but it's not wrong. $\endgroup$
    – garyp
    Oct 18, 2014 at 15:34

The motorcicle will reduce the speed of the truck a little bit, because of the work made by the tension of the rope. The motorcicle will also slow down and actually reverse direction until it is the same than that of the truck. At that point they all will keep moving at the same speed without interacting any longer (the rope tension is zero). Momentum is conserved and kinetic energy somehow dissipated, because it is not an elastic interaction (even if the rope is elastic, otherwise the system will keep oscillating back and forth, although moving forward as a whole, which is another possible answer, depending on your assumptions).


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