# What allows a pull-back toy car to drive further than it was pushed?

Imagine you have a pull back toy car. Its back part is on $$x_0$$. You push it down and move it in the back direction to the point $$y$$ (not marked): Then you leave the car to move away: Then you mark the final position by $$x_1$$: Let's say, that the distance you have pushed the car is $$d_1$$ and the distance the car has travelled is $$d_2$$.

As you can see, the car has travelled more than before (same as $$d_1 < d_2$$). If you don't believe it, try it yourself. Why has that happened? The law of conservation of the energy tells us, that energy can't be created out of anything.

What has happened?

• upsbatterycenter.com/blog/pull-back-toy-motor-work And interesting question btw...! Mar 22, 2020 at 19:52
• Great drawing skills! I approve. And nice question too. Mar 22, 2020 at 20:41
• I think some context is missing for this question. E.g. what's the flywheel you refer to in the first sentence? And what is making the car move? Mar 23, 2020 at 5:24
• @DavidZ: it looks like this is the kind of car toy you push back, pressing firmly to the ground. It stores energy as you push it and then you let i got and wheeeeee it rushes forward. At least this was the toy we got with my brother for Christmas in 1978.
– WoJ
Mar 23, 2020 at 8:24
• Since there's been no clarification about that, I removed the reference to flywheels. The question seems to be perfectly understandable without it. Mar 24, 2020 at 21:19

Are you are expecting that if you roll the car backwards $$30\ \mathrm{cm}$$ then release it, it should move forward $$30\ \mathrm{cm}$$? Why? Most toy cars wouldn't move at all.

If you put a stone in a catapult, pull it back $$30\ \mathrm{cm}$$ then release it, it goes forward much further than $$30\ \mathrm{cm}$$. If you did this in empty space the stone would keep going indefinitely.

Energy hasn't been created out of nothing. You have done work against the catapult, storing elastic energy. When you release the catapult the stored elastic energy is transformed into the kinetic energy of the stone, which is dissipated as heat and sound as the stone flies through the air and hits a target. If there is no air resistance or friction, and nothing impedes the stone, its kinetic energy remains constant forever – its speed doesn't change, it goes infinitely more than $$30\ \mathrm{cm}$$.

The toy car is the same. Instead of an elastic band, it contains a spring. Pushing down engages a gear wheel. As you push the toy car backwards you wind up the spring quickly using a relatively large force. You do work, elastic energy is stored in the spring. When the car is released, it springs back up and a different gear wheel is engaged. Now the spring unwinds itself slowly, supplying a much smaller force to the toy car. (See Note.) Elastic energy is transformed into the kinetic energy of the car, which is dissipated by friction. The car loses its kinetic energy gradually; it slows down and stops. If there was no friction the car would keep going indefinitely on a flat surface.

It is not the distances which you need to compare but the work done, which is force times distance. You give the car elastic potential energy by pushing with a large force over a short distance. The much smaller force of friction takes that energy away over a much longer distance after it has been transformed into kinetic energy.

Suppose the friction force is $$0.1\ \mathrm N$$ and you push the car backwards with a force of $$5.1\ \mathrm N$$ through a distance of $$30\ \mathrm{cm}$$. Then you have done $$5.1\ \mathrm N \times 0.3\ \mathrm m = 1.53\ \mathrm{Nm}$$ of work. Friction works in both directions so $$0.1\ \mathrm N \times 0.3\ \mathrm m = 0.03\ \mathrm{Nm}$$ of the work you do is wasted pushing against friction. The remaining $$1.50\ \mathrm{Nm}$$ of energy gets stored in the spring. When the car is released it is transformed into the kinetic energy of the car. The friction force of $$0.1\ \mathrm N$$ slows the car. You can expect the car to go a distance of $$15\ \mathrm m$$ before stopping because $$0.1\ \mathrm N \times 15\ \mathrm m = 1.5\ \mathrm{Nm}$$.

The car goes $$50$$ times further forwards than you moved it backwards. But you haven't created any energy. In fact, some energy was lost pushing against friction. Only $$1.50\ \mathrm{Nm}$$ of the $$1.53\ \mathrm{Nm}$$ of energy which you supplied was used to move the car forward.

Note: When the spring is fully unwound it is disengaged from the wheels so that the car rolls forward freely instead of winding the spring back up. That's like the catapult which releases the stone; otherwise the stone would stretch the elastic again and keep oscillating until its kinetic energy was used up.

• A better example might be a bicycle wheel. Turn the bike upside down, as if you were changing a tire, and give the front wheel a little shove, moving the rim maybe 1/20 of a meter. It will keep rotating for quite a while (at least if your bearings are in good shape and your brakes aren't rubbing), with the rim travelling a distance of many meters. Mar 23, 2020 at 16:56
• I think you need to bold the point that the gearing to the spring is different when releasing. I'd never concidered this before, and didn't realise it was using different gearing in reverse, to forward (otherwise you coud push it forward, and it'd shoot backwards too!) Mar 23, 2020 at 17:51
• @jamesqf Thanks. Yes that is a simpler example than the catapult but it is more difficult to compare that with the motion of the car. Mar 23, 2020 at 20:27
• I've played with cars that have incorrectly made flywheels that releases the energy in a quick burst; they barely go anywhere because the force is too high; it instantly overcomes the wheel's static friction against the surface and can even flip the car over. Mar 24, 2020 at 3:12
• @djsmiley2kStaysInside: only powering the wheels in one direction could be achieve with a simpler ratchet (which it already has so you can't over-unwind the spring). Separate gearing matters because of practical considerations like limited traction when only its own weight is pressing it down. And because accelerating with the full pull-back force for only ~30cm would make it go out of control from small variations in traction / balance leading to turning. Plus turning all that energy into kinetic right away, then coasting, wouldn't work well. Mar 24, 2020 at 14:12

It works kind of like the sketch below. When you push down and backwards on the car, a high ratio gear meshes that winds the spring, so a few inches backward push turns the car wheels back a few turns and winds the spring several turns. When you release the car, the body lifts. This un-meshes the first gear and meshes a low ratio gear instead, so several unwinding turns of the spring results in many forwards turns of the wheels. Ignore the belt drive I have inserted between the gear usually both gears mesh with just one drive wheel - but that would have made my sketch messy and harder to understand. Maybe a bit like this in reality. • Good description of what's happening inside the car, but the gearing ratio is completely irrelevant to the fact that the car can move forward farther than you pull it back. You can achieve the same thing just by wrapping a rubber band around an axle with no gears at all. Mar 23, 2020 at 14:04
• some cars don't have the push down action and there's some kind of a ratchet like setup that does the gear engagement instead Mar 24, 2020 at 9:59
• @NuclearWang The different gearing ratio is not completely irrelevant. Without it, the car would stop powering itself at the point where we started pushing it backwards and then just roll a bit further until it stops. However, it is easy to observe that this is not what happens, so the simpler answer would just seem wrong and require an immediate follow-up question.
– JiK
Mar 24, 2020 at 15:16
• @JiK I agree the gearing ratio helps to overcome the real mechanical inefficiencies of a cheap toy car to give it more "zip". But in principle, if the car could coast with very low friction, you could even reverse the gearing ratio and have it work - the drive would disengage before where it started, but the car would continue to roll an arbitrarily far distance. Mar 24, 2020 at 15:29
• Regarding gear ratios. In my experience (a few years out of date as my son is past the age where toy cars were a thing), there are two types of pull-back car. In one, the pull-back stores energy in the spring and then releases it fairly quickly when the car is released. In these cars, most of the forwards motion is just coasting, after the energy has been transferred back through the wheels. In the second, the gear ratio is significant as the car is driven for a significant distance.The second version is more fun as the car can climb obstacles incommensurate with its apparent forward momentum. Mar 25, 2020 at 0:55

tl;dr The toy car can go forward longer because it's not being resisted by an equal resistive force. By contrast, pendulums are resisted by an equal resistive force, so they'll go no further than you pull them back.

# It's not a pendulum.

Imagine a pendulum hanging still at the center. That's like the toy car.

If you pull back the pendulum, that's like winding up the toy car. And if you release it, it'll launch forward, like the toy car.

The pendulum will go no further past the center than you pulled it back, because it's storing up gravitational potential energy as it goes along. The toy car, by contrast, isn't storing energy back up in a second spring to launch backwards in the opposite direction; it's just letting its kinetic energy carry it.

• It's more like a pendulum where the bob is released at the bottom of the back-swing and thus continues to roll a long way (no longer trying to go back up the gravity well) Mar 25, 2020 at 12:09

I think a few details are hidden in unclarity of the question. Very nice drawing skills indeed though!

### Detail 1

1. A flywheel is a wheel with mass that starts rotating, and stores its energy as kinetic energy (rotating energy).
2. A torsion spring stores its energy as mechanical energy by winding up the spring. This potential energy.

I think the car you talk about, which goes forward when pulled back, does not use a flywheel, but actually uses a torsion spring. as shown in the link provided by Joshua Ronis in the comments.

### Detail 2

The car you sketched moves along a horizontal surface. That means it does not gain any gravitational potential energy when it moves forward (translates). (It does lose a bit/all of its energy on: air-friction(mainly), rolling friction(mainly), creating noise(small), and I think temperature radiation (small).

So for example, if there was no friction anymore once the car has accelerated in forward direction, the car would just keep on travelling forward forever, (a bit like satellites in high orbits that appear to go on forever, since they experience very little friction (when all they needed was an initial push by a rocket when they get up there {in reality "getting up there and giving the initial push" is usually mingled for energy efficiency though}).

Therefore, the car can indeed travel beyond its initial point of push back. What that implies is that the friction forces it experiences are lower than the forces that are generated by the flywheel/torsion springs potential energy(over the distance up to the starting point). The forces that can be created by the flywheel/torsion springs (over the distance up to the starting point) must be lower than the forces you put on it with your hand when rolling it backwards (due to the conservation of energy and real-life {mechanical} energy translation losses).

This could be mathematically described with: $$s=\frac{1}{2}\cdot a\cdot t^2$$ where:

• $$s$$ = distance travelled by car in $$m$$
• $$a$$ = acceleration in $$\frac{m}{s^2}$$ (coming from $$f-d=m\cdot a$$)
• $$f-d$$ = the accelerating force in newtons created from the flywheel/torsional spring - the drag the car experiences from friction.
• $$m$$ = mass of the car in $$kg$$
• $$t$$ = the time in seconds

Which can be rewritten to: $$s=\frac{1}{2}\cdot \frac{f-d}{m}\cdot t^2$$ hence if the force $$f$$ is large enough, and the drag $$d$$ is small enough, $$s$$ will become arbitrarily large if time becomes large enough. (In reality, $$f$$ is a function of time that goes to 0 in a nonlinear way).

• If there were friction, then one couldn't wind it up by rolling it on the floor. Mar 23, 2020 at 21:28
• Thank you, I included a nuance that ensures the no-friction scenario is discussed only after the car has accelerated in forward direction. Since the acceleration forward happens after winding it up, the description allows friction while winding the car up.
– a.t.
Mar 25, 2020 at 13:06

The spring is what stores the energy that you transform from your mechanical energy by pushing back the car.

Now the flywheel needs to be very heavy, actually heavier then the car itself, thus, when you release the car, the spring transforms back the stored potential energy onto the flywheel, starting it to roll.

Now why does the flywheel roll more forward then backwards? It is because it does have inertia. When you push the car back, and load the energy into the spring, you do not use the flywheel (and its inertia) to move the car back at all, you just use your mechanical force.

When you release the car, the flywheel is actually rolling and its inertia is what drags the car forward until the flywheel loses this inertia caused by the slowdown because of friction on the axle (caused really by gravity).