For our physics (high school level) group project, we were asked to modify a selected toy so that it produces an energy output of at least or roughly 110% (10% more) compared to the original toy.

Our toy is a pull-back toy (you can find it by searching "McDonald's Happy Meal Snoopy Toy" and it's the one with the girl on the sledge). Its mass is 234g and we will be modifying the toy by removing the female character on it. This will reduce the mass to 213g and this should theoretically increase the speed and in turn the energy output.

We tested the toy by placing it on a table with a height of 2m. We marked a starting line, and pulled the toy back 30cm to another marked line, repeating thrice (3 times) and then releasing. We repeated the whole process 3 times and the average distance travelled was 2.23m. The average time travelled (braking naturally) was 5.1s. Therefore:

$$V = \frac{distance}{time}$$ $$V = \frac{2.23}{5.1}$$ $$V \approx 0.437m/s$$

As part of the task, we were required to calculate the gravitational potential energy (GPE) and kinetic energy (KE) of both toys (control/original and modified), all calculations in Joules (not Newtons) to justify our modification. Here are our calculations:

GPE of control toy:

$$GPE = mgh$$ $$GPE = 0.234 \times{9.8} \times{2}$$ $$GPE = 4.5864J$$

KE of control toy:

$$KE = \frac{mv^{2}}{2}$$ $$KE = \frac{0.234 \times{0.437^{2}}}{2}$$ $$KE = 0.022343373J$$

GPE of modified toy:

$$GPE = mgh$$ $$GPE = 0.213 \times{9.8} \times{2}$$ $$GPE = 4.1748J$$

The question I have is: how do I calculate the KE this toy would have? We are required to calculate this without actually conducting tests to find the real result. I can already complete the following:

KE of modified toy:

$$KE = \frac{mv^{2}}{2}$$ $$KE = \frac{0.213 \times{?^{2}}}{2}$$

The issue is the velocity. I only have the values for mass and height, and the result I get using the derived formula from the law of conservation of energy is the same as the GPE value. Is this what is to be expected or not? And if it isn't, what formula do I use?

Given that formula ($V = \sqrt{2gh}$), I can calculate the following:

$$V = \sqrt{2gh}$$ $$V = \sqrt{2 \times{9.8} \times{2}}$$ $$V = \sqrt{39.2}$$ $$V \approx 6.261m/s$$

So subsituting this into the kinetic energy formula, this is what I get:

$$KE = \frac{mv^{2}}{2}$$ $$KE = \frac{0.213 \times{6.261^{2}}}{2}$$ $$KE = \frac{0.213 \times{39.2}}{2}$$ $$KE = 4.1748J = GPE$$

Solution (solved by @Steeven on 6 September 2016 at 8:05:19am GMT+0):

Using the formula for conservation of momentum, we can calculate the following:

$$\sum{p_{before}} = \sum{p_{after}}$$ $$p_{before} = p_{after}$$

And therefore $p_{before} = p_{after} ⇔ m_{1} v_{1}=m_{2} v_{2}$ so we can solve for $V_{2}$ as below:

$$V_{2} = \frac{m_{1}v_{1}}{m_{2}}$$ $$V_{2} = \frac{0.234 \times{0.437}}{0.213}$$ $$V_{2} \approx 0.48m/s$$

And the next step is to input this value into the kinetic energy formula as follows:

$$KE = \frac{mv^{2}}{2}$$ $$KE = \frac{0.213 \times{0.48^{2}}}{2}$$ $$KE = 0.0245376J$$

Now to calculate the increase in energy output, just divide modified toy KE by original control toy KE then convert to percentage:

$$Increase = (\frac{KE_{2}}{KE_{1}} - 1) \times100\%$$ $$Increase = (\frac{0.0245376}{0.022343373} - 1) \times100\%$$ $$Increase \approx 9.82\%$$

And 9.82% is quite close to 10%, and is probably close enough (please leave your opinion).

  • 1
    $\begingroup$ Welcome to the Physics Stack Exchange! Please be aware that this not a homework help site, but you'll find that a good number of the members may be willing to help you, especially when you've put some work into your problem as you have. Since this is a homework problem, I've edited your tags to include. For reference you should do the same for any future questions that are homework related. $\endgroup$
    – docscience
    Sep 6, 2016 at 2:42
  • $\begingroup$ BTW it appears you are on the right track. But here's a hint why experimental data may not match your calculated results: there's something missing in the model. Can some of the energy go somewhere else during the fall? $\endgroup$
    – docscience
    Sep 6, 2016 at 2:47

1 Answer 1


You can't use energy conservation when mass is removed. This has required energy that you are not taking into account.

But you can use conservation of momentum. And this is usual, when one speed is unknown, exactly as you have it here.

All moving objects have momentum $p=mv$, and conservation of momentum says:

$$\sum p_{before} =\sum p_{after} $$

So in you case:

$$ p_{before} = p_{after} \quad\Leftrightarrow\quad m_1v_1=m_2v_2$$

The old and new mass as well as old speed are known, so new speed can be calculated.

  • $\begingroup$ Thanks for that @Steeven and I assume p represents momentum? So therefore: old mass x old velocity = new speed x new velocity Is this the correct assumption to make? $\endgroup$ Sep 6, 2016 at 8:15
  • $\begingroup$ As the answer says, $p$ is momentum, and all momentum before equals all after. Not really an assumption - just another basic conservation law which is very useful to know. $\endgroup$
    – Steeven
    Sep 6, 2016 at 8:28

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