How to calculate KE using only mass and height
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. 483
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 <PERSON> 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. 1023