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11+ ENTRANCE EXAMS
Asked by Mike

A toy car slides down a slope. If the top of the slope is 2m higher than the foot of the slope, how fast will the car be moving when it reaches the foot? (Assume that all of its g.p.e is transformed to k.e Conservation of e?

The key to this question lies in the information within the bracket. Some things you must understand before you answer this question are, 1. Energy cannot be created or destroyed, it can only be transferred 2. Kinetic energy stores and GPE stores. What do I mean by point 1. Well if energy could be created or destroyed, we would never know how much energy the car has at the bottom! The second thing you need to understand is that when it is at the top of the slope, it is at a height and so will have GPE. How much you ask? Well you can work it out using the formula GPE = mgh. In this question they haven't given us the mass and you will see why soon. Next, you must know that KE depends on the speed of the car. The higher the speed, the higher the kinetic energy. Now you must think where does it get this kinetic energy from? Is there anyone applying a force after it has started rolling? NO! When it starts rolling down, the gpe is being CONVERTED into ke. How much of it is converted? All of it! Therefore, if GPE = KE then we can say mgh=½mv². Finally we can work out the speed which in this equation is v. If we multiply both sides by 2 we get 2mgh=mv². Now if we divide both sides by m, we get 2gh= v². Can you see there is no m left in the equation. That means the velocity at the bottom of the ramp doesn't depend on the mass! You can check this yourself by sliding a car down a ramp, and adding weights on. You should notice the speed remains roughly the same. Now finally back to the equation, if we share root both side, we end up with v=sqrt(2gh) (g on earth is 9.81). Now if we substitute h as 2 then we get an answer of 6.26 meters per second. I understand that was a very long explanation but if you understood everything above, you are close to mastering this topic! I hope you found this helpful and I am willing to answer any question you may have.

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