Alex Boorman Ph10 - -

Aim: To find out what affects the height to which a ball bounces.

Variables:

Height from which the ball is dropped

Mass of the ball

Material ball is made from

External factors, i.e. changing air density, temperature

The surface onto which the ball falls

Prediction – reasons for variable control:

Variables that will be altered:

The height the ball is dropped from will affect the height the ball bounces to due to the energy chain the ball goes through as it is dropped and bounces up again. The energy chain is as follows:

Therefore as energy cannot be created or destroyed the energy the ball starts with must be directly proportional to the energy the ball finishes with, at the top of its bounce, and so if the ball starts with more energy it must therefore finish with more. As in both cases the main form of energy is GPE it follows that the higher the ball is dropped from, the higher the ball will bounce.

- The ball starts with more GPE
- As there is more GPE more energy is converted into KE (The ball is going faster, KE=1/2mv2, v is greater therefore KE is greater by a larger amount). More thermal energy is also produced.
- As there is more KE, more energy is converted into elastic potential energy
- As there is more elastic potential energy, more energy is converted back into KE energy
- As there is more KE energy, more work would need to act upon the ball in order to stop it in the same distance. As gravity remains the same the amount of work acting on the ball remains the same (apart from slightly more air resistance due to the ball travelling faster and so hitting more air particles per second, but the effect of this is negligible) and the ball travels further before stopping. Therefore the ball goes higher.
- The higher the ball goes, the more GPE it ends up with. Therefore the ball ends up with more GPE

In short:

GPE=Mass (kg) × Gravitational Field Strength (N/Kg) × Height (m)

If we let mass = m

Gravitational Field Strength = g

Height = h1

An increase in h1, assuming g and m stay constant, results in an increase in m × g × h1 which results in an increase in GPE.

A decrease in h1, assuming g and m stay constant, results in a decrease in m × g × h1 which results in a decrease in GPE.

Therefore any change to the height from which the ball starts with affects the height to which it bounces to.

This is correct unless the ball shows signs of reaching terminal velocity.

The drag force increases as the ball goes faster. This is because as the ball goes faster it passes through a greater volume of air each second. It therefore hits more air particles each second and so the force of drag is bigger the faster the ball goes.

As the ball is accelerating due to gravity, at 9.81m/s2 it is constantly getting faster and therefore the drag force gets bigger and bigger. Drag is a squared function of velocity and therefore as the ball drops drag increases a greater amount each second. Once the drag force equals the gravitational force all forces are equal and acceleration stops. The ball has reached its terminal velocity and cannot fall any faster (unless dropped in a vacuum).

If a ball reaches terminal velocity at 20cm from the floor when dropped from 2m, it will reach terminal velocity at 10cm from the floor when dropped from 1.9m. This means it will be travelling the same speed when it hits the ground. This means that KE is the same for both balls when each ball hits the ground.

If the KE is the same as they hit the ground the energy stored in the ball as elastic potential energy will be the same also. If the elastic potential energy is the same then the same amount of energy is converted back into KE and so the balls leave the floor at the same speed. The balls leave the floor at the same speed with the same amount of KE and so both balls reach the same height and end up with the same amount of GPE at the top of their bounces.

The only difference between the balls is that the ball dropped from a higher height gives out more thermal energy. This is because the ball starts with more GPE. The balls finish up with the same amount of energy and the only energy given out is thermal energy. After the ball reaches terminal velocity, no more GPE is converted into KE is the ball cannot get any faster. Instead, as the ball is still always loosing GPE so long as it is still falling, all GPE is converted into thermal energy. Therefore the ball dropped from the higher height must give out more thermal energy in order to end up with the same amount of energy as the other ball.

Therefore the height the ball bounces will be proportional to the height that the ball is dropped from up to a certain point, where the ball begins to show signs of reaching its terminal velocity before it reaches the ground. Above this point the height all balls will bounce to will not be directly proportional to the height they are dropped from, but the increase in the height they bounce to will increase more slowly in proportion to the height they are dropped from compared to the increase between lower heights before the ball shows signs of approaching its terminal velocity before it hits the ground. I do not believe that any ball will reach terminal velocity in this experiment seeing as the maximum height that they can be dropped from is 2m and, as the ball is quite smooth, I do not believe that it will have enough time to accelerate to its terminal velocity before it hits the floor.

Prediction curve:

Variables that will be kept constant:

N.B:

h1 = The distance between the bottom of the ball before it is dropped and the ground.

h2 = The distance between the bottom of the ball at the top of its arc after bouncing and the ground.