Different Surfaces
Balls bounce different heights off of different surfaces. I believe this is because a soft surface absorbs a lot more energy from the ball, thus giving the ball less energy to bounce back with. In contrast, a hard surface would not absorb a lot of energy from the ball, so the ball would have more energy to bounce up. If my theory is true, a ball dropped from a fixed height (i.e. 0.80 metres) would bounce higher from a solid surface (i.e. wood) than from a soft surface (i.e. sponge), and also the ball would bounce for a longer duration of time on a solid surface. I could not test this theory because the time was limited.
Why a Squash Ball Loses Height When it Bounces
Energy Transfer Diagram
KEY
Blue box = Potential Energy
Green box = Kinetic Energy
Red arrow leading to red box = Wasted Energy
1
This is the point at which the ball is being held in the air, it has only got Gravitational Potential Energy because it is still and is being held in the air. The higher the ball is being held the higher the amount of Epg it will have at this point
For example
If the ball was held at 80 cm (or 0.80 m), we would do 0.024 (mass) * 0.239 (Gravity) * 0.80, which gives us the Gravitational Potential Energy; 0.005 joules.
2
This is when the ball is moving through the air (because we let go of it). At this point, the ball has Kinetic Energy. However, some energy is lost as due to friction between the molecules in the air, and the surface of the ball.
3
This is during the time where the ball is in contact with the floor. There are really three stages here, I will show them below:
i) ii) iii)
In ‘i’, the ball has hit the ground, and because of inertia, the ball tries to keep moving and can’t because the ground beneath it is solid. This causes the ball to change to a sort of ‘oval’ shape, this change of shape causes some energy to be lost as heat and the kinetic energy to become Elastic Energy. Also, the ball hitting the ground will cause some energy to go on as sound and some will be sent through the surface as a wave.
In ‘ii’, the ball is still, and has no energy other than Elastic Energy; it is exactly between ‘i’ and ‘iii’.
In ‘iii’, The Elastic Energy is being converted to Kinetic Energy, and causes the ball to go from the ‘oval’ shape, back to its original shape, and bounces off of the ground. The Elastic Energy in the ball is now becoming Kinetic Energy again and the reshaping of the ball causes some more energy to be lost as heat.
4
Here the ball is going back up after bouncing off of the ground. The ball has Kinetic Energy, and again some energy is lost as heat due to friction between the air and the ball.
5
At this stage, the ball is stationary in the air because gravity has prevented it from rising any further. However, the ball is not as high as it was when it was dropped; this is because some energy was lost as heat. This stage links back to stage two repeatedly, until all of the energy from the ball has been lost, at which point it will become stationary on the ground.
Prediction
With this in mind, I am predicting that the higher the ball is dropped from, the higher it will bounce (due to increased energy). However, I predict that the ball will never reach the height it fell from because of energy which is lost as heat from friction and sound when it hits the ground.
Calculating Epg
The formula mgh (or Mass * Gravity * Height) will show the amount of Gravitational Potential Energy (Epg) the ball has at this stage. The mass of our ball was 0.024 kg, which is constant (it doesn’t change). The gravity here on Earth is 10N per kg of mass, for our ball this would mean 0.24N, another constant. The height from which we drop the ball is a variable.
Therefore to work out the Epg of the ball at any given height we would use the formula 0.024 * 0.24 * Height. We can shorten this to 0.006 because mass and gravity are constant.
For example, if we wanted to know how much Epg the ball had when held at 1.00 m, we would do 0.006 * 1.00, which is 0.006.
The reason for calculating Epg is so that later on the kinetic energy (Ek) of the ball can be calculating, in turn allowing the velocity of the ball upon impact to be calculated.
Method
First of all, two metre sticks were placed vertically against a wall, one above the other, creating a makeshift ‘double-metre stick’, this was held against the wall. Next, the ball was held so that the bottom of it was aligned with the height (e.g. 1.00 m). Meanwhile, another member of the team laid on the floor, facing the metre stick. The ball was then released when the member on the floor was ready. When the ball bounced up the member on the floor noted it down. This was repeated five times for each drop height (0.8m, 1.0m, 1.2m, etc up to 2.0m). After each drop height was done five times, the ball was heated to 40 degrees Celsius in a water bath.
Our variable was the drop height of the ball. We chose the range 0.6m – 2.0m because it achieves a good set of results, while not taking too much time after dropping from each height five times.
We dropped the ball five times from each height and then obtained an average to try and get a good range of results, and also to eliminate anomalous results from our graphs.
Results Table
Analysis
The results in this table show that the Epg increases when the ball is held higher up. It also shows that the ball bounces higher when the drop height is higher, and that the ball will never bounce to the same height it was dropped from. One other thing my table shows is that the higher the drop height, the higher the speed of the ball on impact with the ground. This proves everything I predicted to be correct, and also correlates with my energy transfer diagram, which is what I based my prediction on.
Ball Speed
Epg = Ek on impact. To work out the velocity (speed) of the ball on impact we would use the formula v=√Ek ÷ ½m. First we need to know the value of Ek which is dependant on Epg (If Epg = 0.006J, Ek = 0.006J). For 1.0m we would find the square root (√) of Ek (at 1.0m Ek = 0.006J). The square root of 0.006 is 0.078. So we have v=0.078 ÷ ½m. ½m = 0.012. So v=0.078 ÷ 0.012. 0.078 ÷ 0.012 = 6.5. Therefore speed at impact of a ball dropped 1.0m = 6.5 m/sec.
Evaluation
There are just two anomalies, they are at 1.4m and 1.6m, they is quite far from the line of best fit. I believe the cause was human error – perhaps in the inaccuracy of trying to see how high the ball was in a fraction of a second.
If I had the chance to repeat this investigation, I would improve the procedure by improving the measuring system, perhaps by using a digital video camera to record how high the ball bounced and then playing it back frame by frame on a computer because it is very hard to see where the ball is in a fraction of a second with human eyesight.
I would increase the range of results to be from 0.2m – maybe 5.0m, because it would give a much larger range, in which perhaps the rule of the ball bouncing higher when dropped from higher would be incorrect.