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What Factors Affect the Bounce of a Squash Ball.

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Physics Coursework

What Factors Affect the Bounce of a Squash Ball


        For my physics coursework, I have been asked to investigate the factors which affect the way in which a squash ball will bounce. I looked into a few different factors, including; Heat of the ball, landing surface of the ball, and in depth: the height from which the ball is dropped.

        To observe how the height a ball is dropped from affects the height the ball bounces back to, we put a meter stick against a wall, with another one directly above it (two meters tall). We then held the ball so the bottom of it aligned with the height we were planning to drop it from (.80 m, 1.00 m, etc) and released it. Meanwhile, another member of our team was lying on the floor facing the meter stick, and observed, from ground level, how high the ball rebounded.

We dropped the ball at .2 m intervals between 0.6 and 2.0 m (0.6, 0.8, 1.0, 1.2, etc). We dropped the ball from the same height five times in order to account for anomalous (or ‘misfit’) results, and to keep it a fair test. One more thing we did to try and enforce a fair test was to heat the ball to 40 degrees Celsius after every five drops.

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        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.


        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.


        At this stage, the ball is stationary in the air because gravity has prevented it from rising any further.

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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.


        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.

Alex Powell    Physics    Mr. Smith

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