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Plumb Line Mechanics Experiment

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David Hudson Mechanics 2 Coursework Plumb Line Mechanics Experiment Making Assumptions and relating them to both the model and the experiment To be able to use the mechanics theory that I know, the following assumptions must be made, listed in order of importance. * All motion is vertical - Although the horizontal motion could be taken into account, it cannot really be measured with the equipment we have, and since it isn't wanted anyway it is best just to redo trials that result in some horizontal motion. This requires two further assumptions: o There is no spin - spin causes the movement of the ball in other directions. o The floor is perfectly horizontal - As it not being so would cause the ball to bounce in other directions (although in practise this can't be helped anyway) * There is no air resistance - As I cannot calculate air resistance using the theory I currently know, and the acceleration would no longer be constant. Since the ball is small, spherical and heavy, it should encounter little air resistance and be only lightly affected by it. * The ball is uniform and perfectly spherical - The ball could appear to have different values of e because of having different densities or small deformations at different points on the surface, although the chance of these remaining unnoticed is minimal. ...read more.


e=the coefficient of restitution V2=U2+2AS U=eV(2as) 0=2ase2+2AS A=-a 0=2ase2-2aS 2aS=2ase2 S=se2 ? The fact that e is known to be a constant means that S/s must e2=S/s also be a constant. It can therefore be written as S=ks showing e=V(S/s) that the graph of S against s must be linear. Part 2: Time taken for 3 bounces T=time taken for the three bounces (s) T=t0+2t1+2t2 tn=time for ball to go from top of a bounce to the ground (s) u=speed at the top of a bounce (always 0) sn=top of a bounce (m) a=acceleration due to gravity (ms-2) sn=utn+1/2atn2 sn=1/2gtn2 2sn/g=tn2 tn=V(2sn/g) T=t0+2t1+2t2 T=V(2s0/g)+2V(2s1/g)+2V(2s2/g) ? S=se2 sn+1=sne2 sn=s0e2n T=V(2s0/g)+2V(2s0e2/g)+2V(2s0e4/g) T=(1+2e+2e2)V(2s0/g) Since the graph of T against s0 will be of the form y=kVx it will have the following shape: In other words it will be a slightly curve towards the x-axis. Task 1:Find e Using the dashed lines on the graph, I wish to find the gradient of the line. The vertical line is 10.09cm long and the horizontal line is 8.12cm long. Since the two axes have the same units and scale the figures don't need to be adjusted. The gradient of the line, and hence the value of S/s, is therefore 8.12/10.09 or 0.805. ...read more.


However, this slight difference does not seem to have any importance to the overall results. The two graphs have the general shapes that the model predicted and so are explained by the model. Revision of the process I believe that the greatest contributor to the imprecision of the experiment was the error in timing, so if it were feasible(i.e. cheap) I would use some kind of electronic equipment that detected when the ball bounced or was released. I would also have thought it better so use some simple mechanical device to drop the ball to avoid giving the ball any energy, and it may have been better to test the value of acceleration due to gravity rather than just giving it an assumed value. After all, the position of the moon has enough of a gravitational effect to cause the changing of the tides, so it may have had some noticeable effect on task 2 of the experiment, when gravity was significant. I think testing the value of g beforehand might have helped to reduce the difference between the experimental and predicted results for task 2. I am certain that better timing equipment would reduce the size of the error bars in task 2, and I think that using some mechanical device to release the ball would slightly reduce the error bars in both tasks. ...read more.

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