Mechanics 2 Coursework - Assumptions related to both the model and the experiment.

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Mechanics 2 Coursework


Mechanics 2 Coursework

Assumptions related to both the model and the experiment

Assumptions that are made for both the model and the experiment:

  • The gravity, g, is taken to be 9.8ms-1
  • The ball is taken to be a particle
  • The accuracy of the timing is subjective – the start and end times are based upon the testers’ judgement.
  • Readings are taken from the bottom of the ball and not the centre.
  • The downward direction is taken to be positive, therefore any rebounds are negative.
  • Air resistance is not taken into account.
  • Values are taken to 2 d.p.

Manipulating the model

Firstly, a theoretical equation needs to be obtained linking e (the elasticity of the ball), h1 (the downward height) and h2 (the rebound height).  This can be obtained using the equations of linear motion.

When dropping the ball:

s = h1                a = g                u = 0                v = ?

        Using         v2 = u2 + 2as

                v2 = 0 + (2gh1)

                v =  2h1g

When rebounding the ball:

s = -h2                a = -g                u = ?                v = 0

        Using         v2 = u2 + 2as

        

                u2 = v2 – 2as

                u2 = -2 x –h2 x –g

                u = -  2h2g        

Using v = -eu, we can find an equation in terms of e.

                2h1g = -e(-  2h2g)

                2h1g = e  2h2g

                     e =

                     

     e =

From the above model, it can be seen that the gradient will give us a value of e2.  Therefore, to obtain mean, maximum and minimum values for e, the hand-drawn graph can be used to obtain such gradients.

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Secondly, we must establish an equation for the time taken for the ball to come to rest.

If we use the following model for the time taken for the ball to bounce, an equation can be established.

                                        

                                        For the first bounce, the following can be said:

                                        u = ev                a = -g

                                        v = -ev        t = ?

                                        Using the equation of linear motion v = u + at

                                        -ev = ev – gt

                                        

                                        

t =

For the second bounce, the following can be said:

u = e2v        a = -g                v = -e2v        t ...

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