The purpose of this assignment is to investigate the motion of an object as it slides down a slope and then falls freely to the ground.

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                                Whoosh down the Slide

Introduction:

        The purpose of this assignment is to investigate the motion of an object as it slides down a slope and then falls freely to the ground. A standard model for this situation is set up and is accessed critically the relationship between this model and the experimental results.

        In order to carry out the experiment successfully, several assumptions have to be made beforehand.

The assumptions are as follow:

  • The block is modelled as a particle when it slides down the slope and as a projectile as it comes off the piping.
  • The piping is smooth i.e. it’s frictionless.
  • Air resistance is neglected because the speed of the block is too small for it to have an affect on it.
  • The force of gravity is constant. g = 9.8
  • The energy is conserved.  . The potential energy (at the top of the slope) is converted into the kinetic energy(when the block reaches the end of the slope). Hence the velocity can be calculated in this way.  i.e. PE = KE;  

PE = mgh

KE = ½ mv²                        ∴mgh = ½ mv²  

  • The block has constant acceleration as it slides down the slope, so constant acceleration equations can be applied to the block.

These assumptions are relatively important to the model, because bearing these assumptions in mind, I can work out some predictions using the mechanics theory. They are also important to the experiment, because if there are any errors in the results they may be able to explain by using these assumptions.

Diagram of the set-up:

Using Mechanics Theory:

As the block acts as a projectile as it comes off the piping, its velocity can be split into components, horizontally and vertically.

The block has constant acceleration as it slides down the slope because all the forces acting on it are balanced. Therefore by applying a constant acceleration equation on the block, an expression of time, t, can be found in terms of its velocity, V.

           s = ut + ½ at²

           d  =  Vcosθ t +  ½ ×0 × t²

            h      Vsinθ        ½ ×g ×t²

           

           d = Vcosθt

            t =     d __

                  Vcosθ

h = Vsinθ t + ½ gt²               (substituting t in the equation)

   = Vsinθ   __d__   + ½ g       d__   ²

                   Vcosθ               Vcosθ

   = dVsinθ +  __gd²___

       Vcosθ      2V²cos²θ

h = dtanθ +     gd²___    

                    2V²cos²θ

In the assumptions, I’ve mentioned that energy is conserved, so an expression for V can be found using  PE = KE

                              mgh = ½ mV²

                        mglsinθ = ½ mV²                 ( the mass cancels out)

 

∴ V² = 2g lsinθ

Substituting V in the previous equation gives:

h = dtanθ + __gd²___

                   2V²cos²θ

     = dtanθ + ____  gd²     ___

                     2 (2glsinθ)cos²θ

     = dtanθ + _____gd²__                            ∴h = dtanθ + ____d²____

                       4glsinθcos²θ                                                 4lsinθcos²θ

        

( where h and θ are constant)

Using this equation, the distance d that the block would travel from a fixed distance l at an angle of θº at height h, could be calculated.

e.g. if l = 0.5m      θ = 30º     h = 0.9m

 

              h = dtanθ + ____d²____

                               4lsinθcos²θ

           0.9 = dtan30º + ________d²           ____

                                    4×0.5×sin30º×(cos30º)²

           0.9 = 0.577d + __d²__

                                      0.75

       0.675 = 0.433d + d²

   

d² + 0.433d – 0.675 = 0

So d can be solved since it’s a quadratic equation, but only the positive value of d is to be taken because the block couldn’t have travelled a negative distance.

           d = 0.633m  ( 3d.p)

In theory, this is the distance that the block should have travelled from the end of the pipe to where it lands in the tray of sand.        

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Therefore I can now produce a set of predictions for distance d using the values of h and θ in my experiment. Then a comparison can be made between the theoretical results and the experimental results.

To make the calculations simpler, the equation is to be rearranged into a simpler form.

                            h = dtanθ + ____d²____

                                             4lsinθcos²θ

          4lsinθcos²θh = dtanθ×4lsinθcos²θ + d²

         

In my ...

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