Pulling Forwards = χ Cos θ

Pulling Down = χ Cos α

- No Swaying – In the mathematical model I am going to assume that the falling mass does not sway. This uses the same concept as the rope not being parallel to the trolley. If the mass sways, the falling mass is not using its full potential.

Pulling Down = m

Pulling Sideways = m Cos θ

- Negligible Air-Resistance – This is due to the unique construction of the trolley; low frame, compact design and no extended parts or objects disrupting the aero-dynamics.

Conduct

To mimic the real life situation of the motion of a trolley on a plane I am going to use a trolley of mass ranging from 498g to 1498g, which will be run upon a set of smooth tracks. To accelerate the trolley a light inextensible string will be attached to the trolley, which will then be run over a smooth pulley. At this end of the string masses ranging from 20g – 80g will be attached which will accelerate the trolley. The mass of the trolley will also be changed. The length of the track will always be kept at 1 metre and the time taken for the trolley to travel the metre will be recorded. While conducting the experiment I realised that clamp holding the pulley covered 1cm of the track. Therefore when carrying out the experiment I released the trolley from 1.1m along the track, giving the trolley it’s 1m course to run.

Accuracy

To ensure accurate and reliable results a set of fixed rules must be followed. The length of the track will always be kept to 1 metre. Also three separate readings will be recorded when measuring the time taken for the trolley to travel the fixed metre. Furthermore I am going to ensure that the track is flat, i.e. it is not slanted up, down or to a side, else gravity will also be acting upon the car.

Mathematical Model

To create the mathematical model I am going to use Newton’s second law, which states, ‘The change in motion is proportional to the force’. For objects with constant mass, as is the case with this experiment, this can be interpreted, as the force is proportional to the acceleration.

Resultant force = mass * acceleration

This is written: F = ma

The resultant force and the acceleration are always in the same direction.

If I use the equation of Newton’s second law F = ma and transpose it into the form y = mx + c where the gradient of the graph is gravity.

F = ma

mg – T = ma T = Ma (Substitute into mg – T = ma)

mg – Ma = ma

mg = ma + Ma

mg = a (m+M)

a = g (m/m+M)

a = g (m/m+M) + 0

y = m x + c

This graph should pass through the points (0,0).

To work out acceleration for the mathematical model using the above formula.

Mass of trolley (M) = 498g

Mass of weight (m) = 20g

Distance = 1m

a = g (m/m+M) + 0

a = 9.81 (20/20+498)

a = 0.38 ms-2

All the accelerations have been worked using the above technique and have been presented in the table of results below.

Experimental Results

To work out the acceleration for the actual experiment I am going to use the equations of motion, namely s = ut + ½ at2.

Transpose s = ut + ½ at2 for ‘a’.

u = 0 Therefore ‘ut’ can be taken out of the equation.

s = ½ at2 (*2)

2s = at2

a = 2s/t2

To work out the acceleration of the actual experiment using the above formula.

Mass of trolley (M) = 498g

Mass of weight (m) = 20g

Distance = 1m

Time = 2.06s

a = 2s/t2

a = 2*1/2.062

a = 2/4.2436

a = 0.47 ms-2

All the accelerations have been worked using the above technique and have been presented in the table of results below.

The accelerations will be plotted on the ‘y’ axis and (m/m+M) will be plotted on the ‘x’ axis. Below is a table of results of (m/m+M).

On the following pages are the graphs of acceleration against (m/m+M).

Analysis

As can be seen from the graphs the mathematical model, models the actual experiment fairly well until the ‘m’ (mass of weight) is increased such that the trolley is travelling too fast to ensure accurate timing. Consequently on all three graphs the line of best fit starts from the origin and then gradually veers away from the mathematical model.

On the graph of results for M = 498g, it is observable that the actual experiment models the math model reasonably well, until ‘m’ is 60g. Thereafter, for ‘m’ = 70g & 80g, the trolley is travelling too fast to ensure precise timing hence the big error bars. Therefore I have not taken those two results into consideration when drawing the line of best fit through the points. Furthermore when working out the acceleration for the experimental results I had to square the timing, (i.e. t2) hence doubling the error in timing.

The other two graphs of M = 998g & 1498g, there are no anomalous results. I think the reason for this is, because of the increased weight of the trolley; the trolley will clearly be travelling slower, hence giving more accurate and reliable timing.

The gradient of the line in all the graphs should be in theory 9.81, but this clearly is not the case. Thus I am going to work out the gradient of the lines and compare it with the math model and observe how well the two compare with each other.

Gradient of line for M = 498g

1.25 – 0.9/0.1 – 0.072

= 0.35/0.028

=12.5

Gradient of line for M = 998g

0.65 – 0.4/0.045-0.032

= 0.25/0.013

= 19.23

Gradient of line for M = 1498g

0.46 – 0.3/0.04 – 0.026

= 0.16/0.014

= 11.43

As can be seen from the above results the math model did fairly well to model the real life situation of two connected particles.

Evaluation

The model I designed does not match the results I obtained in the experiment. This is because either I overlooked some variable quantities or the initial assumptions were flawed. On the other hand it may have been the procedure, which was at fault. In any case all these must be investigated into further. Each assumption ought to be scrutinized independently to deduce whether it is viable with regards to the experiment, in that, some assumptions were unnecessary and others were not made.

I think that if the experiment had been conducted in a vacuum and I used air–tracks the experiment would have been a lot more successful.