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The target of my coursework is to find the amount of g acting on a trolley by comparing the angle of the slope, the distance of the slope and the speed it was travelling as it passes through a light gate

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Charlie Vale RGP year 12        Physics b (advancing physics)

Physics coursework

Unit g493


The target of my coursework is to find the amount of g acting on a trolley by comparing the angle of the slope, the distance of the slope and the speed it was travelling as it passes through a light gate (50 cm from standing point of trolley). The distance between the slope and light gate is a constant of 50cm; there are two corresponding variables; the angle, and height of the slope. The first variable is the height of ramp / angle of the slope, of which the distance the trolley travels will remain the same, as will its starting velocity.

As well as measuring the velocity of the trolley as it travels through the light gates, I will also conduct another similar experiment; accept the height of the slope will remain constant at at 8 cm, and the angle will be constant at 9.2°, but the mass of the trolley will be gradually increased by 1 gram at a time.


I predict that as the angle and the height of the slope increases, there will be a natural increase in the trolleys velocity, and the easier it is for gravity to increase its velocity.

Furthermore I believe as mass is added, and all other variables remain constant, there will be less significant frictional effects on the trolley, as it increases in momentum.

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acceleration ms^-6

























Knowing that at 90° the size of g should be around 9.4, I can now use my primary data and use equations to try and figure out a more accurate answer which I can compare to each angle.

Working out G for each angle

Using the tables I can conclude I have the following data:

u = initial velocity
v = final velocity
a = acceleration
m = mass of trolley
Ѳ = angle of the slope
s = length of the slope.

Equations I know:

Component of weight acting on the trolley =mg sin Ѳ
Equation of Motion: v^2 = u^2 + 2 a s

And acceleration=g

And u=0

Therefore: v^2 = 2 (g sin Ѳ) s

final equation: g = v^2 / (2 * sinѲ * s)

Using the Equation I have worked out, I shall work out g for the tables shown previously.

Angle (°)

acceleration ms^-2






























The reason for the graph showing g to be curved may be due to friction, and that is the reason for my estimate of g to be 9.4 and not 9.8, if friction is taken into account then I believe my estimate may be closer to the true value.

Friction=mass of trolley (N) cos Ѳ = 0

Height of point of release before trolley moves is 4mm

Sin^-1 = 0.229183729= Ѳ

0.7cos Ѳ=0.69478 N

Flim = µR

µ= 0.1

New g equation including friction=

g = 0.8881^2/[(sin(9.2000)) – (0.1 x cos (9.2000))]


This result is much over my estimate, but perhaps it was less than 4mm to more the trolley and this error is due to human error.


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Experiment 2 conclusion:

In conclusion the increase in mass, gave the trolley more momentum, and as a result meant friction had less of an effect on the trolley, furthermore this allowed the trolley to increase in acceleration as more weight was added.


If I was to repeat the experiment again, and re-asses the results there are several things I would change:

Instead of releasing the trolley by hand I would find another method of releasing it to guarantee it starts from the same point and starts with zero velocity, potentially use electromagnets to hold it at the same spot and then the trolley can be released with the push of a button.

I would also use more precise measurements, for example, I would measure the height of the ramp to the nearest half a mm rather than to the nearest cm, as it would give me more precise results.

I would similarly figure out a more effective method for finding the true value for g when including friction, as my friction estimates seem to be much too high.

Finally, I am pleased with my overall results, some estimates, and answers are slightly wrong,, but I believe those are purely down to human error.

Websites used for mechanical equations:



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