I will investigate the change of velocity and acceleration of a laterally moving object attached by a string and pulley to a dropped object, when the mass of the dropped object is changed.MethodApparatus** Trolley* Piece of Card* Runway* Weights* 2 Light Gates* Weights* Pulley* Computer* LogIt 9000* Insight Timing* Appropriate cablesDiagramSet up procedureThis is how I will set up the experiment:1. Set up the apparatus as above, with Light gate A at 45 cm from where the centre of the piece of card (not trolley) will start to move from, and Light gate B at 45 cm further.2.

Make sure that there is enough space between the pulley and light gate B for the piece of card to go through the light gate. If not, move the starting point of the trolley back, and the light gates accordingly.3. Plug Light gate A and Light gate B into the LogIt, and plug that into the computer.4.

Start up Insight Timing on the computerThen I am ready to startFair TestI will make this a fair test by limiting the key factors:- Weight of trolley – I will use the same trolley every time to ensure the same weight- Distance travelled – I will keep this constant by releasing the weights from the same height and releasing the trolley form the same place every time. I will also not move the light gates- Friction/Air Resistance – Unfortunately, these cannot be avoided, however they will be minimal due to the equipment used, and will be relatively constant as I will not change apparatus.SafetyThere is little safety to be considered.

No harmful substances are being used, neither are flames or solvents. I will need to take precautions when increasing the dropping mass, and make sure that all the weights are securely fixed. The main point would be to stop/catch the accelerated trolley before it falls off, especially at higher speeds.Execution ProcedureThis is how I will execute the procedure1.

The starting point should be the centre of the piece of card. I will have 10g as the dropping weight2. I will start recording on Insight Timing, and let the weight drop, making the trolley and, more importantly, the piece of card move.3.

I will make sure someone catches the trolley before it falls off (see safety)4. I will stop insight timing and move the trolley back to the start position5. I will then add another 10g to the weight and repeat from 2. until I have measure the velocity at a mass of 100g.I will repeat this procedure two more times, so that I have 3 repeatsTheoryThere are three points to be considered in the theory. The first is how the mass should affect the velocity.

The second is how the height should affect the velocity. The third is how the mass will affect the acceleration.First let me discuss the first point.

This test is based on converting gravitational potential energy into kinetic energy.Gravitational potential energy (g.p.e.) is mgh, and kinetic energy (k.e.) is 1/2mv2.If we consider the mass of the trolley to be m1, and the dropping mass to be m2, this gives us g.

p.e. as m2gh, as only the dropping weight has g.

p.e. The letter ‘g’ is the gravitational constant, and is approximated as 9.81. The letter ‘h’ is the height.

Equally, k.e. is 1/2(m1+ m2)v2, as both the dropping mass and the trolley will move.By Law of Conservation of Energy, energy can neither be created nor can it be destroyed, however it can be converted from one form to another. Therefore, if the g.

p.e. is not ‘wasted’ in any other way (e.g. friction converting it into heat energy), it must all be converted into k.e.

This means that g.p.e.

= k.e., therefore:m2gh = 1/2(m1+ m2)v2If I didn’t have to consider m1, this could be re-written as:v2 = 2m2ghm2This would mean that the m2s will cancel out, proving that the mass of an object has no effect on its falling speed (as confirmed by Galileo). This is not true, however, when there is a constant added to the denominator.

This makes the formula:v2 = 2m2ghm1 + m2This does not allow for the masses to cancel out, therefore the velocity (of both) must depend also on the falling mass.The second point to consider is the height. Due to the setup, the vertical drop will be converted into both vertical and horizontal movement. This means that the height can be represented on the horizontal as the movement of the trolley. Since the speed is measured as the average speed of the card as it moves, the starting point should be the centre of the piece of card.The two light gates are therefore at 45 cm and 90 cm. This means that in the formula, h will be 0.45 and 0.

9 (as h is measured in metres).I should also consider the effect the height will have on the velocity. The formula is:v2 = 2m2ghm1 + m2Since the h is in the numerator, it must directly and positively influence the velocity, i.e. as it increases, so will the velocity.As for the third point, acceleration, a different formula is needed. In this instance, it is:v2 = v02 + 2axv0 is the starting velocity and v is the final velocity (at light gate B). a is the acceleration and x is the distance moved.

In this case, as light gate B is 90 cm from the starting point, x = 0.9. Also, since the trolley is not moving at the start, v0 is 0As we want to know the acceleration and v0 is 0, this can be rearranged and simplified as:a = v22hSince we know v2 to be:2m2ghm1 + m2we can replace the v2 in the acceleration formula to conform. Also After some simplifying, this becomes:a = m2g I(m1 + m2)Therefore, as the mass increases, so will the acceleration.PredictionThere are really two predictions to be made.

One is on the subject of velocity, the other on acceleration.For velocity, I predict that as the dropping mass will increase, so the velocity will also increase. I predict that there will be a linear relationship between v2 and m2, therefore, I predict that the relationship between v and m2 will have to be parabolic, on its side.

I also predict that, due to the height being lower at light gate A than B, the velocity will be higher at B than A. I predict the graph to look like this:For acceleration, I predict that dropping mass will increase, so the acceleration will also increase. Since m2 is in both the numerator and denominator, I predict that the graph will be hyperbolic, on its side. This is because, as the m2s increase, the effect of the added m1 will decrease, almost to a point where it is relatively insignificant, as m2 is so large. This would mean that the m2s increase to a theoretical point where m1 is irrelevant, and they cancel out, making the acceleration stay the same, as g (9.

81). This would make the graph hyperbolic with the asymptote as 9.81:A – AnalysingResultsRepeat 1Repeat 2MassVelocity AVelocity BTimeAccelerationMassVelocity AVelocity BTimeAccelerationkgm/sm/ssm/s/skgm/sm/ssm/s/s0.000.000.

000.000.000.000.000.000.000.

000.010.270.381.350.080.010.

270.381.360.080.020.450.650.800.

260.020.470.670.770.250.

030.580.830.620.

400.030.600.860.

610.430.040.690.960.

520.520.040.701.000.520.590.050.

771.030.470.540.

050.761.120.470.770.060.841.170.

430.760.060.

801.210.440.910.

070.911.190.400.690.070.911.

320.401.040.080.

971.380.371.100.080.971.

410.371.190.

091.011.470.

361.300.091.021.460.361.220.

101.021.480.

351.290.101.111.530.341.23Repeat 3MassVelocity AVelocity BTimeAccelerationkgm/sm/ssm/s/s0.000.

000.000.000.000.010.310.

461.200.130.020.480.720.770.310.

030.600.900.610.480.040.701.

040.530.650.

050.781.150.480.770.060.841.230.

440.880.070.911.370.

411.110.080.961.

420.391.170.091.021.500.

371.310.101.

071.590.351.49From these results I can work out an average set of results:MassVelocity AVelocity BTimeAccelerationkgm/sm/ssm/s/s0.000.000.000.

000.000.010.

280.411.300.100.020.470.

680.780.270.030.590.860.610.440.

040.701.000.520.580.050.771.

100.470.690.

060.831.200.440.850.070.

911.290.400.

950.080.971.

400.381.150.091.

021.480.361.

280.101.071.530.351.34AnalysisThere are several pieces of information that can be extracted from the above table. The first two are the changes in Velocity at points A and B when the mass increases.

The second is the change in acceleration as the mass increases. Also these values can be compared to the theoretical to show how outside factors (like friction) have affected the results.The first that I will look at is the two velocities.Here, I have put on a graph the two point’s velocities against the changing mass. As you can see, the velocities both increase with the mass, however they increase less every time.

Also, it shows how the velocity at B, further on (therefore at a higher height) is larger than the velocity at A (at a smaller height).Theory backs both these remarks. Firstly, I shall consider the height’s influence. The formula for the velocity (squared) is:v2 = 2m2ghm1 + m2As h increases, its position makes it increase the amount in the numerator, therefore directly increasing the velocity. Therefore, as B’s height is larger than A’s, the velocity at B will always be higher that A. A notable exception is when m2 = 0, as then the height will have no effect as the numerator, and therefore velocity will have to equal 0.

Secondly, I shall consider the influence of the dropping mass. In the formula, removing the influence of the m1 would make the m2s cancel out and make the velocity the same. However, this is not so. Since the denominator involves an addition, and the numerator does not, this mean that the velocity must increase as the mass increases.

As to why the graph curves, it would be easier to consider a simpler fraction, namely:x Ix+1When x = 1, the fraction is 1/2. When x = 2, the fraction is 2/3. The difference between these is 1/6.

When x = 3, the fraction is 3/4. The difference between the last two is now 1/12; half of what it was before. When x = 4, the fraction is 4/5. Now, the difference between the last two is 1/20; again, nearly half. If the fraction was plotted on a graph, this downwards trend in differences would make the line curved, as it would increase less every time.This does not change with multipliers, neither in the nominator nor denominator, as they only change the scale, not the shape, neither does it change when the added constant is changed. This means that the same rule applies to the formula for velocity.As for comparing this graph to the prediction, the easies way would be an overlay of the two:This shows the prediction and real graph, both scaled to fit each other.

The lines from the prediction are not exactly the same as in the prediction, as I have scaled them to fit the graph. As you can see, they follow each other nearly exactly. This, coupled with the fact that my theory agreed with the graph of the results, means that my prediction must have been correct.The second graph that I will look at is the accelerationIn this graph, the trend is not so apparent. Only one definitive piece of information can be derived from this graph: that the acceleration increases as the mass increases.Before I consider the shape of the graph, I will first explain the above point. The formula for acceleration against mass is:a = m2g I(m1 + m2)As I have described in my analysis of the velocity graph, this means that as m2 increases, a must also increase.

However, the fact that this increase should be smaller every time is not so apparent. To explain this, I will look at the theoretical graph on the same scale:This graph, save for a few anomalies in the actual graph, is exactly the same. Therefore there are two options: the theoretical graph is incorrect or the graph is incorrectly drawn.

My suspicion is that it is the latter. I believe that the graph will turn, however not yet, i.e. that the scale is too small. If I do the graph again, with the mass going not up to 0.

10 but to 1.00, the graph looks like this:The graph now does indeed curve, as I had predicted. I believe therefore that had I continued the experiment up to a dropping mass of 1 kg instead of 0.

1 kg, my actual graph would also curve. This, however, only speculation, and there is no way to tell without actually doing this. I can therefore only conclude that the acceleration increases as the mass increases. This does, conform with my prediction, however it does not prove that the graph would be hyperbolic with the asymptote at 9.81, as the graph does not go that far.E – EvaluatingStandard DeviationMassVelocity AVelocity BTimeAccelerationkgm/sm/ssm/s/s0.000.000.

000.000.000.010.

020.050.090.030.

020.020.040.020.030.030.010.

040.010.040.040.

010.040.010.070.050.010.060.

010.140.060.

020.030.010.080.070.

000.090.010.230.080.010.020.

010.050.090.010.020.010.050.100.

050.060.010.14In this experiment, a very fair result could be achieved due to the three repeats. In fact, as this table shows, the standard deviation was never higher than 0.23, and only three times exceeding 0.10.

Obviously, this could be further improved by adding more repeats.Another way to improve the results would be to better standardise the place form which the trolley was released, as on the day this was only roughly estimated, leading to anomalies. Also, although this would be only theoretical, it would have been better to decrease the friction in the wheels, runway, and pulley and even air resistance.

Factors such as these lead to energy losses reaching 54%.Energy Statistics at AMassEnergykgInput (J)Output (J)Loss (%)0.000.000.00N/A0.010.

040.0254%0.020.090.0636%0.

030.130.0930%0.040.180.1326%0.050.

220.1626%0.060.260.1928%0.070.

310.2424%0.080.350.2723%0.090.400.3023%0.100.440.3423%Energy Statistics at BMassEnergykgInput (J)Output (J)Loss (%)0.000.000.00N/A0.010.090.0452%0.020.180.1232%0.030.260.2025%0.040.350.2724%0.050.440.3325%0.060.530.4123%0.070.620.4823%0.080.710.5719%0.090.790.6419%0.100.880.7120%The input in these energy losses has been calculated as the theoretical input of the appropriate mass over the appropriate distance, namely:Input = mghThe output has been calculated as:Output = v2(m1+m2)2The loss percentage has been calculated as:Loss % = Input-OutputInputObviously, at a mass of 0, input and output are 0, and therefore the loss is not calculable, as it would include division by 0.These tables show that friction and air resistance created a very large margin of error for the experiment, especially at lower masses. The real and theoretical lines are compared on the next two graphs, of the velocity at A and at B, respectively. The frication and air resistance factor would explain the difference in scale, but similarity of the lines, which would other wise have to be explained with very coincidental anomalies.It is difficult to identify anomalies on a curved graph, however, as the graph shows, the real lines quite closely follows the theoretical lines, apart from the aforementioned difference in scale. On the first graph, there is one slight ‘bump’, which could be described as anomalous, at 0.06 kg, and a slightly smaller curve than predicted at 0.01 kg, however these are the worst examples of anomalies on the graph, and are minute. On the second graph, again there is a slightly smaller curve than predicted at 0.01 kg, there is again a small ‘bump’, this time at 0.07 kg, and the point is a bit lower than it should at 0.10 kg, however again the anomalies are minute.Much worse anomalies come on the graph of acceleration. Problems due to friction and air resistance are obviously present, as the real graph is below the theoretical graph; however the real line is also, for lack of a better word, ‘wobbly’. The unsteadiness of the line makes it difficult to identify which points are actually anomalous, however by drawing a straight line of best fit (also shown), I approximate these to be at 0.01, 0.08, 0.09 kg.The increased size of the anomalies may be due to the fact that the acceleration is calculated by using three measurements, velocity at A, velocity at B and the time, all of which can have slight mistakes which add up.If factors such as friction and air resistance could not be eliminated, it might have been beneficial to measure them instead, and adjust the final results accordingly. Therefore, as further work, it would have been helpful to find out the co-efficient of friction of the runway, i.e. find the force needed to move the trolley (measured using a Newton meter, by pulling the trolley along the runway) and divide this by the force exerted upon the trolley by the runway (this being its weight, 0.5 kg, multiplied by the gravitational constant, 9.81, making it 4.905 N). Air resistance could also be mesaured, by Stokes law which says that air resistance force is proportional to density of the air times the cross sectional area of the object times the square of the velocity of the object.Investigation on converting Gravitational Potential energy into horizontal and vertical Kinetic Energy