The diagram below shows a slide. An object slides a distance L down the slide, and then shoots out at the end to fall through a vertical distance H before hitting the ground. Obviously, the greater L the further the horizontal distance D that it lands away from where it left the slide.

L?H Plastic boxDTo investigate how D is related to L for one particular angle of inclination, I am goingto do an experiment to see if this relationship turns out in practice to be as predicted.The apparatus that is required is:* A Wood ramp(slope);* A Aluminium weight(cube);* A plastic box;* A piece of card board;* A meter rule;* A protractorBefore I do the experiment, I need to make some assumptions which are relating to both the model and the experiment:When the experiment is conducted, it must be insured that all of the apparatus is attached securely to ensure that nothing comes apart. General laboratory rules must be recognised to ensure safety throughout.The variable that will be investigated is the range that the object is projected.The independent variable is the height up the slope that the block is released from. It is measured using a metre ruler.The dependant variable is the distance from the slope in the horizontal direction that the block will travel.When the object leave the slope, it might rotate then land on the ground.

In fact, my calculation should consider the rotational energy of the object. However, this will make the calculation very complicated and it is beyond our syllabus, so the rotational energy has been ignored. To minimise this error, I need to be careful that do not use something will round along the slope or has strange shape.

To simplify this experiment, air resistance has been ignored as a factor in the range of the projectile. As it is a projectile(after leave the wood ramp) that will be measured, the only other factor working on it are gravitational forces. The horizontal motion of the projectile is independent of the vertical motion, and that is why the falling motion of the block does not need to be considered.A aluminium block will be used, as it is a fairly dense object. This has benefits because although we are ignoring the effect of air resistance in the calculations, we should minimise the affect that it has. A lighter object, or one that has a larger surface area (i.e.

is less dense), would show more affect from air resistance. I discussed with my group mates then decided to use a aluminium cube with dimension 3cm x 4cm x 5cm to do the experiment.Unsure surface of the wood ramp, this is causing the fiction vary during the object is sliding down the slope or before sliding. Therefore, the work done against friction will vary.

This will cause the most significant error in my experiment and I will explain why later.Energy transfer:L?The gravitational potential energy of the object gained at the top of the slope transfer to the gravitational potential energy of the object at the bottom, the kinetic energy at the bottom of the ramp and the work done against friction.–> Ept = Epb + Ekb + W.DChange the angle ?, will also change the gravitational potential energy of the object gained at the top of the slope, therefore the kinetic energy at the bottom and the friction will be different.

Determining the work done against friction:mgcos ?f =???Rmgsin ?Omg?From the diagram above I known when the block just slides, the components of weight parallel to the slope should be equalled, which means:mgsin ? = ??R = ??mgcos ?–> ????mgsin ? /mgcos ?–> ????tan ?I can determine the angle which the block just slides, do this by gradually increase the angle(about 1 second 5 degrees). Then I can get a angle ? when the block just slides by putting a big protractor beside the wood ramp(see picture below).Alternative way to get the angle is by measuring the horizontal and vertical distance of the object to the point O. tan? = vertical distance / horizontal distance.Because I have got that excellent equipment which is available in the physics laboratory, It makes the process faster and more accurate, so I will choose my 1st method to do it.I did 5 time, which give me ? = 20.0ï¿½,21.

8ï¿½,18.7ï¿½,22.2ï¿½,19.2ï¿½.(all to 3s.f)And I average those 5 values then gives ? = 20.4ï¿½(to 3.

s.f)?????tan ? = tan20.4ï¿½=0.372 (to 3s.f)Rearrange the equations:From the “energy transfer” I got a equation:Ept = Epb + Ekb + W.D–> Ept = Epb + Ekb + Ltan???R–> Ept = Epb + Ekb + Ltan?mgcos?–> mg(H+Lsin?) = mgH + 1/2mv2 + Ltan?mgcos?–> v2=2gLsin? – 2gLtan?cos?–> v= (2gL(sin? – tan?cos?))^1/2Therefore, I had a equation involves velocity, ? and ?. As long as I known the ? and ?, I will be able to calculate the velocity which the object just leave the slope fall towards the ground.

vh ? ?v vvHDvh = vcos? ; vv = vsin?? D = vh * t ? t = D / vhConsider the vertical component,S = ut + 1/2 ut2 –> H = vv t + 1/2 gt2–> H = vv D / vh + 1/2 g D2 / vh2–> H = vsin? D/ vcos? + 1/2 g D2 / v2cos2?–> H = D tan? + g D2 / (2 cos2?(2gL(sin? – tan?cos?)))–> H = D tan? + D2 / (4 cos2?L(sin? – tan?cos?))–> D2 / (4 cos2?L(sin? – ?cos?)) + D ? – H =0(Noticed that, this is a quadratic equation.)I can calculate the horizontal distance D by substitute the ? and ? values in then solve the quadratic equation. This will give me the predicted value of D.Let 1/(4 cos2?L(sin? – ?cos?))=a; ?= tan? =b; -H=cThe quadratic equation becomes aD2 + bD +c = 0 .I will produce a table in Microsoft Excel, showing the ? in degrees, ? in degrees, L in metres, H in metres and a, b and c.

Then as long as I got the values for a, b and c, because it is a quadratic equation, so I will be able to calculate the value of D by using a formula to find the roots of a quadratic equation:- (-bï¿½(b2-4ac)^(1/2))/2a.And in this case, I will only need the positive solution of the quadratic equation, which will be give by using the formula (-b+(b2-4ac)^(1/2))/2a.(The calculations are done by Microsoft Excel, and the calculated values a, b and D are all in 5 decimal places)?(ï¿½)?(ï¿½)L(m)H(m)abcD3020.40.50.2823.

746830.37190-0.2820.229173020.40.40.2824.683540.

37190-0.2820.208873020.40.30.2826.244720.37190-0.

2820.184803020.40.20.2829.367080.37190-0.2820.

154793020.40.10.

28218.734170.37190-0.

2820.11316?(ï¿½)?(ï¿½)L(m)H(m)abcD3520.40.50.2822.770710.37190-0.

2820.258903520.40.40.2823.463390.37190-0.

2820.236663520.40.30.

2824.617850.37190-0.2820.210113520.40.20.

2826.926780.37190-0.2820.176703520.40.10.

28213.853550.37190-0.

2820.12988?(ï¿½)?(ï¿½)L(m)H(m)abcD4020.40.50.2822.380690.37190-0.

2820.274814020.40.40.

2822.975860.37190-0.2820.

251634020.40.30.2823.967810.37190-0.2820.223824020.

40.20.2825.951720.

37190-0.2820.188664020.40.10.

28211.903440.37190-0.

2820.13909The predicted results are shown above, then I am going to draw the graph of “D against L”.Before that, I did the lower bound and upper bound values for the predicted value of D. I calculated the lower bound by multiplied 0.95, upper bound by multiplied 1.

05 because I assume the error is ï¿½5%.?(ï¿½)D(ï¿½)Lower boundUpper bound300.229170.217710.24062300.208870.198420.21931300.

184800.175560.19404300.154790.147050.

16253300.113160.107510.11882?(ï¿½)D(ï¿½)Lower boundUpper bound350.258900.

245950.27184350.236660.224830.24850350.210110.

199600.22062350.176700.167870.

18554350.129880.123390.13638?(ï¿½)D(ï¿½)Lower boundUpper bound400.274810.261070.

28856400.251630.239050.

26421400.223820.212630.23501400.188660.179230.

19809400.139090.132130.14604two dotted lines in each graph are lower bound and upper bound respectively.I will do some comparisons between these model graph and the actual graphs.

The practical element:The method that will be undertaken is firstly to set up the apparatus as shown in the diagram in “Introduction”. The height above the ground, H, should be measured using a metre ruler and this should be kept constant throughout the experiment. The angle at which the wood ramp is at should also be kept constant and should be held in place by clamps, so that the edge of the slope is parallel with the edge of the plastic box(which the wood ramp is standing on it). A piece of card board should be placed on the floor running away from the plastic box. Practice runs of placing the object at different length on the slope should be done to find roughly the place where the piece of card board should be placed.

For the experiment, the object should be held in place on the ramp at 5 different lengths. 50cm, 40cm, 30cm, 20cm and 10cm to the edge of the slope. It can then be released and as the object strikes the piece card board it should leave a mark on it. The range(distance) can then be measured from the edge of the table to the centre of the mark left by the clear trail. The experiment can then be repeated 3 times for each length on the ramp to reduce the effect of anomalous results.

The experiment will be conducted using H = 0.283m and at a angle 30ï¿½to make a start. Then will do 35 and 40 degrees. 5 different lengths for each angle.This will hopefully obtain a good set of results.

I then got the results as below:?(ï¿½)?(ï¿½)L(m)H(m)D1(m)D2(m)D3(m)D(average)3020.40.50.2820.2330.2280.

2350.232003020.40.40.

2820.2090.2030.2040.

205333020.40.30.2820.1760.1730.1850.

178003020.40.20.2820.1580.1550.

1610.158003020.40.10.2820.

1430.1450.1520.14667?(ï¿½)?(ï¿½)L(m)H(m)D1(m)D2(m)D3(m)D(average)3520.40.50.2820.

2390.2380.240.

239003520.40.40.

2820.2290.2210.2190.223003520.40.30.2820.

2040.2050.2050.

204673520.40.20.2820.180.1790.

180.179673520.40.10.2820.1310.1340.

1330.13267?(ï¿½)?(ï¿½)L(m)H(m)D1(m)D2(m)D3(m)D(average)4020.40.50.2820.

2140.220.2210.218334020.40.40.2820.2110.

210.2120.211004020.40.30.2820.

1930.190.1910.191334020.40.

20.2820.1720.1690.160.167004020.

40.10.2820.1290.1310.1330.13100Possible variation:Unsure surface of the block and the wood ramp varied lost energy;Twisting affected the energy transfer;Errors in reading the mark on the piece of card board.Graph:I will draw 3 graphs for each angle, then compare them to the model graphs see what the differences are.

For the first and second graphs, ?=30ï¿½and ?=35ï¿½, most part of the ‘actual lines’ lie within the error bounds. However, the last graph ?=40ï¿½, the whole ‘actual line’ is not in the error bounds.I guess the reason for this is the contact surface between the block and the wood ramp. When the angle is about 35 degrees, there is a best contact of the block to the wood ramp surface, therefore the friction I obtained is very close to the real value in just this case, so my graph for this one is the best in those three. When the angle increased to 40 degrees, the contact of block to the wood ramp might become less, the friction I got would has some error compare to the actual value, this obviously affected the landed distance, so the graph for this looks very odd.

Sources of error:There must have been some amount of varied friction during the movement of the block moving down the slope. There must also have been air resistance of the block dropping during its trajectory. Both of these were ignored in the calculations. To minimise both types of friction, a dense and smooth object was used as this would lower the affect of both air resistance and resistance with the slope. The slope was also looked at to check that it was smooth and that there were no obstacles or blemishes on the surface to ensure a smooth run.The ruler was not set exactly vertical throughout the experiment, giving false readings.

To try to minimise this, set squares were used to keep the ruler vertical, and clamps were used to try to hold it in place so that it was the same for every attempt.The object was released in a way that tried to minimise the amount of spin in any direction apart from that of down the slope. Excessive spin or spin in an incorrect direction would alter the acceleration of the object sliding down the slope. To try to avoid this the block was released carefully using a piece of wood to hold it rather than a human hand releasing it,.

This meant that it started to slide in the correct way from the start of its descent down the slope.The piece of card board that was used to make an impression may have moved slightly as the block struck it at an angle. It was either held in place, or sticky-tape was used to ensure that it did not move during impact. Also, to maintain accurate results, the reading was always taken from the centre of the line impression. However, it doesn’t mean that the centre of the line impression is the exactly point the block landed, because it is a block, not only a particle, it has centre of mass, the position of the centre of mass of the block when it landed should be the most accurate point I want, but it is very hard to determine, so I will just take the centre of the line impression to be my readingThe most significant error for my experiment is the value of ?.

I obtained it by measured the angle where the block just slides. Human reading for the angle is not very accurate, and the range for those readings is about 3.5ï¿½(22.2-18.7), so the percentage error is roughly (3.5/20.

4 *100%) 17.2%. It is a huge percentage error.

Not only the error in reading the angle, but also the calculation ????tan ?, because the angle is in decimal place(not accurate), take the tangent to that angle will increase the error again. If the value of ??is not accurate, then the friction is not perfect as well, from the energy transfer equation, Ept = Epb + Ekb + W.D, the kinetic energy I obtained will be affected by the varied friction, the velocity when the block just leave the slope will change, therefore the distance land will vary.

e.g if the calculated ??is less than the actual value, the work done due to friction is smaller, so the velocity is greater than the actual value, then the landed distance obviously would be further away from the edge than the real landed distance. The graph for this case should lie a bit higher than the true graph.Percentage error:The error for the experiment caused by lots of things. Friction measurement, height measurement, length measurement and distance measurement.Take the accuracy of each measurement and divide by the smallest value I have got for each measurement then times 100% to get the percentage errorThe error for angle (Friction): 17.2%After doing the calculation ?????tan ?, the error increased.The error for L (Length): 0.001mPercentage error = 0.001/0.1 * 100% =1.0%The error for H (Height): 0.001mPercentage error = 0.001/0.282 * 100% =0.4%The error for D (distance): 0.001mPercentage error = 0.001/0.129 * 100% =0.8%Total percentage error = 17.2% + 1.0%+0.4%+0.8% =19.4%Guowen Qin 2007-5-10