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# Mechanics 2 Coursework - 'woosh' down the slide

Extracts from this document...

Introduction

Guowen Qin                                                                                                     2007-5-10

Mechanics 2 Coursework

‘woosh’ down the slide

Introduction:

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

Ө

H            Plastic box

D

To investigate how D is related to L for one particular angle of inclination, I am going

to 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 protractor

Before 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.

Middle

vvv

H

D

vh = vcosӨ ;    vv = vsinӨ

∵ D = vh * t     ∴ t = D / vh

Consider the vertical component,

S = ut + ½ ut2→ H = vv t + ½ gt2

→ H = vv D / vh + ½ g D2 / vh2

→ H = vsinӨ D/ vcosӨ + ½ g D2 / v2cos2Ө

→ H = D tanӨ + g D2 / (2 cos2Ө(2gL(sinӨ - tanФcosӨ)))

→ H = D tanӨ + D2 / (4 cos2ӨL(sinӨ - tanФcosӨ))

→ D2 / (4 cos2ӨL(sinӨ - μcosӨ)) + D μ - H =0

(Noticed that, this is a quadratic equation.)

I can calculate the horizontal distance D by substitute the Ө and Ф values in then solve the quadratic equation. This will give me the predicted value of D.

Let 1/(4 cos2ӨL(sinӨ - μcosӨ))=a; μ= tanФ =b; -H=c

I will produce a table in Microsoft Excel, showing the Ө in degrees, Ф 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:-  (-b±(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.

Conclusion

°(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 μ = tan Ф, because the angle is in decimal place(not accurate), take the tangent to that angle will increase the error again. If the value of μ 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 μ 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 error

The error for angle (Friction): 17.2%

After doing the calculation ,μ = tan Ф, the error increased.

The error for L (Length): 0.001m

Percentage error =  0.001/0.1 * 100% =1.0%

The error for H (Height): 0.001m

Percentage error =  0.001/0.282 * 100% =0.4%

The error for D (distance): 0.001m

Percentage error =  0.001/0.129 * 100% =0.8%

Total percentage error = 17.2% + 1.0%+0.4%+0.8% =19.4%

/

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