I am going to investigate the relationship between the velocity of a moving object, and the distance it travels down a ramp, using secondary data obtained by a class experiment.

The apparatus was set up as shown above and illustrates a runaway vehicle down a hill. The light gate was placed at several points along the slope and measured the velocity of the card passing through it.

The trolley, of mass 1000g (1kg), was released 126cm up the slope from front of the card. The palm top then measured the time it took for the whole piece of card to pass through the light gate.

Once this was done the light gate was moved down the slope by 10cm at a time and again recorded the time it took for the card to pass through the light gate. This was carried out for 8 different locations. Each location’s time was repeated to end up with 3 readings. The average of these could then be taken and used as the time it took for the card to pass through the light gate.

To reduce the fiction of the wheel axis on the trolley, I have sprayed it with a lubricant (WD40).

The results I have been given are as follows:

Distance to Light Gate (m)

Velocity (m/s)

1st go

2nd go

3rd go

Mean Velocity

0.8

0.565

0.556

0.556

0.559

0.7

0.532

0.518

0.525

0.525

0.6

0.487

0.484

0.481

0.484

0.5

0.449

0.436

0.437

0.441

0.4

0.395

0.390

0.393

0.393

0.3

0.339

0.338

0.339

0.339

0.2

0.277

0.277

0.274

0.276

0.1

0.186

0.191

0.190

0.189

I have decided to make a preliminary graph to show my expected results.

The graph above shows that as the slope distance increases the velocity of the trolley must increase. This can be seen in the fact that the time it takes to pass through the light gate decreases. It can also be said that there is a greater change in velocity at the start of the ramp than at the end, which can be seen due to the fact that the gradient becomes shallower. This must mean that there is a larger force opposing the trolley as it picks up speed and could be due to air resistance.

In order to calculate velocity and other such information about the trolley I will use the SUVAT equations as well as Newton’s Second Law. This will enable me to make sense of the data that I am provided with, which includes the time it takes for the card to pass through the light gate and the slope distance. Below are shown the formulae I will use.

1. a = (v-u)/ t a = Acceleration

2. v2 = u2 + 2as v = Final Velocity

3. s = ut + 1/2at2 u = Initial Velocity

4. s = 1/2(v + u)t s = Distance

5. v = u + at t = Time

As well as this I can use Newton’s Second Law to Model the Particle, in order to find out friction etc.

Newton’s second law states, ‘The Force, F, applied to a particle is proportional to the mass, m, of the particle and the acceleration produced.’

This can then be represented by the equation F = ma.

In order to model the trolley I must know the acceleration. I will therefore use the SUVAT equations first.

Firstly, I shall work out is the time that the trolley took to reach the light gate by rearranging the equation:

s = 1/2 (v + u)t

Therefore 2s = (v + u)t

t = 2s / v (as initial velocity is always zero)

Therefore for the 0.1m light gate the trolley takes:

t = 0.2 / 0.189

t = 1.06s

I can now do this for all the other light gate positions also.

I can now work out the acceleration of the trolley through the light gate by using the formula:

a = (v – u) / t

For the 0.1m light gate:

a = 0.189 / 1.06 (because u = 0)

a = 0.179ms-2

I will now apply this equation for all the other light gate positions.

Now that I have acceleration for the trolley I can model it going down a slope and find out the model acceleration. This value can then be subtracted from the actual value to give resistance to the path of the trolley.

This is the simplified right-angle triangle from the diagram on the page before. This will make it easier to see what is happening.

The angle theta (?) is the angle of the ramp, but first I will find the missing side using Pythagoras’s Theorem:

1.262 – 0.0262 = 1.587 ï¿½ X = V1.587 = 1.259m

Doing this would give me more ways of finding the angle if I need to.

Now I am going to find angle theta:

sin? = Opp/Hyp = 0.026/1.26 = 0.0206 ï¿½ sin-10.0206 = 1.18ï¿½.

Therefore using the rules of a triangle, ‘z’ must be 180 – 90 – 1.18 = 88.8ï¿½

Using this I can now work out what the acceleration of the model is. This should be greater than the values I obtained using the SUVAT formulas as I am not taking into account friction etc.

Perpendicular to the Slope: mgcos? = N

(1 x 9.8) cos1.18 = N

R = 9.80N

I can now calculate the friction along the slope at various distances down the slope. This is the overall resistance to the driving force of the trolley, so can include air resistance.

Along the Slope: F = mgsin? – ï¿½N (where F is total driving force)

At 0.1m down slope ma = mgsin? – ï¿½N

0.179 = 0.20 – 9.80ï¿½

– 0.021= – 9.80ï¿½

0.0021 = ï¿½

At 0.8m down the slope: F = mgsin? – ï¿½N

ma = mgsin? – ï¿½N

0.195 = 0.20 – 9.80ï¿½

– 0.005 = – 9.80ï¿½

0.0005 = ï¿½

Using this formula I have calculated the friction for all the other distances and it is shown in the results table.

As well this I can calculate the gravitational potential energy and the kinetic energy. I will then be able to take the kinetic energy away from the gravitational potential energy to work out the energy lost in the form of sound, heat etc. Below is shown two examples of each.

Kinetic energy = 1/2mv2

At 0.1m down the slope: k.e. = 1/2 x 1 x 0.5592 = 0.018J

At 0.8m down the slope: k.e. = 1/2 x 1 x 0.1892 = 0.156J

Gravitational Potential Energy = mgh

In order to work out the gravitational potential energy I must use the trigonometry calculated earlier to find the height of the trolley at certain light gates. For simplicity only two examples are shown below. The others are shown in the results table later on.

At 0.1m down the slope: (from 1.26m)

Angle ? = 1.18ï¿½

If sin ? = X / 1.16

Then X = 1.16 sin ?

X = 0.024m

Therefore I can now calculate the gravitational potential energy of the trolley at the 0.1m light gate.

G.p.e = Mass x Gravity x Height (in metres)

= 1 x 9.8 x 0.024

= 0.23J

At 0.8m down the slope:

Angle ? = 1.18ï¿½

If sin ? = Y / 0.46

Then Y = 0.46 sin ?

Y = 9.5×10-3m

Therefore G.p.e = Mass x Gravity x Height

= 1 x 9.8 x (9.5×10-3)

= 0.093J

With my findings, I can now find out the energy lost in the form of sound, heat etc. To do this I will take away the kinetic energy from the gravitational potential energy.

For 0.1m down the slope = 0.234 – 0.018 = 0.216J

For 0.8m down the slope = 0.093 – 0.156 = – 0.063J

See results table for other readings.

From this you can see that by the time it reaches 0.8m away, energy is actually being taken in and as the light gate gets closer, there isn’t time for this effect to happen; therefore the trolley must loose most heat/sound energy when it first starts accelerating.

Error:

Measuring the distance from where the trolley is released was a source of error in the experiment. If the trolley was released slightly in front or behind from where it should have been released from, this would cause the time for the card to pass through the light gate to differ.

As well as this the card may have hit the light gate causing it to drastically slow down and therefore cause the time taken for it to pass through the light gate to increase. This would further have an impact on the rest of the experiment because the shape of the card may have been modified upon collision – E.g. initially the length of the card which the light gate is measuring would be ‘n’cm. If, however, it then crashes, the shape of the card would be distorted. The point at which the light gate is measuring the time for the card to pass through the light gate may then increase as the card has crumpled and so it longer in length and vice versa.

We must also consider the fact that at some point the slope may have slightly slipped, causing the gradient of the slope to change. This would obviously greatly affect the rate of acceleration of the trolley and produce inaccurate results.

There are therefore many sources of error in this procedure which must be accounted for in the experiment.

I can draw many conclusions by examining the graphs I have created from my results:

Graph 1: Comparing Actual Velocity of Trolley as it goes down the slope.

From the graph you can see that as the trolley goes down the hill it picks up speed. The greatest change in velocity occurs at the top of the slope and can be described due to the fact that at this point is has the most G.p.e to convert to k.e.

As the trolley goes down the slope, however, it is clear that the velocity increase is not proportional to the distance it travels. This can be accounted for due to the friction I have calculated. However this remains constant so there must be another factor involved. It must therefore be air resistance as this increases as you go faster. Therefore as the trolley picks up speed, the resultant force decreases until it reaches its terminal velocity.

I chose to do this graph as it is a visual aid in seeing how the trolley changes in speed through its course.

Graph 2: Graph showing how the long it takes for the trolley to go down the slope to various distances.

The graph shows an almost proportional trend in that as the distance increases so does the time it takes for the trolley to go this distance. There is however a slight curve. This can be accounted for in that the trolley has more air resistance as it goes down the slope, so the time it takes must increase. Therefore the rate must slow, so the graph starts to level out. However it seems that the trolley reaches its terminal velocity reasonably quickly as the rate is almost constant.

Graph 3: Graph showing how the Kinetic Energy of the trolley changes as it goes down the slope.

The graph shows as that as the trolley goes further down the slope, its kinetic energy increases. This is very easy to explain in that as it moves down the slope it picks up more speed. The equation for kinetic energy is k.e. = 1/2mv2. The mass of the particle does not change and so the rise in kinetic energy is solely due to the trolley increasing in speed. When it is higher up the slope, it has more gravitational potential energy so it cannot posses as much kinetic. Lower down the slope it has less G.p.e. so it can posses more k.e.

Graph 4: Graph to show how the Gravitational Potential Energy of the trolley changes as it goes down the slope.

The graph shows that as the trolley travels further down the slope it has less gravitational potential energy. This is also easy to explain in that when it is at the top of the ramp it has more height. Since G.p.e. = mgh, the more height it has the more G.p.e. it shall have. As it moves down the slope it is not as high up, so it has less G.p.e.

Graph 5: Comparing G.p.e. with k.e.

This graph basically illustrates the connection between G.p.e. and k.e. It shows that when one increases the other must decrease. Using this graph and plotting interpolation lines and then using the G.p.e. against distance graph one can work out the position of the trolley at a given location.