The experiment consisted of recording the results of a small toy car being allowed to roll down a ramp and then leave the end of the ramp by continuing straight off the drop from the end of the desk.

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Josh Wakeford 12B1

27/04/03

Physics Coursework: Data Analysis

Explanation

The experiment consisted of recording the results of a small toy car being allowed to roll down a ramp and then leave the end of the ramp by continuing straight off the drop from the end of the desk.  I measured the ramp’s length, the height of the ramp, and the distance it covered horizontally.  I also measured the mass of the car.  It was necessary to measure these things so that I can perform calculations later.  All of the information I gathered can be seen in the diagram below.  

To get the results, I will attach ticker tape to the back of the car, so the car will pull the ticker tape through the machine as it moves, and therefore create a dot on the piece of ticker tape every 0.2 seconds.  When I come to analyse this, I will obviously not use every point, I will use probably every one point in five

Predictions

I think that the car is going to gradually accelerate as it goes down the ramp, therefore gradually gaining speed as it goes down.  Then the car will accelerate even more as it drops off the end of the ramp and the friction from the ramp will no longer be present.  Although there will still be air resistance, the friction from the ramp will be gone, and only friction from the air and from the ticker tape will be present.  If we were to assume no resistance, the car’s acceleration would go up to about 9.8ms-2 (although this value for gravity is not exact, gravity varies so much in different areas of the Earth that it is hard to get an exact value, 9.8 is judged to be close enough).  So the acceleration time graph will probably look like the one below:

You may wonder why the acceleration will only go up to 9.8, or close to, when the car goes off the end of the ramp, after all, gravity was acting upon the car before this time wasn’t it?

The acceleration will only go up to gravity – drag when there is no force acting upon the car in a vertically upwards direction.

Ie, the forces on the car will be as follows:

All these predictions assume that

“F = ma”

which is Newton’s second law of motion.

We can see that, when the car is on the ramp, there is a force from the ramp stopping the car from going directly downwards; however, gravity is still affecting the car.  Nevertheless, the gravity can only act directly downwards on the car, not parallel to the slope, so most of the acceleration due to gravity is cancelled out by the drag.

We can easily work out how much gravity will effect the car as it goes down the slope as we have the horizontal length and the vertical height and can therefore work out the angle the ramp is at.  If we take the sin of the angle and multiply it by the mass of the car multiplied by the acceleration due to gravity, we get the total force acting down the ramp due to gravity.  From this we can get the acceleration of the car down the ramp by dividing by the mass.  Now we know this, we can actually find out how long the car would spend on the ramp, and how long it will take to fall to the ground using suvat equations.  However, this all still assumes no drag or friction of any kind, which is obviously impossible, but it will tell us how much friction and drag there is present.

To find the acceleration on the ramp we must first find the angle it is at.

To find the angle θ we must use trigonometry.

We know:

Tan θ = 0.320/1.608

So Tan-1 (0.320/1.608) = θ

Therefore θ = 11.255°

To find the acceleration on the car due to gravity we do sinθ x the normal acceleration due to gravity.

So:

Acceleration = sin 11.255 x 9.8

Acceleration = 1.9m/s (2 sf)

We can also find out how long the car would spend on the ramp, and the time that it would take to hit the ground after leaving the ramp.  We know the length of the ramp, and we know the acceleration and initial speed of the car, so we can work out how long the car would take to go down the ramp, and fall to the ground below.

We work out the length of the ramp using Pythagoras’ Theorem:

Length2 = 0.3202 +1.6082

Length = √(0.3202 +1.6082)

Length = 1.639m

To find the time the car spends on the ramp, we use the equation s = ut + ½ at2

So, if there were no friction, the car would take 1.3 seconds to get to the end of the ramp.  We can also find out how long the car will take to fall to the ground after it has left the end of the ramp.  And we can use a similar method for this, however there is the added complication that the car already has a velocity when it leaves the end of the ramp, and even more complicated, it is not heading in a horizontal direction.  So first we must find its velocity, split it into the horizontal and vertical components, and then find from there what speed it will be heading down at when it leaves the ramp.

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To find the velocity we use v = u + at.

v = 0 +1.9 x 1.3

this gives us v = 2.47 ms-1

However, we must remember that this velocity is heading towards the floor at 11.255°.  Therefore to find the vertical velocity, assuming down in positive, we do

Velocity down = Speed x sin 11.255

V vertically = -0.48 ms-1

Velocity horizontally = Speed x cos 11.255

V horizontally = 2.24 ms-1

We find when the car hits the floor using s = ut + ½ at2 again.  This time we use the ...

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