However, after interpolating values on the graph, it was shown that a y = x² relationship was not possible as when interpolating, as one value doubled, the X value did not quadruple as it would if there was a y = x² relationship.
With each mass addition the net force acting on the trolley increased. To calculate the net force acting after each addition of mass I used the following formula:
F = mg
Where F is the net force, m is the mass of the falling weight and g is the gravitational field strength which was taken to be 10 N kg¯¹
By comparing the net force acting on the system to the time it takes the mass to reach the floor, it is possible to produce a graph and find the impulse acting on the system and the momentum gain of the system.
To calculate the time values I first took the following formula:
s = ut + ½ at²
Where s = the distance that the trolley falls, a is the acceleration and t is the time.
This formula was re-arranged to get:
T = √2s/a
However, the acceleration of the trolley was unknown and so to calculate the time taken values, a formula for acceleration had to be substituted into the first formula.
Here the formula used was:
Force = Mass X Acceleration
This was re-arranged to give:
a = mg / Mass
Where a = acceleration, mg is the mass multiplied by the gravitational field strength (taken to be 10N/kg) is the weight and the Mass value is the total mass acting on the trolley calculated by the trolley mass of 1kg added to the varying mass pulling down on the trolley.
Here this formula was substituted into the first formula where:
A = mg / (M + m) substituted into t = √2s/a
This gave the following formula which could be used to calculate the time taken values.
T = √2s(M+m) / mg, where s is the height that the mass drops, M is the mass of the trolley, m is the mass of the falling weight and g is the gravitational field strength.
Below is a table comparing the net force acting on the system and the time taken for the mass to reach the floor.
Here the graph produced a curve with a negative correlation. This was expected as the more weight pulling down on the trolley, the faster the trolley will move and therefore the shorter the time taken for the mass to reach the ground.
To calculate the impulse acting on the system in Newtons seconds I used the following formula:
Impulse Acting = F X t
Where F is the force acting on the system and t is the time in seconds.
The momentum gain of the system should be equal to the impulse acting on the system and so this allowed me to check the values found.
To calculate the momentum gain of the system I used the following formula:
Momentum Gain = v[M+m]
Below is a table of all of the calculated data from the first test.
Here the impulse acting on the system and the momentum gain of the system values were very close to each other which helped to validate the results.
To calculate the work done on the system and the kinetic gain of the system a second test was used where the mass was kept constant at 400g, and the height of the mass was varied in 10cm intervals.
Here I would expect that as the height increases, the more kinetic energy it will be able to gain and so the greater the speed of the trolley.
Below is a results table and overleaf a graph for this second test.
Here, the graph produced a curve with a positive correlation. The curve showed that as the height increases, the rate of change of speed of the trolley decreases.
To calculate the work done for this second test the following formula was used:
Work Done = F X S
Where F is the net force and S is the distance the mass was dropped.
To calculate the kinetic energy gain of the system the following formula was used:
K.E. Gain = ½ v²[M+m]
Below is a table of this calculated data for the second test made.
Conclusion
By calculating the different data from the results using a spreadsheet I have been able to find numerous relationships.
These include:
- Distance is proportional to Time squared
- Kinetic Energy is proportional to the mass of the falling load
- The Kinetic Energy is proportional to the Gravitational Potential Energy
- Weight of falling load divided by Net Force is proportional to Acceleration squared.
