The greatest distance should be achieved at around 45o with each mass of ball bearing because the horizontal and vertical components of velocity equal each other. Any angle above 45o will have too much vertical velocity and not enough horizontal force, therefore just going up and down, without gaining much distance. Angles less than 45o will have more horizontal force, but not the height, so will not be able to achieve the maximum distance, falling quickly to earth.
Graphs to plot
- Distance-angle with different data series for different ball bearings
e.g.
angle
-
sinθcosθ-distance
e.g.
sinθcosθ
3. Distance-1/mass
e.g.
1/mass
4. Estimated distance and actual distance against angle for a given diameter of ball bearing
This graph will show the difference between the actual results and the calculated distance using the energy (0.1J). It will be useful because, if the energy is correct, it will show how other factors including air resistance and measuring errors affect the results. Therefore the figures for estimated distance will probably be larger, i.e. the curve will be above the actual results as there is no air resistance.
Sample calculation
Using the result:
- Average distance = (1395+1412+1418)/3 = 1408.3mm
-
Volume of ball bearing = [4/3*Pi*r3]
= 4/3 * 3.142 * (0.01/2 – radius)3
= 5.24E-07
-
Mass of ball bearing = density * volume
= 8020 * 5.24E-07 = 0.004Kg
-
Initial Velocities =
-
Initial velocity = √(0.1/0.5*0.004) = 6.9ms-1
-
Initial velocity vertically = 6.9 * sin0.262 = 1.786 ms-1
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Initial velocity horizontally = 6.9 * cos0.262 = 6.666 ms-1
(note: in excel radians have to be used instead of degrees)
-
Total flight time: as we know the displacement (0), acceleration (-9.8) and initial vertical velocity (1.786) the equation s = ut + ½at2 can be used.
(rearranged) time = 2(Uv/gravity)
= 2(1.786/9.8)
= 0.365 seconds
- We assume no air resistance so
Distance traveled = time * initial velocity horizontally
S = 0.365 * 6.666 = 2.43m
Conclusions
My original prediction is supported by the results obtained. Looking at the distance-mass graph it is clear that the smaller the mass the greater the distance it will travel, at a set angle. This is due to the fact the launcher supplies each projectile with the same amount of energy on each release. Therefore smaller masses will achieve a greater initial velocity than larger masses, because of less gravitational force acting on them, and consequently will go further.
From the distance-angle graph it can be seen that between 40-45o the projectile travels the furthest, whatever the mass. At 45o the vertical and horizontal velocity components are equal. 45o is therefore a critical angle and launches above and below this angle will not travel as far. Any angle above 45o will have too much vertical velocity and not enough horizontal force, and will just rise and fall without gaining much horizontal distance. Angles less than 45o will have more horizontal force, but not the height, so will not be able to achieve the maximum distance, falling quickly to earth.
Both the graph cosθsinθ - distance and 1/mass - distance show straight line relationships as expected by my prediction. The first of these graphs is a straight line because the launch angle is directly proportional to the distance travelled. The gradient of the line is 2*(initial velocity)2/gravity. As the mass remains constant the gradient will be constant achieving a straight line from the equation: S[y]=(2U2sinθcosθ[x])/g. The second graph’s straight line is not as well defined as the first graph, but is still easily visible. This shows that distance is proportional to 1/mass, in agreement to the equation:
S[y] = (4Ke) * sinθcosθ.
mass[x] *gravity
The final graph (actual/estimated distance - angle) shows what the actual results should have looked like without any errors or air resistance in the experiment. The estimated results show that the actual results follow the correct trend of increasing distance up to 45o and then decreasing again. However the actual distances are below what was expected. This may have been due to a number of factors (described below) or air resistance, which caused drag on the projectile, reducing its horizontal velocity.
Errors
All the measurements recorded have a small degree of inaccuracy. For example, the distance the projectile travelled was measured with meter rulers to the crater formed in the sandpit, and the accuracy of this measurement is about +/- 1-2mm. This would not affect the results considerably as it is only a small percentage error - 2/2000 = 0.1%. Other small inaccuracies may have occurred with measuring the diameter of the ball bearing (+/- 0.1mm), adjusting the launch angle (+/- 0.5o) and aligning launcher at the same height as the sandpit (+/- 2mm). I feel the biggest error was in measuring the energy of the spring. This is probably why the calculated distance the ball bearing should have travelled is a lot further than the actual distance travelled. If this was the correct energy
provided by the spring air resistance would have reduced the distance travelled by almost 50%! (2380/4859*100 = 49%). Considering the experiment was conducted in a classroom without any wind this seems a very large value.