- #1
JMatch
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This is my physics coursework task and although I have had only minor difficulties in actually doing the work, I thought I would post some of the things that I found to verify.
Basically, I began with saying that the initial speed (u), initial height (h), gravitational constant (g) and the angle of projection (the) all played a role in determining the horizontal displacement (d) of the projected mass.
I varied the initial speed (u) by constructing a ramp and changing the vertical height of the mass 'up' the ramp (H). Keeping h, g constant and [tex]\theta[/tex] = 0, enabled me to find the initial speed as [tex] v = \sqrt{2gH}[/tex] (using conservation of energy).
I then found the time it would take for the mass to hit the floor, [tex]t = \sqrt{(2h)/g}[/tex] and combining this with the value for v enabled me to find out the horizontal displacement (d) in terms of H and h. [tex]d = 2\sqrt{Hh}[/tex]
That was the easy part. Deciding what factors made these theoretical results different from the experimental ones was a little harder.
Obviously air resistance will be a factor, as will the co-efficient of friction on the ramp. Those were the only two factors I could see that would make a difference.
I have a bit of a block when it comes to air resistance. I know I am wrong in thinking that air resistance is constant magnitude but varies in direction, however I am thinking it and it is annoying me.
Now I am forgetting the entire point of the task.. damn you PF!
(It should come to me soon, but any views so far?)
Basically, I began with saying that the initial speed (u), initial height (h), gravitational constant (g) and the angle of projection (the) all played a role in determining the horizontal displacement (d) of the projected mass.
I varied the initial speed (u) by constructing a ramp and changing the vertical height of the mass 'up' the ramp (H). Keeping h, g constant and [tex]\theta[/tex] = 0, enabled me to find the initial speed as [tex] v = \sqrt{2gH}[/tex] (using conservation of energy).
I then found the time it would take for the mass to hit the floor, [tex]t = \sqrt{(2h)/g}[/tex] and combining this with the value for v enabled me to find out the horizontal displacement (d) in terms of H and h. [tex]d = 2\sqrt{Hh}[/tex]
That was the easy part. Deciding what factors made these theoretical results different from the experimental ones was a little harder.
Obviously air resistance will be a factor, as will the co-efficient of friction on the ramp. Those were the only two factors I could see that would make a difference.
I have a bit of a block when it comes to air resistance. I know I am wrong in thinking that air resistance is constant magnitude but varies in direction, however I am thinking it and it is annoying me.
Now I am forgetting the entire point of the task.. damn you PF!
(It should come to me soon, but any views so far?)