Why Isn't My Physics Experiment Calculating Distance Correctly?

[lucky_boy]

PLEASE HELP! Physics coursework!

I have to use the formulas:

v=√2gh

and

t= √((2s) ÷ a)

once i have these i should be able to use 'd=s * t' to find the distance but it does not work.

The experiment is investigating the range of a ball bearing released from an inclined track. The track is on a table 0.84m off the ground and the height from the track to one of the release points (0.6 metres along track) is 0.185m.

This should work but doesn't and any help would be greatfully received. As quickly as possible please.

 

[Hootenanny]

Assuming you have applied them correctly, the formulae are valid. Perhaps if you post what you have attempted we could see where you've made a mistake.

 

[lucky_boy]

v=√2 * 10 * (0.84 + 0.185) = 4.527692569

and

t= √((2 * 0.84) ÷ 10) = 0.40987803

and

d = s * t = 1.855801711

but i have done the experiment and i got an average of 0.5833m

 

[Hootenanny]

You need to resolve the velocity into verticle and horizontal compontents thus;

[tex]V_{x} = V\cos\theta[/tex]

[tex]V_{y} = V\sin\theta[/tex]

Where [itex]\theta[/itex] is the angle between the table and your track. The is no acceleration in the horizontal plane so Vx is constant. The acceleration in the vertical plane is equal to g.

 

[lucky_boy]

Is that ok? I really need this done as quickly as possible, thankyou.

 

[lucky_boy]

I understand that buut there's nothing wrong with my formulas and numbers so what is going on?

 

[Hootenanny]

I understand that buut there's nothing wrong with my formulas and numbers so what is going on?

I don't think you so understand. There's nothing wrong with your formulae, but there is something wrong with the way you are applying them.

v=√2 * 10 * (0.84 + 0.185) = 4.527692569

This is not correct, you are calculating the final velocity rather than the intial (after the ramp). By using the next equation as such;

t= √((2 * 0.84) ÷ 10) = 0.40987803

You are assuming that gravity acts parallel to the balls trajectory, which it doesn't it always acts down. I suggest you get your textbook out and read the chapter on projectile motion. Also the above equation assumes intial velcoity = 0, which is not the case here. You will need to use additional formulae.

I'll show you the first few steps to solving this problem.

(1) Find the velocity (v) of ball when it leaves the inclined plane.
(2) Find the angle ([itex]\theta[/itex]) above the horizontal of the inclined plane (using basic trig).
(3) Use [itex]v_{y} = v\sin\theta[/itex] to find the vertical velocity.
(4) Use a kinetmatic equation to calculate the time taken for the ball to hit the floor.

 

[Arowana]

did you take errors into consideration?

like the ball slides instead of roll??