In this lab, we dropped a ball bearing from a ramp and our motive was to predict where it would land and put the target there. We used the photo-gates to calculate the horizontal velocity of the ball right before it began free-fall. We did this by putting the photo gate right beside each other and then calculating the distance from one photo gate to another. Then we sensor told us how long it took for the metal ball to go through the gates. Eventually we had displacement and time, which gave us velocity. We then used the height of the table to calculate the time of flight. Finally, we multiplied the time of flight by the horizontal velocity to find the range of the ball.
Procedure:
1) Attach the Photo-gates really close to one another, as this will give you a close to instantaneous velocity. Attach the wires to the adapter and the adapter to the main reader.Find the distance between the center of both the gates. and record the time (between gates). Use these numbers to calculate the horizontal speed of the ball.
2) Find the Vertical height of the desk. Use the height to find the time of flight of the ball. (hint: initial Vy=0m/s. Multiply the time of flight by the velocity you found in step one, this should give you the range of the ball. you should place the bulls-eye sheet at this point. |
Our Calculations
Projectile motion performance
- We decided to roll the metal ball down the ramp 8 times through the gates and then we would get the average from those 8 values.
Time between Photo gate time values:
1. 0.008754 s
2. 0.008727 s
3. 0.008814 s
4. 0.008680 s Average out of those 8 times: 0.008840 s
5. 0.008990 s
6. 0.009108 s
7. 0.008740 s
8. 0.008910 s
Now we took a small ruler and measured the distance between the two gates (∆dg) to be 0.0153m
Velocity = ∆dg/∆t
Now we split our findings in two columns: Vertical and horizontal
Horizontal
Trying to find ∆dx
T = ? Vix = v = ∆dg/∆t =1.7308m/s *sub in equation (3) from vertical component with time and the velocity with Vix = ∆dg/∆t Range=(dx) = (∆t • Vix) = (√(2∆dy)/a)( ∆dg/∆t) = 75.27 cm =0.7527 m Therefore, the distance we should place the target away from the island is 0.75m |
Vertical
Viy = 0
∆dy = -0.9267m² T = ? Gravity (a) =9.8m/sΩ≈ç√∫∫˜µµ≤≥÷æ…¬˚∆˙∆˙©ƒ∂ßå Equations: (1) ∆d = Viy∆t + (at2)/2 (2) ∆dy = at2/2 (3) ∆t = √ ((2∆dy)/a) |
Observations:
Analysis:
After having dropped the ball bearing, we noticed that it deviated quite far from what we calculated(0.75m (10 on the bullseye)). This may have resulted from measuring errors that occurred near the beginning of the experiment such as the initial velocity of the ball bearing or the displacement between the two gates. Another problem was that the equation that we used are for ideal conditions which would mean that there is no air resistance. However, in reality there is actually air resistance causing the ball to decelerate or slow down which would cause it to land earlier than anticipated. Therefore, if we were to do the experiment again we would account for the air resistance by moving the bullseye closer causing the ball bearing to land even closer to the ten on the bullseye.