# Ball 1 was rolled down the incline from the top of

• 5
• 100% (11) 11 out of 11 people found this document helpful

This preview shows page 2 - 4 out of 5 pages.

Ball 1 was rolled down the incline from the top of the ruler 10. Distance was measured from the point on the floor directly below the edge of the table where ball 2 was sitting to the point on the floor where it makes first contact 11. Steps 9 and 10 were repeated 10 times and observations were recorded in table 2 and results were summarized F n F n F f Ball 2 initial/Ball 1 final F gh Ball 1 initial F gv F g F g Ball 2 final F g
Observations - Difficult to maintain exactly where ball landed - difficult to ensure ball 2 was directly in line with the path of ball 1 - when collision was not head on, ball one did not completely stop and ball 2 did not travel as far. Both balls had paths in directions slightly off the intended path Results Table 1 (only one ball used) Table 2 (elastic collision with 2 balls) Trials Horizontal distance (m) Trial 1 0.28 Trial 2 0.295 Trial 3 0.275 Trial 4 0.28 Trial 5 0.27 Trial 6 0.28 Trial 7 0.29 Trial 8 0.30 Trial 9 0.28 Trial 10 0.28 Avg horizontal distance = 0.283 m Avg horizontal distance = 0.288 m Vertical height of table h 1 = 0.73 m Vertical height of ruler (height above table) h 2 = 0.055 m Discussion 1. d x = 0.28 m d y = 0.73 m v i = 0 m/s a = 9.8 m/s 2 using kinematics equations for projectile motion, solve for t d y = v i t + 1/2at 2 0.73 = 0(t) + ½(9.8)t 2 0.73 = 4.9t 2 t = 0.148979 t = 0.38978 t = 0.39 s Since the horizontal velocity always remains constant, horizontal distance can be used to find horizontal velocity. This will be velocity as the ball leaves the table V x = d x /t Trials Horizontal distance (m) Trial 1 0.30 Trial 2 0.29 Trial 3 0.30 Trial 4 0.28 Trial 5 0.29 Trial 6 0.26 Trial 7 0.30 Trial 8 0.29 Trial 9 0.29 Trial 10 0.28
• • • 