permits, repeat the process after switching cart-driver and dot-layer (record this data on another piece of paper). Data Tables: (2 marks) N1 - N2 N2 – N3 N3 – N4 N4 – N5 N5 – N6 N6 – N7 N7 – N8 N8 – N9 Time between dots (s) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 Distance between dots (m) Instantaneous Velocity (m/s) N9 – N10 N10 – N11 N11 – N12 N12 – N13 N13 – N14 N14 – N15 N15 – N16 Time between dots (s) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 Distance between dots (m) Instantaneous Velocity (m/s) Note: N stands for the dot number. Therefore, N1 is dot 1, N2 is dot 2, etc. Note: Instantaneous velocities will be calculated and filled in after distances are measured. Note: Assume the click-track has an accuracy of 1.0 s. Analysis: Sample Instantaneous velocity calculation. (2 marks) Average velocity calculation. (2 marks)
Launch angle: ___________° Launch velocity: ___________ m/s Analysis: Sample calculation for finding velocity components of the car. (2 marks) Sample calculations for finding time of flight of the car. (2 marks) Sample calculations for finding range of the car. (2 marks) Evaluation: See the instructions for writing evaluations. (5 marks) ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… P20 Unit A: Kinematics Water Rocket Lab ver 2
Time between dots (s) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 Distance between dots (m) 2.14 1.89 1.69 1.58 2.60 2.69 2.79 2.08 Instantaneous Velocity (m/s) 2.1 1.9 1.7 1.6 2.6 2.7 2.8 2.1 N9 – N10 N10 – N11 N11 – N12 N12 – N13 N13 – N14 N14 – N15 N15 – N16 Time between dots (s) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 Distance between dots (m) 2.69 2.79 2.08 2.16 2.17 1.97 2.11 Instantaneous Velocity (m/s) 2.7 2.8 2.1 2.2 2.2 2.0 2.1 Sample Instantaneous Velocity Calculation: v = d / t v = (2.14 m) / (1.0 s) = 2.1 m/s Sample Average Velocity Calculation: (SLOPE) v = (32.5 m/s – 5.0 m/s) / (14.5 s – 2.5 s) = 2.3 m/s Sample Distance vs. Time Graph: Sample Data for “Uniform Motion” Lab Average # of intervals per second = 100 (approximately) t = 0.0100 s N1 - N2 N2 – N3 N3 – N4 N4 – N5 N5 – N6 N6 – N7 N7 – N8 N8 – N9 Time between dots (s) 0.0100 0.0100 0.0100 0.0100 0.0100 0.0100 0.0100 0.0100 Distance between dots (m) 0.0090 0.010 0.010 0.011 0.009 0.010 0.012 0.011 Instantaneous Velocity (m/s) 0.090 0.10 0.10 0.11 0.090 0.10 0.12 0.11 Slope = velocity = 0.98 m/s Area under graph = 8.0 cm
Sample Graphs for “Acceleration due to Gravity” Dry Lab Sample Graphs for “Hot Wheels Projectile Motion” Lab initial velocity: 3.3 m/s manipulated variable launch angle: 26° controlled variable displacement: 0.87 m responding variable Estimated Sample Data for “The Mass vs. Weight Game” Lab Mass (g) Weight (N) 526 5.2 124 1.2 441 4.4 575 5.6 121 1.2 220 2.2 225 2.2 330 3.2 121 1.2 567 5.6 321 3.2 444 4.3 412 4.1 120 1.2 321 3.1 Sample Data for “Calculating the Coefficient due to Friction” Lab. mtire+masses (kg) 3.950 4.450 4.950 5.450 5.950 6.450 6.950 7.450 7.950 8.450 FN (N) 38.75 43.65 48.56 53.46 58.37 63.27 68.18 73.08 77.99 82.89 FF Kinetic (N) 17 20 22 24 26 27 30 34 37 40
Angle (°) x – acceleration (m/s2) parallel force (mN) y – acceleration (m/s2) perpendicular force (mN) 0 0 0 10 2.3 x 102 10 1.7 39 9.6 2.2 x 102 20 3.4 79 9.2 2.1 x 102 30 5.1 1.2 x 102 8.5 2.0 x 102 40 6.5 1.5 x 102 7.6 1.8 x 102 Sample Calculation for 30°: Parallel Force of Gravity = mgsin(θ) Parallel Force of Gravity = (0.02314 kg)(9.81 m/s2)sin(30°) = 0.114 N = 1.1 x 102 mN Perpendicular Force of Gravity = mgcos(θ) Perpendicular Force of Gravity = (0.02314 kg)(9.81 m/s2)cos(30°) = 0.197 N = 2.0 x 102 mN Sample Data for “The Flying Pig” Lab Data: Radius (m) Time for 5 Revolutions (s) Average Time (s) Period (s) Speed (m/s) Time 1 Time 2 Time 3 1.000 m 16.122 16.307 15.887 16.105 3.221 1.951 Mass of Pig: _500 g______ Length of String: _273.3 cm___ Analysis: Experimental Speed Calculation: v = 2πr/T = 2π(1.000 m) / (3.221 s) = 1.951 m/s Theoretical Speed Calculations: Draw a free body diagram of the forces acting on the pig.
Using a second diagram of the length of string and radius, determine the value of θ in the last question. θ = sin-1 (r/L) θ = sin-1(1.000m /2.733 m) θ = 21.46° Use the free body diagram to determine the centripetal force acting on the pig. Fc = Fg x tanθ Fc = (0.500 kg x 9.81 m/s2) x tan(21.46°) Fc = 1.928 N Using the centripetal force, determine the theoretical velocity of the pig. Fc = mv2/r v = (Fc r / m)1/2 = 1.96 m/s * Note: It seems that better results are yielded by using a larger mass in the experiment. * Sample Calculations for “Conservation of Energy in Hot Wheels” Lab Data: Trial Number photogate 1 time (s) photogate 2 time (s) 1 0.056 0.042 2 0.058 0.041 3 0.063 0.043 4 0.058 0.042 Average 0.059 0.042 h = _84.5 cm_ d = _29.0 cm_ s1 = _5.5cm__ s2 = _9.5 cm__ Analysis:
Actual velocity calculation for average of times through photogate 1 and the average through photogate 2. (2 marks) v = d/t v = (0.055 m)/(0.059 s) = 0.932 m/s vphotogate 1 = _0.932 m/s_ v = d/t v = (0.095 m)/(0.042 s) = 1.61 m/s vphotogate 2 = _1.61 m/s__ Conservation of energy calculation to find theoretical speed at photogate 1. Ep = Ep + Ek mgh = mgh + 1/2mv2 [2(g(0.845 m) – g(0.290 m))] = 3.30 m/s vphotogate 1 = _3.30 m/s__ Conservation of energy calculation to find theoretical speed at photogate 2. Ep = Ek mgh = 1/2mv2 [2g(0.845 m)] = 4.07 m/s vphotogate 2 = _4.07 m/s__ Sample Data for “Conservation of Energy in a Pendulum” Lab d = 3.0 cm t1 = 0.177 s h1 = 21.6 cm t2 = 0.175 s tave = 0.176 s h2 = 22.3 cm t3 = 0.176 s m = 500 g actual maximum velocity (at equilibrium) v = d/t v = 0.030 m / 0.176 s = 0.17 m/s Potential Energy at maximum amplitude: PE = mgh = (0.500 kg)(9.81 m/s2)(0.233 m – 0.216 m) = 0.0834 J Theoretical Velocity from Conservation of Energy Calculation: KE = 1/2mv2 0.0834 J = ½(0.500kg)(v2) v = 0.578 m/s