FINAL PROJECT PHYS

ds Momentum Howard Gilmore 20/06/2021

Activity 1: Elastic Collision with Equal Masses Data Table 1 Table 1A. Cart A before collision. Cart A mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d/t (m/s) v A 0.0581 0.5 Trial 1: 0.31 0.37 1.351 Trial 2: 0.46 Trial 3: 0.35 Table 1B. Cart A after collision. Cart A mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d/t (m/s) v A ’ 0.0581 0.7 Trial 1: 0.39 0.49 1.428 Trial 2: 0.44 Trial 3: 0.65 Table 1C. Cart B after collision. Cart B mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d/t (m/s) v B ’ 0.0659 0.5 Trial 1: 0.96 1.03 0.485 Trial 2: 0.94 Trial 3: 1.19 Calculations for Activity 1. Elastic Collision with Equal Masses Apply the law of conservation of momentum to the two-cart system by calculating the momentum before and after the collision. Helpful equations : Momentum before the collision = ? ? ? ? + ? ? ? ? Momentum after the collision = ? ? ? ? ′ + ? ? ? ? ′ ? ? ? ? + ? ? ? ? = ? ? ? ? ′ + ? ? ? ? ′ Percent difference = | first value − second value first value + second value 2 | x 100% 1. Calculate the momentum of the system before the collision (the left side of the equation) and after the collision (the right side of the equation). B: 0.0581(1.351)+0.0659(0) = 0.1239 A: 0.0581(1.428) + 0.0659(0.485) 2. Calculate the percent difference between the two values. 40.68% 3. Explain any difference in the values before and after the collision. 1 © 2016 Carolina Biological Supply Company

2. Calculate the percent difference between the two values. 62.64 3. Explain any difference in the values before and after the collision. Net External Impulse: 0.083 Cause : Natural Factor (Friction, Gravity, Air Resistance) Activity 3: Elastic Collision: Mass Added to Cart B Data Table 3 Table 3A. Cart A before collision. Cart A mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d/t (m/s) v A 0.0834 0.35 Trial 1: 0.33 0.33 1.06 Trial 2: 0.32 Trial 3: 0.33 Table 3B. Cart A after collision. Cart A mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d/t (m/s) v A ’ 0.0834 -0.28 Trial 1: 1.02 1.38 -.020 Trial 2: 1.35 Trial 3: 1.78 Table 3C. Cart B after collision. Cart B mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d/t (m/s) v B ’ 0.353 0.35 Trial 1: 1.10 1.17 0.30 Trial 2: 1.21 Trial 3: 1.19 Calculations for Activity 3. Elastic Collision: Mass Added to Cart B . Apply the law of conservation of momentum to the two-cart system by calculating the momentum before and after the collision. Helpful equations : Momentum before the collision = ? ? ? ? + ? ? ? ? Momentum after the collision = ? ? ? ? ′ + ? ? ? ? ′ ? ? ? ? + ? ? ? ? = ? ? ? ? ′ + ? ? ? ? ′ Percent difference = | first value − second value first value + second value 2 | x 100% 1. Calculate the momentum of the system before the collision (the left side of the equation) and after the collision (the right side of the equation). 3 © 2016 Carolina Biological Supply Company

B:0.0834+0.353=0.009 A: 0.0834 + 0.353= 0.089 2. Calculate the percent difference between the two values. 163.27% 3. Explain any difference in the values before and after the collision. Net External Impulse : -0.08 Cause : Natural Causes Questions for Momentum : 1. The law of conservation of momentum states that the total momentum before a collision equals the total momentum after a collision provided there are no outside forces acting on the objects in the system. What outside forces are acting on the present system that could affect the results of the experiments? Gravity and friction are the primary forces that effected the results air resistance was also a factor as the cart speeds up the resistance kept it from advancing further without additional force behind it 2. What did you observe when Cart A containing added mass collided with Cart B containing no mass? How does the law of conservation of momentum explain this collision? Cart A vs Cart B – A was heavier therefore air resistance was less against it as its weight carried it, momentum from cart B caused it to travel, when cart A collides with cart B it has a push back. 3. In one of the experiments, Cart A may reverse direction after the collision. How is this accounted for in your calculations? Because cart A bounces back it actually loses distance with generates the negative value we see, anytime distance is lost it’s a loss of value. 4 © 2016 Carolina Biological Supply Company