PHY 150 M7 Momentum Lab Report

Momentum Joel Hernandez 12/7/23

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 0.081kg 0.5m Trial 1: 0.5 0.5 1m/s Trial 2: 0.49 Trial 3: 0.51 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 ’ 0.081kg 0m Trial 1: 0 0 0 Trial 2: 0 Trial 3: 0 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 ’ 0.065kg 0.5m Trial 1: 0.37 0.39 1.28m/s Trial 2: 0.38 Trial 3: 0.42 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 = 𝑚 𝐴 𝒗 𝐴 + 𝑚 𝐵 𝒗 𝐵 Momentum after the collision = 𝑚 𝐴 𝒗 𝐴 ′ + 𝑚 𝐵 𝒗 𝐵 ′ 𝑚 𝐴 𝒗 𝐴 + 𝑚 𝐵 𝒗 𝐵 = 𝑚 𝐴 𝒗 𝐴 ′ + 𝑚 𝐵 𝒗 𝐵 ′ 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). 0.081(1)+0=0.081 0+0.065(1.28)=0.083 2. Calculate the percent difference between the two values. (0.81-0.83)/(0.81+0.83)/2=0.006*100%=0.6% 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. (0.28-0.3)/(0.28+0.3)/2=0.017*100%=1.7% 3. Explain any difference in the values before and after the collision. Due to the major difference in weight between the two carts, cart A conserves the most kinetic energy as it keeps rolling after the collision. 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 0.257kg 0.35m Trial 1: 0.21 0.227 1.542m/s Trial 2: 0.23 Trial 3: 0.24 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 ’ 0.257kg -0.1m Trial 1: 0.1 0.1 -1m/s Trial 2: 0.12 Trial 3: 0.09 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 ’ 0.385kg 0.5m Trial 1: 0.7 0.627 0.797m/s Trial 2: 0.65 Trial 3: 0.72 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 = 𝑚 𝐴 𝒗 𝐴 + 𝑚 𝐵 𝒗 𝐵 Momentum after the collision = 𝑚 𝐴 𝒗 𝐴 ′ + 𝑚 𝐵 𝒗 𝐵 ′ 𝑚 𝐴 𝒗 𝐴 + 𝑚 𝐵 𝒗 𝐵 = 𝑚 𝐴 𝒗 𝐴 ′ + 𝑚 𝐵 𝒗 𝐵 ′ Percent difference = | first value − second value first value + second value 2 | x 100% 3 © 2016 Carolina Biological Supply Company

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). 0.257(1.542)+0=0.396 0.257(-1)+0.385(0.797)=0.05 2. Calculate the percent difference between the two values. (0.396-0.05)/ (0.396+0.05) /2=0.388*100%=38.8% 3. Explain any difference in the values before and after the collision. By adding weight to cart B its velocity is reduced meaning it has less kinetic energy transferer to it. The kinetic energy from cart A is transferred to cart B, but due to the added weight the force reaction is greater sending cart A back and conserving more kinetic energy for cart A. 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? The friction from the foam board, friction from the floor (depending on the surface, carpet in my case), the friction from the faulty front wheel of cart A and the difference in mass. 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? Since cart A keeps rolling forward, it means that it is conserving more of its momentum due to its greater mass. 3. In one of the experiments, Cart A may reverse direction after the collision. How is this accounted for in your calculations? This is accounted in my calculations by the negative distance which resulted in negative velocity for cart A after collicion. 4 © 2016 Carolina Biological Supply Company