PHY 150 M7 Momentum Lab Report_KAbbott

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Apr 3, 2024

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Momentum Keon Abbott 12/8/23
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 0.081 kg 0.5 m Trial 1: .49 0.51 0.9804 Trial 2: .53 Trial 3: .52 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 0.081 kg 0 m 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 0.068 kg 0.5 m Trial 1: .34 0.37 1.3514 Trial 2: .38 Trial 3: .39 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). 1.--- 0.081(0.9804) + 0 = 0.07964 2.--- 0 + 0.068(1.3514) = 0.09205 2. Calculate the percent difference between the two values. 1 © 2016 Carolina Biological Supply Company
| 0.079 0.092 0.079 + 0.092 2 | × 100 = 3.8% 3. Explain any difference in the values before and after the collision. It was observed that most of the energy was conserved from the collision, and I realized that there was stark difference in values before and after the collision. This led me to determine that it is vastly difficult to truly accurately measure time. Activity 2: Elastic Collision: Mass Added to Cart A Data Table 2 Table 2A. 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 0.24 kg 0,5 m Trial 1: .37 0.39 1.2821 Trial 2: .39 Trial 3: .32 Table 2B. 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 0.24 kg 0.2 m Trial 1: .16 0.14 1.4286 Trial 2: .12 Trial 3: .15 Table 2C . 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 0.066 0.5 m Trial 1: .35 0.35 1.4286 Trial 2: .37 Trial 3: .34 Calculations for Activity 2. Elastic Collision: Mass Added to Cart A . 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 2 © 2016 Carolina Biological Supply Company
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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). 1. 0.24(1.28) + 0 = 0.3072 2. 0.24(1.4286) + 0.066(1.4286) = 0.4372 2. Calculate the percent difference between the two values. | 0.3072 0.4372 0.3072 + 0.4372 2 | × 100 = 8.7 % 3. Explain any difference in the values before and after the collision. There is a huge weight difference between the two carts and because of that, the kinetic energy was more conserved by cart A. This determination was made after cart A kept 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 .342 kg .40 m Trial 1: .26 0.29 1.3462 Trial 2: .32 Trial 3: .31 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 .342 kg -.1 m Trial 1: .11 0.13 -0.769 Trial 2: .15 Trial 3: .13 Table 3C. Cart B after collision. 3 © 2016 Carolina Biological Supply Company
Cart B mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d/t (m/s) v B .438 kg .5 m Trial 1: .78 0.74 0.6757 Trial 2: .69 Trial 3: .75 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% 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). 1. 0.342(1.3462) + 0 = 0.4618 2. 0.342(-0.769) + 0.438(0.6757) = 0.0329 2. Calculate the percent difference between the two values. | 0.4618 0.0329 0.4618 + 0.0329 2 | × 100 = 43.3% 3. Explain any difference in the values before and after the collision. One of the observations I made is that when weight was added to Cart B, it reduced the velocity, which in exchange reduced the kinetic energy that was transferred. The kinetic energy is transferred from Cart A to Cart B, but because of the additional weight on Cart B it created more force reaction. This in turn caused Cart A to propel backwards which conserves more kinetic energy for Cart A. Questions for Momentum : 4 © 2016 Carolina Biological Supply Company
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? There are two outside forces that I can think of that would have affecte the results of the experiment. One of them was the human nature of making errors with the results of the times. It is not a computer-based result and we’re depending on our eyes and our reaction time when we start and end a trial. Another factor would be friction which then could cause the loss of momentum. 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? For the fact that Cart A had more mass than Cart B, it was able to continue to roll forward after collision. When this happened, Cart A was able to conserve its momentum. 3. In one of the experiments, Cart A may reverse direction after the collision. How is this accounted for in your calculations? Cart A did have a negative distance after colliding with Cart B which resulted in a negative velocity. 5 © 2016 Carolina Biological Supply Company
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