PHY 101L Module Six Lab Report

PHY 101L Module Six Lab Report Name: Dmitrius Washburn Date: 2/13/2024 Complete this lab report by replacing the bracketed text with the relevant information. Activity 1: Elastic Collision with Equal Masses 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 0.0641 kg 0.5 m Trial 1: 0.47 s 0.46 s 1.09 m/s Trial 2: 0.47 s Trial 3: 0.45 s 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 ’ 0.0641 kg 0 m Trial 1: 0.17 s 0.16 s 0 Trial 2: 0.14 s Trial 3: 0.16 s 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 ’ 0.0632 kg 0.5 m Trial 1: 0.40 s 0.41 s 1.22 m/s Trial 2: 0.43 s Trial 3: 0.41 s 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). Before Collision = 0.069869 kg*m/s After Collision = 0.077104 kg*m/s 2. Calculate the percent difference between the two values. 6.67% 3. Explain any difference in the values before and after the collision. Before the collision had a lower momentum total than after the collision. This is most likely because cart B had a higher final velocity as it traveled a faster time over the same distance than cart A had traveled. Therefore the elastic collision of the bumper against the band actually gave more force to cart B. This may also be because the bumper band in front of the card was able to absorb energy from cart A and amplify it with the recoil of the bumper band around cart B. Plus any forces acting against the carts that weren’t accounted for such as friction or air resistance. Activity 2: Elastic Collision: Mass Added to Cart A 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 0.1843 kg 0.5 m Trial 1: 0.37 s 0.35 s 1.43 m/s Trial 2: 0.34 s Trial 3: 0.34 s 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 ’ 0.1843 kg 0.20 m Trial 1: 0.46 s 0.46 s 0.43 m/s Trial 2: 0.47 s Trial 3: 0.44 s 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 ’ 0.0632 kg 0.5 m Trial 1: 0.33 s 0.34 s 1.47 m/s Trial 2: 0.35 s Trial 3: 0.34 s

Cart A mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d / t (m/s) v A ’ 0.0892 kg -0.10 m Trial 1: 0.20 s 0.20 s -0.5 m/s Trial 2: 0.21 s Trial 3: 0.18 s 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 ’ 0.3244 kg 0.35 m Trial 1: 0.87 s 0.85 s 0.41 m/s Trial 2: 0.85 s Trial 3: 0.84 s 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 | × 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). Before Collision = 0.148964 kg*m/s After Collision = 0.088404 kg*m/s 2. Calculate the percent difference between the two values. 31.35% 3. Explain any difference in the values before and after the collision. Before the collision the momentum is higher than after the collision. This could be in part because the velocity of cart A is actually negative after the collision compared to before the collision, therefore affecting the amount of momentum after the collision. Plus any forces acting against the carts that weren’t accounted for such as friction or air resistance.

Lab Questions: Activities 1–3 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? Friction, air resistance, human error, improper set up of the system. 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 continued to move forward rather than stopping or moving in reverse when it collided with Cart B which had no mass. The law of conservation states that Cart A continue to moving in the direction it was going because there are no external forces to stop or reverse its motion. The law of conservation ensures that its momentum remains unchanged by the collision with Cart B. 3. In one of the experiments, Cart A may reverse direction after the collision. How is this accounted for in your calculations? The distance it traveled is marked as negative because it isn’t going in a positive direction along the system. This inevitably caused the velocity to also be negative to reflect it moving against the original flow of the system.