PHY 101L Module Six Lab Report

PHY 101L Module Six Lab Report Name: Thomas Hubert Date: 10/07/2023 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 0.067 0.50 0.66 0.69 0.7246 0.73 0.68 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 ’ 0.067 0.03 0.00 0.00 0.00 0.00 0.00 Table 1C: Cart B After Collision Cart A mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d / t (m/s) v B ’ 0.067 0.50 0.53 0.693 0.7215 0.87 0.68 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 = 0.067 kg (0.7246 m/s) + 0.067 kg (0.7215 m/s) = 0.09689 kg*m/s After collision = 0.067 kg (0.00 m/s) + 0.067 kg (0.7215 m/s) = 0.04834 kg*m/s 2. Calculate the percent difference between the two values. Percent difference = |(0.09689 - 0.0483)/((0.09689 + 0.0483)/2)| x 100% = 66.93% 3. Explain any difference in the values before and after the collision. Most of the momentum was not conserved, this could be from friction and 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 0.193 0.50 0.57 0.506 0.9881 0.48 0.47 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 ’ 0.193 0.30 0.76 0.86 0.3488 1.01 0.81 Table 2C : Cart B After Collision Cart A mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d / t (m/s) v B ’ 0.193 0.50 0.41 0.40 1.25 0.38 0.41 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 :

Cart A mass, m (kg) Distance, d (m) Time, t (s) Average time, t (s) Velocity = d / t (m/s) v B ’ 0.094 0.35 1.46 1.43 0.2448 1.33 1.51 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: (0.094 x 0.7056) + (0.343 x 0.2448) = 0.1503 After Collision: (0.094 x 0.0806) + (0.343 x 0.2448) = 0.0764 2. Calculate the percent difference between the two values. Percent difference = |(0.1503 - 0.0764)/((0.1503 + 0.0764)/2)| x 100% = 65.2% 3. Explain any difference in the values before and after the collision. There was almost no loss of momentum. “ A perfectly elastic collision is a collision in which the total kinetic energy of the system of colliding objects is conserved.” ( Momentum and Collisions Review , n.d.) 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?

Gravity and friction are forces that can affect the results. “…if there is some external force that acts on all of the objects (gravity, for example, or friction), then this force changes the momentum of the system as a whole” (Moebs et al., 2016) There could also be imperfections on the surface of the table, and air resistance. 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? When Cart A collided with Cart B, Cart A slowed down, and Cart B gained speed. Cart A transferred some of its kinetic energy to Cart B. This caused Cart B to accelerate, while Cart A slowed down. 3. In one of the experiments, Cart A may reverse direction after the collision. How is this accounted for in your calculations? The reverse in direction is accounted for by having a negative value for the distance.