6-1 lab report

PHY 101L Module Six Lab Report Name: Zachary Prat Date: 10/3/2023 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 0.0668 kg .35 m .35 s .316 1.10 m/s .29 s .31 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 ’ 0.0668 0.01 .29 .263 0.03 m/s .25 .25 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 ’ 0.07 .5 m .34 s .333 1.50 m/s .32 s .34 s Calculations for Activity 1: Elastic Collision with Equal Masses ? ? ? ? + ? ? ? ? = ? ? ? ? + ′ ? ? ? ? ′ 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). .0668 * 1.10 + .070 * 0 = 0.073 .0668 * 1.10 + .070 * 1.50 = 0.073 + .105 = 0.178

2. Calculate the percent difference between the two values. 0.073 - 0.178 = -0.105 0.073 + .178 = 0.251 -0.105/0.251/2 -20.9 % 3. Explain any difference in the values before and after the collision. The loss comes from energy being conserved at the point of collision between the carts. 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 0.1915 kg .5 m .64 s .62 .806 m/s .63 s .59 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 ’ 0.1915 kg .2 m .40 s .39 .51 m/s .37 s .40 s 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 ’ .07 kg .5 m .22 s 0.236 2.18 m/s .25 s .24 s Calculations for Activity 2: Elastic Collision: Mass Added to Cart A Helpful equations : Momentum before the collision = ? ? ? ? + ? ? ? ? Momentum after the collision = ? ? ? ? + ′ ? ? ? ? ′ ? ? ? ? + ? ? ? ? = ? ? ? ? + ′ ? ? ? ? ′ 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). .1915 * (.806) + .07 * (0) = 0.154

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). .0935 * (.2397) + .3323 * (0) = 0.224 .0935 * (-.330) + .3323 * (.507) = -0.198 6. Calculate the percent difference between the two values. .42/.026/2 = 8.07 % 7. Explain any difference in the values before and after the collision. Due to b carts extra mass, verse a cart, it took on a lot of the energy input and there didn’t have as much kinetic energy to be pushed to the line quicker. 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? I began to notice that the board was sagging towards the middle of the experiment. To keep it consistent, and following instructions, I had to use another book in the middle to prevent the sag from effecting the cart to much. While the angle stayed at 10 degrees, the book in the middle as extra support could allow for the cart to have less resistance rolling down the ramp and skew the results towards a faster time. 8. 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? With the spring atached to the b cart taking on the force from a cart, that was transferred from the cart resting at a static position, with potential energy, to kinetic energy. With out my “Gem Trail of Washington” book blocking the path, it would have kept moving so long as nothing further impeded its path. 9. In one of the experiments, Cart A may reverse direction after the collision. How is this accounted for in your calculations?

On a horizontal plain, a forward motion is simply shown by a positive number, while anything in the opposite way of travel is show with a negative number.