Lesson_8_Problem_Set4 (2)

PHY 111 Lesson 8 Problem Set 4 Name ___________________________ Momentum Section _________ Directions: Several of the following problems are missing some of the given variable information. Your instructor will provide the values through email in the “From Your Instructor” section of the weekly lesson or in a course announcement. Use the values provided by your instructor to answer all of the questions. Show all your work . Make sure to use appropriate labels and units, and highlight your final answers if they are numerical. 1. A toy car with mass m 1 travels to the right on a frictionless track with a speed of 3 m/s. A second toy car with mass m 2 travels with a speed of 6 m/s to the left on the same track. The two cars collide and stick together. What are the cars’ speeds after the collision? (4 pts) Let m 1 = 0.36 kg and m 2 = 0.74 kg Initial momentum of the first car (P initial 1 ): P 1 = m 1 * 3 m/s P 1 = 0.36 kg * 3 m/s = 1.08 kg·m/s Initial momentum of the second car (P initial 2 ): P initial2 = m 2 * (-6 m/s) P initial2 = 0.74 kg * (-6 m/s) = -4.44 kg·m/s Total initial momentum (P TI ): P TI = P initial 1 + P initial2 P TI = 1.08 kg·m/s - 4.44 kg·m/s = -3.36 kg·m/s The conservation of momentum: P initial total = (m1 + m2) * V final -3.36 kg·m/s = (0.36 kg + 0.74 kg) * V final -3.36 kg·m/s = 1.10 kg * V final

V final: V final = -3.36 kg·m/s / 1.10 kg ≈ -3.05 m/s So, the cars' speeds after the collision are approximately -3.05 m/s to the left. 2. A student fires a 0.07 kg arrow at an object with mass m that is initially at rest on a frictionless surface. The speed of the arrow before the collision is 90 m/s. The speed when the arrow emerges from the object is v. What is the resulting velocity of the object? (4 pts) Let m = 6.1 kg and v = 0.78 m/s Initial momentum of the arrow (P IA ): P IA = m * v P IA = 6.1 kg * 0.78 m/s = 4.758 kg·m/s After the collision: P final = (m arrow + m) * v Now plug in the values: 4.758 kg·m/s = (0.07 kg + 6.1 kg) * v 4.758 kg·m/s = 6.17 kg * v Now, solve for v: v ≈ 4.758 kg·m/s / 6.17 kg ≈ 0.771 m/s So, the velocity of the object is approximately 0.771 m/s. b) Is the collision between the arrow and the object elastic or inelastic? Include evidence to support your answer. (5 pts) Initial kinetic energy (K inital ) = 0.5 * 6.1 kg * (0.78 m/s) 2 = 1.7987 J Final kinetic energy (K final ) = 0.5 * (0.07 kg + 6.1 kg) * (0.771 m/s) 2 ≈ 2.2509 J K inital is less than K final , kinetic energy is not conserved. The collision is inelastic.

VCx ≈ -0.821 m/s VCy ≈ 2.704 m/s b) What is the velocity of particle C (magnitude and direction)? (4 pts) Magnitude (vC): vC = √(VCx 2 + VCy 2 ) vC = √((-0.821 m/s) 2 + (2.704 m/s) 2 ) vC ≈ √(0.674 + 7.331) ≈ √8.005 ≈ 2.83 m/s Direction (θ): θ = arctan(VCy / VCx) θ = arctan(2.704 m/s / -0.821 m/s) (VCx is negative) θ ≈ arctan(-3.290) θ ≈ -72.58° Magnitude of particle C's velocity is 2.83 m/s, and the direction is approximately -72.58 degrees 5. An inattentive car driver crashes into their neighbor’s mailbox. a) Compare the force of the car on the mailbox to the force of the mailbox on the car. Explain your reasoning. (2 pts) The force of the car on the mailbox and the force of the mailbox on the car are equal in magnitude and opposite in direction. b) Compare the change in momentum experienced by the car to the change in momentum experienced by the mailbox. Explain your reasoning. (2 pts) The change in momentum experienced by the car and the mailbox will be equal in magnitude but opposite in direction.

c) Compare the change in velocity experienced by the car to the change in velocity experienced by the mailbox. Explain your reasoning. (2 pts) The change in velocity experienced by the car and the mailbox will depend on their respective masses and the duration of the collision.