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MSU PHYS101 - Week 2 Assignment Solutions

MSU PHYS101 - Week 2 Assignment Solutions - Week 2...

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Week 2 Assignment Questions - Phys 101 Online Energy and Conservation Laws (All references are from the text book: Ostdiek, V. & Bord, D. (2008). Inquiry into physics, 6 th ed. Belmont,CA: Thomson.) 1. A runner with a mass of 60kg accelerates from 0 to 9 m/s in 3 s. Fin the net force on the runner using the alternate form of Newton’s second law. (ref: pg 88) F = MA = M(∆V/∆T) where F is the force, M is mass, A is acceleration , ∆V is the change in velocity, and ∆T is the change in time. F = 60kg ( 9m/s / 3s) = 180 N 2. A 4000kg truck runs into the rear of a 1000kg car that was stationary. The truck and car are locked together after the collision and move with speed 10m/s. what was the speed of the truck before the collision? Compute how much kinetic energy was “lost” in the collision? (ref: pp. 89-93; 99-100; 111-114) a) First we must use momentum conservation to figure out the cars speed before collision. M 1 V 1 + M 2 V 2 = (M 1 + M 2 ) V 3 where M 1 is the mass of car, M 2 is the mass of truck, V 1 is the velocity of car prior to collision, V 2 is the velocity of truck prior to collision, and V 3 is the velocity of both locked vehicles. Since the car was stationary, V 1 = 0 Therefore: M 2 V 2 = (M 1 + M 2 ) V 3 4000kg V2 = (4000kg + 1000kg) 10m/s V 2 = 12.5m/s b) Now that we have the velocities prior to the collision, we can calculate the KE’s KE before = ½ M 2 V 2 = ½ 4000kg (12.5m/s) 2 = 312500J KE after = ½ (M 1 +M 2 ) V 3 = ½ (4000kg + 1000kg)(10m/s) 2 = 250000J KE change = KE after – Ke before = 250000J – 312500J = -62500J
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3. Two persons on ice skates stand face to face and then push each other away. Their masses are 50kg and
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