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(a) If m, = 2.0 g, m, = 1.0 g, and v = 5.5 m/s, then %3D m-m2 m, + m2. 1.0 g 5.5 5.5 3.0 g m/s) = m/s, 3.0 and 2m, 4 m/s) = 5.5 mi+ m2. 3.0 g Your response differs from the correct answer by more than 10%. Double check your calculations. 3.0 m/s.

(a) If m, = 2.0 g, m, = 1.0 g, and v = 5.5 m/s, then %3D m-m2 m, + m2. 1.0 g 5.5 5.5 3.0 g m/s) = m/s, 3.0 and 2m, 4 m/s) = 5.5 mi+ m2. 3.0 g Your response differs from the correct answer by more than 10%. Double check your calculations. 3.0 m/s.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Tutorial Exercise A 2.0-g particle moving at 5.5 m/s makes a perfectly elastic head-on collision with a resting 1.0-g object. (a) Find the speed of each particle after the collision. (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g. (c) Find the final kinetic energy of the incident 2.0-g particle in the situations described in parts (a) and (b). In which case does the incident particle lose more kinetic energy? Step 1 The initial velocity of the target object is zero (v2i = 0). Let m, be the mass of the first particle, v,, its initial velocity, and v2r its final velocity. Let m, be the mass of the target object and vr its final velocity. From conservation of momentum before and after the collision, we have the following equation. m V1f + m2v2f = m,V1 + 0 For a perfectly elastic head-on collision, we have the following relationship between the objects' final velocities and initial velocities derived from conservation of momentum and conservation of kinetic energy. -- ("x- v2) V1i - V2i Because v2, = 0, we can solve for the final velocity of the second object. V2f = V1f+ V1/ Substituting equation (2) above into equation (1) and solving for v,, we have m, - m, Vif = m, + m2 Substituting this result into equation (2) and simplifying, we have 2m1 V2f = \m, + m2 (a) If m, = 2.0 g, m, = 1.0 g, and v = 5.5 m/s, then m, - m2 1.0 g/ 5.5 Vif = m, + m2 m/s) = 5.5 m/s, %3D 3.0 g 3.0 and ms) = ( 2m1 4 5.5 \m, + m2 3.0 g Your response differs from the correct answer by more than 10%. Double check your calculations. 3.0 m/s. Submit Skip (you cannot come back)
Transcribed Image Text:Tutorial Exercise A 2.0-g particle moving at 5.5 m/s makes a perfectly elastic head-on collision with a resting 1.0-g object. (a) Find the speed of each particle after the collision. (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g. (c) Find the final kinetic energy of the incident 2.0-g particle in the situations described in parts (a) and (b). In which case does the incident particle lose more kinetic energy? Step 1 The initial velocity of the target object is zero (v2i = 0). Let m, be the mass of the first particle, v,, its initial velocity, and v2r its final velocity. Let m, be the mass of the target object and vr its final velocity. From conservation of momentum before and after the collision, we have the following equation. m V1f + m2v2f = m,V1 + 0 For a perfectly elastic head-on collision, we have the following relationship between the objects' final velocities and initial velocities derived from conservation of momentum and conservation of kinetic energy. -- ("x- v2) V1i - V2i Because v2, = 0, we can solve for the final velocity of the second object. V2f = V1f+ V1/ Substituting equation (2) above into equation (1) and solving for v,, we have m, - m, Vif = m, + m2 Substituting this result into equation (2) and simplifying, we have 2m1 V2f = \m, + m2 (a) If m, = 2.0 g, m, = 1.0 g, and v = 5.5 m/s, then m, - m2 1.0 g/ 5.5 Vif = m, + m2 m/s) = 5.5 m/s, %3D 3.0 g 3.0 and ms) = ( 2m1 4 5.5 \m, + m2 3.0 g Your response differs from the correct answer by more than 10%. Double check your calculations. 3.0 m/s. Submit Skip (you cannot come back)
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