Conceptual Physics Practice Page Chapter 6 Momentum Answers Conceptual Physics Practice Page Chapter 6 Momentum Answers This document provides answers to the practice problems found in Chapter 6 of Conceptual Physics covering the topic of momentum The document will be structured as follows 1 Briefly introduces the concept of momentum and its significance 2 Section 1 Momentum and Impulse Concept Explanation Explains the definition of momentum and the relationship between impulse and change in momentum Practice Problem Answers Detailed solutions for practice problems related to calculating momentum and impulse 3 Section 2 Conservation of Momentum Concept Explanation Explains the principle of conservation of momentum and its applications Practice Problem Answers Detailed solutions for practice problems involving collisions and explosions applying the conservation of momentum 4 Section 3 Momentum and Energy Concept Explanation Explains the relationship between momentum and energy including kinetic energy and workenergy theorem Practice Problem Answers Detailed solutions for practice problems that explore the connection between momentum and energy 5 Conclusion Briefly summarizes key takeaways from the chapter and provides guidance for further learning Section 1 Momentum and Impulse Concept Explanation Momentum is a measure of an objects mass in motion It is calculated as the product of an objects mass m and its velocity v Momentum p mass m velocity v Impulse is the change in momentum of an object It is calculated as the force applied F 2 multiplied by the time interval over which the force acts t Impulse J force F time interval t The impulsemomentum theorem states that the impulse acting on an object equals the change in its momentum J p Practice Problem Answers Problem 1 A 1000 kg car traveling at 20 ms collides with a stationary car of equal mass If the collision lasts 01 seconds what is the average force exerted on the moving car during the collision Solution Calculate the initial momentum of the moving car p mv 1000 kg 20 ms 20000 kgms The final momentum of both cars after the collision is zero as they are stationary Therefore the change in momentum is 20000 kgms Using the impulsemomentum theorem the average force is calculated as F pt 20000 kgms 01 s 200000 N Problem 2 A 05 kg ball traveling at 10 ms is caught by a baseball glove If the glove stops the ball in 02 seconds what is the average force exerted on the ball by the glove Solution Calculate the initial momentum of the ball p mv 05 kg 10 ms 5 kgms The final momentum of the ball is zero Therefore the change in momentum is 5 kgms Using the impulsemomentum theorem the average force is calculated as F pt 5 kgms 02 s 25 N Section 2 Conservation of Momentum Concept Explanation The principle of conservation of momentum states that the total momentum of a system remains constant in the absence of external forces This implies that in a closed system momentum cannot be lost or gained only transferred between objects In collisions the total momentum before the collision equals the total momentum after the collision In explosions the initial momentum of the system is zero and the final momentum of the fragments must also sum to zero Practice Problem Answers 3 Problem 3 A 10 kg bowling ball moving at 5 ms collides with a 2 kg pin initially at rest After the collision the bowling ball moves at 3 ms What is the velocity of the pin after the collision Solution Calculate the initial total momentum of the system p m1v1 m2v2 10 kg 5 ms 2 kg 0 ms 50 kgms Calculate the final momentum of the bowling ball p mv 10 kg 3 ms 30 kgms Apply the conservation of momentum 50 kgms 30 kgms 2 kg v Solving for the velocity of the pin v 50 kgms 30 kgms 2 kg 10 ms Problem 4 A 5 kg bomb at rest explodes into two fragments One fragment has a mass of 2 kg and a velocity of 10 ms What is the velocity of the other fragment Solution The initial momentum of the bomb is zero since its at rest The final momentum of the system must also be zero Calculate the momentum of the first fragment p mv 2 kg 10 ms 20 kgms To balance this momentum the second fragment must have an equal and opposite momentum 20 kgms The velocity of the second fragment is v pm 20 kgms 3 kg 667 ms the negative sign indicates the fragment moves in the opposite direction Section 3 Momentum and Energy Concept Explanation Kinetic energy is the energy of motion It is related to momentum by the equation KE p2 2m The workenergy theorem states that the net work done on an object equals the change in its kinetic energy Momentum is conserved in all collisions but kinetic energy is not necessarily conserved Practice Problem Answers Problem 5 A 01 kg ball traveling at 5 ms collides with a stationary wall and bounces back with a velocity of 4 ms What is the change in kinetic energy of the ball Solution Initial kinetic energy KE 05 01 kg 5 ms2 125 J Final kinetic energy KE 05 01 kg 4 ms2 08 J 4 Change in kinetic energy KE 08 J 125 J 045 J Problem 6 A 2 kg block sliding on a frictionless surface at 4 ms collides with a stationary 1 kg block After the collision the 2 kg block moves at 2 ms What is the velocity of the 1 kg block after the collision Is kinetic energy conserved in this collision Solution Using conservation of momentum the velocity of the 1 kg block can be calculated as 6 ms Initial kinetic energy KE 05 2 kg 4 ms2 16 J Final kinetic energy KE 05 2 kg 2 ms2 05 1 kg 6 ms2 20 J Kinetic energy is not conserved in this collision as it has increased Conclusion Understanding momentum is essential for comprehending the motion of objects particularly in collisions and explosions The principle of conservation of momentum provides a powerful tool for analyzing such interactions By mastering the concepts of momentum impulse and their relationship to energy you gain a deeper understanding of the laws governing motion in the universe For further exploration delve into more complex collisions involving multiple objects the concept of elastic and inelastic collisions and the application of momentum in rocket propulsion