Explanation Given data: Mass of the first block is 11.2 kg. Mass of the second block is 21.7 kg. Spring constant is 1330 N m . Compression length of the spring is 0.141 m . Formula used: Write the expression for kinetic energy. K E = 1 2 m v 2 (1) Here, KE is the kinetic energy. m is the mass. v is the velocity. Write the expression for momentum. L = m v (2) Here, L is the momentum. Write the expression for spring force. F spring = 1 2 k x (3) Here, F spring is the spring force. k is the spring constant. Explanation: Kinetic energy of the first mass is, Substitute K E 1 for KE , m 1 for m and v 1 for v in equation (1), K E 1 = 1 2 m 1 v 1 2 (4) Final kinetic energy of the first mass is, Substitute K E f 1 for KE , m f 1 for m and v f 1 for v in equation (1), K E f 1 = 1 2 m f 1 v f 1 2 (5) Initial kinetic energy of the first mass is, Substitute K E 01 for KE , m 01 for m and v 01 for v in equation (1), K E 01 = 1 2 m 01 v 01 2 (6) Kinetic energy of the second mass is, Substitute K E 2 for KE , m 2 for m and v 2 for v in equation (1), K E 2 = 1 2 m 2 v 2 2 (7) Final kinetic energy of the second mass is, Substitute K E f2 for KE , m f2 for m and v f2 for v in equation (1), K E f2 = 1 2 m f2 v f2 2 (8) Initial kinetic energy of the second mass is, Substitute K E 02 for KE , m 02 for m and v 02 for v in equation (6), K E 02 = 1 2 m 02 v 02 2 (9) Initial spring force is, Substitute x 0 for x in equation (3), F spring = 1 2 k x 0 2 (10) Final spring force is, Substitute x f for x in equation (3), F spring = 1 2 k x f 2 (11) The total mechanical energy of the system consists of the kinetic energy and the spring force. The total initial kinetic energy of both the objects and the spring force is same as the total final kinetic energy of both the objects and the spring force. 1 2 m 1 v f1 2 + 1 2 m 2 v f2 2 + 1 2 k x f 2 = 1 2 m 1 v 01 2 + 1 2 m 2 v 02 2 + 1 2 k x 0 2 (12) Since the blocks are initially at rest, the initial velocities are zero, that is, v 01 = 0 m s and v 02 = 0 m s Final displacement of the spring is zero, since the spring comes finally to rest, that is, x f = 0 m . Substitute 0 m s for v 01 , 0 m s for v 02 and 0 m for x f in equation (12), 1 2 m 1 v f1 2 + 1 2 m 2 v f2 2 + 1 2 k ( 0 m ) = 1 2 m 1 ( 0 m s ) 2 + 1 2 m 2 ( 0 m s ) 2 + 1 2 k x 0 2 1 2 m 1 v f1 2 + 1 2 m 2 v f2 2 = 1 2 k x 0 2 (13) Momentum of the first object is, Substitute L 1 for L , m 1 for m and v 1 for v in equation (2), L 1 = m 1 v 1 (14) Final momentum of the first object is, Substitute L f1 for L 1 and v f1 for v 1 in equation (14), L f 1 = m 1 v f1 Initial momentum of the first object is, Substitute L 01 for L 1 and v 01 for v 1 in equation (14), L 01 = m 1 v 01 Momentum of the second object is, Substitute L 2 for L , m 2 for m and v 2 for v in equation (2), L 2 = m 2 v 2 (15) Final momentum of the second object is, Substitute L f2 for L 2 and v f2

10th Edition

ISBN: 9781118486894

Author: David Young, Shane Stadler

Publisher: WILEY

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Chapter 10, Problem 84AP

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Chapter 10, Problem 83AP

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