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# You and your friends are doing physics experiments on a ISBN: 9780321675460 31

## Solution for problem 74P Chapter 8

University Physics | 13th Edition

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Problem 74P

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 80.0 kg, is given a push and slides eastward. Abigail, with mass 50.0 kg, is sent sliding northward. They collide, and after the collision Sam is moving at 37.0o north of east with a speed of 6.00 m/s and Abigail is moving at 23.0o south of east with a speed of 9.00 m/s. (a) What was the speed of each per-son before the collision? (b) By how much did the total kinetic energy of the two people decrease during the collision?

Step-by-Step Solution:

Solution 74P Problem (a) Step 1: Mass of Sam m = 80.01 Mass of Abigail m = 50.0Kg 2 Final velocity of Sam v = 6.001/s (north of east) = 37.0° Final velocity of Abigail v = 9.00 m/s (south of east) 2 = 23.0° Step 2: To find initial velocities of the two persons. Initial velocities of the two persons are u 1 and u r2spectively. After collision the momentum is conserved along both x and y axis. The two person go with different velocities and in different directions. The directions are represented as and . To find the initial velocities of both persons. Initial Momentum of Sam P1 P 1cos + P ’cos2-----(1) Initial Momentum of Abigail P = P ’.sin - P ’sin -----(2) 2 1 2 Where P1 momentum before collision (m u ) 1 1 P - momentum before collision (m u ) 2 2 2 P1 - momentum of Sam after collision (m v ) 1 1 P2 - momentum of Sam after collision (m v ) 1 1 Step 3: From (1) m 1 1m v1 1 + m v cos1 1 m 1 c1s + m v 2os2 u1 m 1 ---(3) u = 80* *cos 37 + 50 *0 *os 23 1 80 u1 9.97 m/s Initial speed of Sam is 9.97 m/s Step 4: From (2) m 2 2m 1 1n - m 1 1n m 1 1in m 2 2in u2 m 2 ----(4) u = 80* *sin 37 *0 *0 sin 23 2 50 u2 2.26 m/s Initial speed of Abigail is 2.26 m/s Problem (b) Step 1: To find decrease in kinetic energy of two persons after collision Kinetic energy before collision 1 2 1 2 KE Im u2 1 1 + m2u 2 2 KE = 1 80 9.97 + 2 1 50 2.26 2 I 2* * 2 * * KE I4103.73 J or 4104 J Step 2: Kinetic energy after collision 1 2 1 2 KE Fm v2 1 1 + m2v 2 2 KE = 1 80 6 + 2 1 50 9 2 F 2 * * 2 * * KE F3465 J

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##### ISBN: 9780321675460

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