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Asteroid Collision. Two asteroids of equal mass in the

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 31E Chapter 8

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 31E

Asteroid Collision.? Two asteroids of equal mass in the asteroid belt between Mars and Jupiter collide with a glancing blow. Asteroid A , which was initially traveling at 40.0 m/s, is deflected 30.0° from its original direction, while asteroid B, which was initially at rest, travels at 45.0° to the original direction of A (?Fig. E8.31?). (a) Find the speed of each asteroid after the collision. (b) What fraction of the original kinetic energy of asteroid A dissipates during this collision?

Step-by-Step Solution:

Solution 31E Step 1 of 11: (a) Find the speed of each asteroid after the collision. The speed of the each asteroid after collision can be calculated using law of conservation of momentum. Now let us consider the case(1) before collision, the mass of the both the asteroid A and B is M, but asteroid A is initial moving with velocity along x-axis with v = 40 m/s and the other asteroid B is at rest, so its speed will be v = 0 as A1 B1 shown in the figure below, Step 2 of 11: Now let us consider the case(2) after collision, after collision the asteroid A 0 makes an angle 30 with x-axis and moves with speed v A2 . Similarly after 0 collision asteroid B makes an angle 45 with x-axis and moves with speed v B2, as shown in the figure below Step 3 of 11: Now let us consider the law of conservation of momentum, as momentum is a vector. The component along the axis will be conserved. Let us consider the components of momentum along x-axis before collision and after collision, That is, Along x-axis momentum before collision = Along x-axis momentum after collision P A1x+ P B1x= P A2x+ P B2x Using P=mv Mv A1x + Mv B1x= Mv A2x+ Mv B2x vA1x + vB1x= vA2x + vB2x Step 4 of 11: As asteroid B is at rest before collision,B1x=0 in above equation v = v + v A1x A2x B2x Substituting the components along x-axis from figure above, 0 v A1x= vA2os(30 ) + v cB2( 45) Using v A1= 40 m/s 40 m/s = v ( ) + v ( ) 2 A2 2 B2 2 80 m/s = 3 v A2 + 2 vB2 ……….1 Step 5 of 11: Similarly now let us consider the components of momentum along y-axis before collision and after collision, That is, Along y-axis momentum before collision = Along y-axis momentum after collision P A1y+ P B1y = P A2y+ P B2y Using P=mv Mv A1y+ Mv B1y = Mv A2y+ Mv B2y v + v = v + v A1y B1y A2y B2y Step 6 of 11: As asteroid B is at rest before collision, B1y=0 and Asteroid A no component of speed along y-axis v A1y=0 0 = v + v A2y B2y Substituting the components along y-axis from figure above, 0 = v sin(30 ) + v sin( 45) A2 B2 0 = v ( ) v ( )2 A2 2 B2 2 0 = v A2 2 v B2 v A2 2 v B2= 0………..2

Step 7 of 11

Chapter 8, Problem 31E is Solved
Step 8 of 11

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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