Home > Uncategorized > physics giancoli 6th edition

physics giancoli 6th edition

January 19th, 2010 admin Leave a comment Go to comments

Two train cars A and B have different masses:m(A)=15,000kg,m(B)=13,000kg on the same track..?

(Physics Giancoli 6th edition 2004, chapter 7, similar to example 7-3 using conservation of momentum)

The velocity of A, v(A)=28m/s and car B is initially at rest. v(B)=0m/s. The two cars collide, find the velocity of the coupled cars after the collision.

In the real world the collision is inelastic, that is the kinetic energy before collision is not equal to the kinetic energy after collision. Use the data in “PART 1″ to calculate the difference in kinetic energy before and after the collision.

Thanks guys!

Let us thank the principle of conservation of momentum. Before collision, the total momentum was
15000*28+13000*0 = 420000 kgm/s
After collision let the coupled move with a velocity v. Then the final momentum is 28000*v
By the conservation of momentum, 28000 v = 420000
So v = 15 m/s
Let us compute the KE before collision and after collision. The familiar expression 1/2 m v^2 is to be used.
1/2 {15000*784+13000*0} = 5880000 J this is the KE before collision.
After collision 1/2 * 28000*225 = 3150000 J
So certainly there seems a loss of KE. It is 5880000–3150000 = 2730000 J

Giancoli solutions: Chapter 5 Problem 2, 6th Edition, or Chapter 5 Problem 1, 5th Edition

Share and Enjoy:
Categories: Uncategorized Tags: physics giancoli 6th edition chapter 4, physics giancoli 6th edition chapter 4 answers, physics giancoli 6th edition ebook, physics giancoli 6th edition solutions, physics giancoli 6th edition solutions manual
Comments (0) Trackbacks (0) Leave a comment Trackback
  1. No comments yet.
  1. No trackbacks yet.