ch09-p081

# ch09-p081 - 16 9 v 1 &amp;amp;gt; m 1 v 1 . 81. Using Eq....

This preview shows page 1. Sign up to view the full content.

v 2 ff = m 2 m 3 m 2 + m 3 v 2 f = m 2 3 m 2 2 3 v 1 i = 2 9 v 1 i v 3 ff = 2 m 2 m 2 + m 3 v 2 f = 2 m 2 3 m 2 2 3 v 1 i = 4 9 v 1 i . (a) Setting v 1 i = 4 m/s, we find v 3 ff 1.78 m/s. (b) We see that v 3 ff is less than v 1 i . (c) The final kinetic energy of block 3 (expressed in terms of the initial kinetic energy of block 1) is K 3 ff = 1 2 m 3 v 3 2 = 1 2 (4 m 1 ) © § ¹ · 16 9 2 v 1 i 2 = 64 81 K 1 i . We see that this is less than K 1 i . (d) The final momentum of block 3 is p 3 ff = m 3 v 3 ff = (4 m 1 ) ©
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 16 9 v 1 &gt; m 1 v 1 . 81. Using Eq. 9-67 and Eq. 9-68, we have after the first collision v 1 f = m 1 m 2 m 1 + m 2 v 1 i = m 1 3 m 1 v 1 i = 1 3 v 1 i v 2 f = 2 m 1 m 1 + m 2 v 1 i = 2 m 1 3 m 1 v 1 i = 2 3 v 1 i . After the second collision, the velocities are...
View Full Document

## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online