Homework7-F08 - Homework7 Chapt 10 11 1 the last part of Introduction of Collision Most real collisions are somewhere between elastic and perfectly

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework7 Chapt. 10 & 11 1/ the last part of Introduction of Collision Most real collisions are somewhere between elastic and perfectly inelastic. This is indicated by the elasticity E of the collision, which measures the difference in the velocities of the particles after the collision compared with the difference in velocities before the collision. For instance, in a perfectly inelastic collision, the two particles stick together after colliding. The elasticity of such a collision is E = 0, because the difference in velocities between the particles is 0 after they collide. Technically, the elasticity is defined by the relation , where and are the initial and final velocities of particle 1, and and are the initial and final velocities of particle 2. In this problem, the formula is simplified by our definition of and the hypothesis . So, using for the final velocity of particle 1 and for the final velocity of particle 2, we obtain the simpler formula . This final form will be most useful to you in solving Part F. Part F If the two particles with equal masses m collide with elasticity E = 0.650 , what are the final velocities of the particles? Assume that particle 1 has initial velocity and particle 2 is initially at rest. Look at the applet to be sure that your answer is reasonable. Give the velocity of particle 1 followed by the velocity of particle 2, separated by a comma. Express the velocities in terms of . 1/ read the problem, and understand what is elasticity E 2/ Since from definition: 2 1 2 1 ( ) f f i i V V E V V - = - - , in this problem, V(2i) = 0, E = 0.650, so the equation goes to 2 1 1 0.65 ) f f i V V V - = and from the momentum conservation law 1 1 2 i f f mV mV mV = + ie. 1 1 2 i f f V V V = + , solve these two equations, you will get V 1f =0.175v, and V 2f =0.825v. 10.5 10.5 The figure shows a ball’s before-and-after pictorial representation for the three situations in parts (a), (b) and (c).
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Solve: The quantity K + U g is the same during free fall: f gf i gi K U K U + = + . We have (a) ( ) 2 2 1 1 0 0 2 2 2 2 2 1 0 1 1 1 2 2 2 [(10 m/s) (0 m/s) ]/(2 9.8 m/s ) 5.10 m mv mgy mv mgy y v v g + = + = - = - × = 5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground. (b)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/18/2010 for the course PHYS ? taught by Professor Zhou during the Spring '10 term at Georgia State University, Atlanta.

Page1 / 7

Homework7-F08 - Homework7 Chapt 10 11 1 the last part of Introduction of Collision Most real collisions are somewhere between elastic and perfectly

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online