Download presentation

Presentation is loading. Please wait.

Published byRebecca Ritchie Modified over 4 years ago

1
Ball A Ball B Ball A is moving toward Ball B. Ball B is stationary Ball A and Ball B have the same mass!

2
When A strikes B, Ball B moves upward and Ball A moves downward. Whatever momentum A has before the collision A and B share after they collide vAvA v B v A KE = KE v 1 2 + v 2 2 = v 1 2 + v 2 2 If v 2 = 0 then, v 1 = v 1 2 + v 2 2

3
vAvA v B v A vAvA v B v A vAvA v B v A v B = Cos 1 v A v A = Sin 1 v A

4
vAvA v B v A v Bx v By v Ax v Ay v Bx = Cos v B v By = Sin v B v Ax = Cos v A v Ay = Sin v A v A = v Bx + v Ax 0 = v By + v Ay

5
vAvA v B v A v Bx v By v Ax v Ay p = p mv A = mv A Cos 2 + mv B Cos 1 m 1 = m 2 v A = v A Cos 2 + v B Cos 1

6
1 2 vAvA v B p = p On y axis 0 = Sin 1 v B + Sin 2 v A Sin 2 = -Sin 1 v B /v A v a vAvA v B v A v Bx v By v Ax v Ay

7
Two balls of equal mass strike each other. Ball A moves at 5.20 m/s and strikes ball B, which is stationary. After the collision ball A travels above the original path at an angle of 21.0 0. What is the p vs of each ball after the collision? 1. v A = Cos 1 v A v A = Cos 21.0 0 x 5.20 m/s v A = 4.85 m/s 2. v A = v A 2 + v B 2 v B = v A 2 - v B 2 v B = 1.88 m/s 3. Sin 2 = -Sin 1 v A /v B 2 = -Sin21 0 x 4.85m/s 1.88m/s = 68 0

Similar presentations

Presentation is loading. Please wait....

OK

Collisions and Conservation of Energy

Collisions and Conservation of Energy

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google