HW08-solutions - lee (cyl327) HW08 gentle (56245) 1 This...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lee (cyl327) HW08 gentle (56245) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Two identical balls (labelled A and B) move on a frictionless horizontal tabletop. Initially, ball A moves at speed v A, = 10 m / s while ball B is at rest ( v B, = 0). The two balls collide off-center, and after the collision ball A moves at speed v A = 6 m / s in the direction A = 53 from its original velocity vector: 10 m / s A before B after 6 m / s A 0 m/s b b 53 Which of the following diagrams best repre- sents the motion of ball B after the collision? 1. after 4 m / s B 37 b 2. 4 m / s B after b 3. 0 m/s B b after 4. after 8 m / s B 37 b correct 5. 8 m / s B after b 6. after 6 m / s B 53 b 7. after 8 m / s B 53 b 8. 6 m / s B after b 9. after 4 m / s B 53 b 10. after 6 m / s B 37 b Explanation: Because we are given both the speed and the direction of ball A after the collision, the problem can be solved in terms of momentum conservation only. Indeed, there are no (horizontal) external forces acting on the two balls, so their net (horizontal) momentum vector is conserved during the collision: vector P net = mvector v A, + mvector v B, [before] = mvector v A + mvector v B [after] and since vector v B, = vector (ball B is initially at rest), vector v A + vector v B = vector v A, . lee (cyl327) HW08 gentle (56245) 2 Pictorially, this means vector v A vector v B A B vector v A, and in components, v A cos A + v B cos B = v A, , v A sin A- v B sin B = 0 . Solving for the two components of ball Bs velocity, we find v B,x v B cos B = v A,- v A cos A (1) = (10 m / s)- (6 m / s) cos53 = 6 . 4 m / s , v B,y v B sin B = v A sin A (2) = (6 m / s) sin53 = 4 . 8 m / s , and hence v B = radicalBig v 2 B,x + v 2 B,y = radicalBig (6 . 4 m / s) 2 + (4 . 8 m / s) 2 = 8 m / s , B = arctan v B,y v B,x = arctan 4 . 8 m / s 6 . 4 m / s = 37 . Solving an Elastic Collision: If we know that the collision in question is perfectly elastic and it is then we do not need to be given both the speed v A and the direction A of the ball A after the collision: Any one of these two parameters would determine the other. The reason for this is conservation of kinetic energy in an elastic collision: K net = mv 2 A, 2 + mv 2 A, 2 [before] = mv 2 A 2 + mv 2 B 2 [after] and therefore v 2 A + v 2 B = v 2 A, . (3) At the same time, eqs. (1) and (2) above imply v 2 B = v 2 B,x + v 2 B,y = parenleftBig v A,- v A cos A parenrightBig 2 + parenleftBig v A sin A parenrightBig 2 = v 2 A,- 2 v A, v A cos A + v 2 A (the cosine theorem), and hence v 2 A + v 2 B- v 2 A, = (4) = 2 v 2 A- 2 v A, v A cos A ....
View Full Document

Page1 / 11

HW08-solutions - lee (cyl327) HW08 gentle (56245) 1 This...

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