PH 211 Winter 2014

1. PH 211 Winter 2014 Friday February 28

2. FINAL EXAM • We are one section, so not a group final! • Monday March 17 at 18:00 (6pm) • Rooms TBA

3. Midterm pickup Office hours Tuesday 2:00-3:30

4. Google moderator Which topic would you like to discuss in class on Monday? http://www.google.com/moderator/#16/e=20eb3b

5. Google moderator Which topic would you like to discuss in class on today? http://www.google.com/moderator/#16/e=21293d

6. Impulse = Change in momentum

8. Average force A 4 kg ball hits a block with a speed of 7 m/s and bounces back with the same speed. The whole interaction between the ball and block lasted 2 ms. What was the average force by the block on the ball?

9. A 4 kg ball hits a block with a speed of 7 m/s and bounces back with the same speed. The whole interaction between the ball and block lasted 2 ms. What was the average force by the block on the ball? • 14N • 28N • 56N • 14kN • 28kN • 56kN 121 of 207 1

10. Average force The force of the ball on the block is always equal and opposite to the force of the block on the ball. From J=mvfinal – m vinit we have J = (4)(7)-(4)(-7)=56 Ns Also, J=Faverage T gives Faverage = J/T = 56/0.002 = 28 kN.

11. A 10 g bullet travelling 500 m/s hits a 300 g wooden block and is stuck inside the block. It entered 20 cm. Assume that the force on the bullet was constant while slowing down. How long did it take for the bullet to stop inside the block? PRESS ENTER AFTER RESPONSE! In seconds!

12. This question can always be answered using basic equations for motion with constant acceleration, x(t)=x0+v0 t + ½ a0 t2 , and velocity equation.

13. This question can always be answered using basic equations for motion with constant acceleration. • True • False 0 of 207 1

14. Interaction time T No friction with floor! x=0

15. Interaction time T x=0

17. Velocity equations

18. This result looks simple. What is the case? It is just an accidental occasion. It follows directly from the ratio of the impulses on both objects. It follows directly from conservation of momentum. It follows directly from the fact that there are no net normal forces perpendicular to the path of the bullet.

19. This result looks simple. What is the case? • It is just an accidental occasion. • It follows directly from the ratio of the impulses on both objects. • It follows directly from conservation of momentum. • It follows directly from the fact that there are no net normal forces perpendicular to the path of the bullet. 0 of 207 1

20. (0.010)(500)+(0.300)(0)=(0.310)vfinal vfinal = 16.1 m/s

21. Distance equations T=0.8 ms

22. Average velocity????

23. This answer is soooo simple….. • For all physics problems there is always a quick answer for those “smart” folks. • If you want to be a good engineer you need to be able to solve problems in the fastest manner. • Short answers relate to fundamental physics principles. • This is why I hate physics. • Making sure that your answer is right is the most important.

24. This answer is soooo simple….. • For all physics problems there is always a quick answer for those “smart” folks. • If you want to be a good engineer you need to be able to solve problems in the fastest manner. • Short answers relate to fundamental physics principles. • This is why I hate physics. • Making sure that your answer is right is the most important. 0 of 207 1

25. Change of reference frame! • Block is accelerating frame, problems with F=ma??? • Constant acceleration makes it special. • From block’s perspective: bullet is still accelerating at a constant rate, but not equal to F/m. • Constant (negative!) acceleration from vinit to zero, average velocity indeed vinit/2.

26. Problem solving expectations. You should be able to construct the four equations describing the motion of the block and bullet. You should recognize that the velocity equations are not needed and can be replaced by conservation of momentum. You are not expected to reason that a change of reference frame leads to simpler equations.

27. Notice there is nothing about our equations that makes them 1-d, they are full vector equations, and can be extended to any number of dimensions. However, as with any vector equation, we need to work each set of components out independently: Pi = Pf becomes Pix = Pfx and Piy = Pfy…

28. Example 9.41: A firecracker in a coconut blows it into three pieces. Two of equal mass fly south and west, speed v0. The other piece has twice the mass. Find the velocity of the other piece.

30. Make a diagram to determine something about the solution: • A 20 g ball of clay traveling east at 3 m/s collides with a 30 g ball of clay traveling north at 2 m/s. What are the speed and direction of the resulting 50 g ball of clay? • The resulting ball must be stationary • The resulting ball must move roughly north-east • The resulting ball must move roughly north of north-east • The resulting ball must move roughly south of north-east

31. N NNE NE SNE E

32. A 20 g ball of clay traveling east at 3 m/s collides with a 30 g ball of clay traveling north at 2 m/s. What are the speed and direction of the resulting 50 g ball of clay? • The resulting ball must be stationary • The resulting ball must move roughly north-east • The resulting ball must move roughly north of north-east • The resulting ball must move roughly south of north-east 0 of 207 1

34. Direction NE Magnitude of P: 0.06 √2=0.084 Nm/s Speed: P = Mtotal v  v = 0.084/0.05 = 1.68 m/s