1. 1 Physics 7B - AB Lecture 5 May 1 Recap on vectors Momentum Conservation Model - Elastic/Inelastic collisions - Use of Momentum Chart - Collision and Impulse - Force diagram

2. 2 Quiz 1 average 9.18 Quiz 1 Re-evaluation Request Due May 8 (next Thursday) Quiz 2 graded and being returned this week, Solution+Rubrics on the web site

3. 3 To describe the motion of objects, we use several vector quantities such as… Position vector R e.g. R initial, R final Displacement vector ∆R = R final – R initial Velocity vector v = dr/dt Acceleration vector a = dv/dt Force vector F How are these vectors related to each other??

4. 4 To describe the motion of objects, we use several vector quantities such as… Position vector R vs Displacement vector ∆R ∆R = R final – R initial Displacement vector ∆R vs Velocity vector v = dr/dt Velocity vector v = dr/dt vs Acceleration vector a = dv/dt Acceleration vector a = dv/dt vs Force vector F How are these vectors related to each other??

5. 5 Conservation of Momentum Example Rifle recoil Before shooting (at rest)

6. 6 Conservation of Momentum Example Rifle recoil Before shooting (at rest) p i,total = p i,bullet + p i,Rifle = 0

7. 7 Conservation of Momentum Example Rifle recoil After shooting Before shooting (at rest) p i,total = p i,bullet + p i,Rifle = 0 p f,total = p f,bullet + p f,Rifle = 0

8. 8 Conservation of Momentum Example Rifle recoil After shooting p f,bullet p f,Rifle Before shooting (at rest) p i,total = p i,bullet + p i,Rifle = 0 p f,total = p f,bullet + p f,Rifle = 0

9. 9 Conservation of Momentum Example Rifle recoil After shooting v f,bullet v f,Rifle |p f,bullet | = |p f,Rifle | m bullet |v bullet | = m Rifle |v Rifle | p f,bullet p f,Rifle

10. 10 Momentum of the closed system (= Rifle + bullet) is conserved,i.e., p i, total = p f, total

11. 11 What if our system = bullet (only)? Momentum of the closed system (= Rifle + bullet) is conserved,i.e., p i, total = p f, total

12. 12 When does momentum of something change?? … when a force F acts on the something during a time interval e.g. A bat hits a baseball change in momentum is called: Impulse Impulse Is related to the net external force in the following way: Net Impulse ext = ∆ p = ∫ ∑ F ext (t)dt Approximate a varying force as an average force acting during a time interval ∆t Net Impulse ext = ∆ p = ∑ F ave.ext x ∆ t

13. 13 What if our system = bullet (only)? During the gun powder explosion Exploding gun powder/Rifle system exerts force on the bullet. The bullet exerts force on the Exploding gun powder /Rifle system. (Newton’s 3rd law = Every action has an equal and opposite reaction) F Exploding gun powder/Rifle on the bullet = – F bullet on Exploding gun powder/Rifle Moementum of the closed system (= Rifle + bullet) is conserved,i.e., p i, total = p f, total

14. 14 What if our system = bullet (only)? During the gun powder explosion Net Impulse ext on the bullet = ∆ p bullet = ∫ ∑ F ext (t)dt Approximate a varying force as an average force acting during a time interval ∆t Net Impulse ext on the bullet = ∆ p Bullet = ∑ F ave.ext x ∆ t = = F ave.Exploding gun powder/Rifle on the bullet x ∆ t

15. 15 What if our system = bullet (only)? After shooting v bullet p bullet P f, bullet = p i, bullet + ∆p bullet ∆ p bullet = ∑ F ave.ext x ∆ t = F ave.Exploding gun powder/Rifle on the bullet x ∆ t ∆p bullet is non zero because the bullet system interacted with the (external) gun powder/Rifle system

16. 16 What if our system = Rifle (only)? After shooting v Rifle p Rifle Net Impulse ext on the Rifle = ∆ p Rifle = ∫ ∑ F ext (t)dt Approximate a varying force as an average force acting during a time interval ∆t Net Impulse ext on the bullet =∆ p Rifle = ∑ F ave.ext x ∆ t = = F ave.bullet on the Exploding gun powder/Rifle x ∆ t

17. 17 What if our system = Rifle (only)? After shooting v Rifle p Rifle p f, Rifle = p i, Rifle + ∆p Rifle ∆ p Rifle = ∑ F ave.ext x ∆ t = F ave.bullet on Exploding gun powder/Rifle x ∆ t ∆p Rifle is non zero because the Rifle/Exploding gun powder system interacted with the (external) bullet system

18. 18 Conservation of Momentum Railroad cars collide A 10,000kg railroad car A, traveling at a speed of 24m/s strikes an identical car B, at rest. If the car lock together as a result of the collision, what is their common speed afterward? Before collision vAivAi p i,tot = p i,A + p i,B = p i,A After collision v B i =0 At rest v A+B f AB A+B

19. 19 Conservation of Momentum Railroad cars collide A 10,000kg railroad car A, traveling at a speed of 24m/s strikes an identical car B, at rest. If the car lock together as a result of the collision, what is their common speed afterward? Before collision vAivAi p f,tot = p f,A+B After collision v B i =0 At rest v A+B f AB A+B p i,tot = p i,A + p i,B = p i,A p A+B f

20. 20 Collisions Inelastic collisions: Elastic collisions: Momentum conserved regardless Total energy conserved regardless In inelastic collisions, KE is transferred to other types of energy initialfinal initial final Definitely inelastic Maybe elastic or inelastic (Need calculation/energy-bubbles to determine)

21. 21 Alice Bob (Vector magnitudes not to scale, directions shown accurately) You are a CSI at the scene of a car crash. The drivers, Alice and Bob, are unharmed but each claims the other was speeding. Stuck together after impact Conservation of Momentum Use of Momentum Chart

22. 22 Alice Bob (Vector magnitudes not to scale, directions shown accurately) Looking at the car make, you discover that Bob’s car has twice the mass of Alice’s car. As shown, the cars travelled at roughly a 45 degree angle from the point of impact. The ground is flat. Stuck together after impact Conservation of Momentum Use of Momentum Chart Question: Who was going faster at the time of collision? (a) Alice(b) Bob(c) No way to know

23. 23 Solution 1. Write in the directions we know from the problem (lengths are not to scale at this point) 0 2. We know this is a closed momentum system. 3. Use last row to find direction of initial momentum. 4. This means that |p i, Alice | = |p i, Bob | so that the system initial momentum is at 45 o Go back and make sure these are the same length

24. 24 Solution 0 Same length! !!!

25. 25 Recap (what just happened?) 0 The final momentum of the system is pointing, therefore the initial momentum of the system must point. We only get the initial momentum right if the magnitude of the momentum for Alice and Bob are the same. As m B > m A we know v B < v A. Initial time: just before collision Final time: just after collision (friction negligible)

26. 26 Conservation of Momentum Question: Falling on a sled An empty sled is sliding on a frictionless ice when Dan drops vertically from a tree above onto the sled. When he lands, the sled will; (a) Speed up (b) Slow down (c) Keep the same speed

27. 27 Conservation of Momentum Question: Falling on a sled An empty sled is sliding on a frictionless ice when Dan drops vertically from a tree above onto the sled. When he lands, the sled will; (a) Speed up (b) Slow down (c) Keep the same speed

28. 28 Collisions and Impulse how not to break a leg Question: Why is it a good idea to bend your knee when landing after jumping from some height?

29. 29 Collisions and Impulse how not to break a leg Question: Why is it a good idea to bend your knee when landing after jumping from some height? Hint: Net Impulse ext = ∆ p = ∑ F ave.ext x ∆ t

30. 30 Collisions and Impulse how not to break an ankle + =

31. 31 Next week May8 Quiz4(20min) will cover: Today’s lecture Activities and FNTs from DLM9 and Activities from DLM10 Bring Calculator! Closed-book, formulas will be provided.

32. 32 Be sure to write your name, ID number & DL section!!!!! 1MR 10:30-12:50 Dan Phillips 2TR 2:10-4:30Abby Shockley 3TR 4:40-7:00John Mahoney 4TR 7:10-9:30Ryan James 5TF 8:00-10:20Ryan James 6TF 10:30-12:50John Mahoney 7W 10:30-12:50Brandon Bozek 7F 2:10-4:30Brandon Bozek 8MW 8:00-10:20Brandon Bozek 9MW 2:10-4:30Chris Miller 10MW 4:40-7:00Marshall Van Zijll 11MW 7:10-9:30Marshall Van Zijll