Quiz 1 average 9.18 Quiz 1 Re-evaluation Request Due May 8 (next Thursday)

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

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

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

4. To describe the motion of objects, we use several vector quantities such as… • Position vector Rvs Displacement vector ∆R • ∆R = Rfinal – Rinitial • 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. Conservation of MomentumExample Rifle recoil Before shooting (at rest)

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

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

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

9. Conservation of MomentumExample Rifle recoil After shooting vf,bullet vf,Rifle p f,bullet p f,Rifle |pf,bullet | =|pf,Rifle| mbullet|vbullet|= mRifle|vRifle|

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

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

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 Impulseext = ∆ p = ∫ ∑ Fext(t)dt Approximate a varying force as an average force acting during a time interval ∆t Net Impulseext = ∆ p = ∑ Fave.ext x ∆ t

13. Moementum of the closed system (= Rifle + bullet) is conserved,i.e., pi, total = pf, total 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) FExploding gun powder/Rifle on the bullet = – Fbullet on Exploding gun powder/Rifle

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

15. What if our system = bullet (only)? After shooting vbullet pbullet Pf, bullet= pi, bullet+ ∆pbullet ∆ pbullet = ∑ Fave.ext x ∆ t = Fave.Exploding gun powder/Rifle on the bullet x ∆ t ∆pbullet is non zero because the bullet system interacted with the (external) gun powder/Rifle system

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

17. What if our system = Rifle (only)? After shooting vRifle pRifle pf, Rifle= pi, Rifle+ ∆pRifle ∆ pRifle = ∑ Fave.ext x ∆ t = Fave.bullet on Exploding gun powder/Rifle x ∆ t ∆pRifle is non zero because the Rifle/Exploding gun powder system interacted with the (external) bullet system

18. Conservation of MomentumRailroad 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? vAi vBi =0 At rest Before collision A B pi,tot= pi,A+ pi,B= pi,A vA+Bf After collision A+B

19. Conservation of MomentumRailroad 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? vAi vBi =0 At rest Before collision A B pi,tot= pi,A+ pi,B= pi,A vA+Bf After collision A+B pA+Bf pf,tot= pf,A+B

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

21. Stuck togetherafter impact Conservation of Momentum Use of Momentum Chart Alice 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. Bob (Vector magnitudes not to scale, directions shown accurately)

22. Question: Who was going faster at the time of collision? (a) Alice (b) Bob (c) No way to know Stuck togetherafter impact Conservation of Momentum Use of Momentum Chart Alice 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. Bob (Vector magnitudes not to scale, directions shown accurately)

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

24. !!! Solution Same length! 0

25. 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 mB > mA we know vB < vA. 0 Recap (what just happened?) Initial time: just before collisionFinal time: just after collision (friction negligible)

26. Conservation of MomentumQuestion: 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; Speed up Slow down Keep the same speed

27. Conservation of MomentumQuestion: 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; Speed up Slow down Keep the same speed

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

29. Collisions and Impulsehow 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 Impulseext = ∆ p = ∑ Fave.ext x ∆ t

30. Collisions and Impulsehow not to break an ankle + =

31. Next weekMay8 Quiz4(20min) will cover:Today’s lecture Activities and FNTs from DLM9 and Activities from DLM10Bring Calculator!Closed-book, formulas will be provided.

32. Be sure to write your name, ID number & DL section!!!!!