1. Lecture 12 Momentum & Impulse

2. Lecture 12 Goals: Momentum & Impulse Momentum & Impulse  Solve problems with 1D and 2D Collisions  Solve problems having an impulse (Force vs. time) Energy Energy  Understand the relationship between motion and energy  Define Potential & Kinetic Energy  Develop and exploit conservation of energy principle

3. Inelastic collision in 1-D: Example l Inelastic means that the objects stick. l A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V : What is the momentum of the bullet with speed v ? v V beforeafter x

4. A perfectly inelastic collision in 1-D Example What is the momentum of the bullet with speed v ?  Key question: Is x-momentum conserved ? v V beforeafter x BeforeAfter

5. Exercise Momentum Conservation A. Box 1 B. Box 2 C. same l Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. l The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks.  Which box ends up moving fastest ? 1 2

6. Exercise Momentum Conservation l The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks.  Which box ends up moving fastest ?  Notice the implications from the graphical solution: Box 1’s momentum must be bigger because the length of the summed momentum must be the same.  The longer the green vector the greater the speed 1 2 BeforeAfterBeforeAfter Ball 1 Ball 2 Box 1 Box 2 Box 1+Ball 1 Box 2+Ball 2

7. A perfectly inelastic collision in 2-D l Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). vv1vv1 vv2vv2 V beforeafter m1m1 m2m2 m 1 + m 2 l If no external force momentum is conserved. l Momentum is a vector so p x, p y and p z 

8. A perfectly inelastic collision in 2-D vv1vv1 vv2vv2 V beforeafter m1m1 m2m2 m 1 + m 2 x-dir p x : m 1 v 1 = (m 1 + m 2 ) V cos  y-dir p y : m 2 v 2 = (m 1 + m 2 ) V sin  l If no external force momentum is conserved. l Momentum is a vector so p x, p y and p z are conseved 

9. Elastic Collisions l Elastic means that the objects do not stick. l If no external force, then momentum will always be conserved l Start with a 1-D problem. BeforeAfter

10. Billiards l Consider the case where one ball is initially at rest. p p a  ppbppb F PPa PPa  before after The final direction of the red ball will depend on where the balls hit. v v cm

11. Billiards: All that really matters is conservation momentum (and energy Ch. 10 & 11) l Conservation of Momentum x-dir P x : m v before = m v after cos  + m V after cos  y-dir P y : 0 = m v after sin  + m V after sin  p p after  ppbppb F P P after  before after

12. Force and Impulse (A variable force applied for a given time) l Gravity: At small displacements a “constant” force t Springs often provide a linear force (- k  x) towards its equilibrium position (Chapter 10) l Collisions often involve a varying force F(t): 0  maximum  0 Δρ l We can plot force vs time for a typical collision. The impulse, Δρ, of the force is a vector defined as the integral of the force during the time of the collision.

13. Force and Impulse (A variable force applied for a given time) F Δρ l Δρ reflects momentum transfer t titi tftf tt Δρ Impulse Δρ = area under this curve ! (Transfer of momentum !) Impulse has units of Newton-seconds ΔρΔρΔρΔρ

14. Force and Impulse Δρ l Two different collisions can have the same impulse since Δρ depends only on the momentum transfer, NOT the nature of the collision. tt F t F t tt same area F  t big, F small F  t small, F big

15. Average Force and Impulse tt F t F t tt F av  t big, F av small F av  t small, F av big F av

16. Exercise Force & Impulse A. heavier B. lighter C. same D. can’t tell l Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second. Which box has the most momentum after the force acts ? FF light heavy

17. Boxing: Use Momentum and Impulse to estimate g “force”

18. Back of the envelope calculation (1) m arm ~ 7 kg (2) v arm ~7 m/s (3) Impact time  t ~ 0.01 s Question: Are these reasonable?  Impulse  p ~ m arm v arm ~ 49 kg m/s  F ~ Δρ/  t ~ 4900 N (1) m head ~ 6 kg  a head = F / m head ~ 800 m/s 2 ~ 80 g ! l Enough to cause unconsciousness ~ 40% of fatal blow l Only a rough estimate!

19. سبحان الله During "collision" with a tree a head ~ 600 - 1500 g How do they survive? Jaw muscles act as shock absorbers Straight head trajectory reduces damaging rotations (rotational motion is very problematic ) Woodpeckers

20. Home Exercise l The only force acting on a 2.0 kg object moving along the x- axis. Notice that the plot is force vs time. l If the velocity v x is +2.0 m/s at 0 sec, what is v x at 4.0 s ?  p = m  v = Impulse m  v =  + + m  v = (-8)1 N s + ½ (-8)1 N s + ½ 16(2) N s m  v = 4 N s  v = 2 m/s v x = 2 + 2 m/s = 4 m/s

21. Energy l We need to define an “isolated system” ? l We need to define “conservative force” ? l Recall, chapter 9, force acting for a period of time gives an impulse or a change (transfer) of momentum l What if a force acting over a distance: Can we identify another useful quantity?

22. Energy F y = m a y and let the force be constant y(t) = y 0 + v y0  t + ½ a y  t 2   y = y(t)-y 0 = v y0  t + ½ a y  t 2 v y (t) = v y0 + a y  t   t = (v y - v y0 ) / a y Eliminate  t and regroup So  y = v y0 (v y - v y0 ) / a y + ½ a y (v y 2 - 2v y v y0 +v y0 2 ) / a y 2  y = (v y0 v y - v y0 2 ) / a y + ½ (v y 2 - 2v y v y0 +v y0 2 ) / a y  y = ( - v y0 2 ) / a y + ½ (v y 2 - +v y0 2 ) / a y  y = ½ (v y 2 - v y0 2 ) / a y

23. Energy And now  y = ½ (v y 2 - v y0 2 ) / a y can be rewritten as:  ma y  y = ½ m (v y 2 - v y0 2 ) And if the object is falling under the influence of gravity then a y = -g

24. Energy - mg  y= ½ m (v y 2 - v y0 2 ) - mg  y f – y i ) = ½ m ( v yf 2 -v yi 2 ) Rearranging to give initial on the left and final on the right ½ m v yi 2 + mgy i = ½ m v yf 2 + mgy f We now define mgy as the “gravitational potential energy” A relationship between y-displacement and change in the y-speed

25. Energy l Notice that if we only consider gravity as the external force then the x and z velocities remain constant To ½ m v yi 2 + mgy i = ½ m v yf 2 + mgy f Add ½ m v xi 2 + ½ m v zi 2 and ½ m v xf 2 + ½ m v zf 2 ½ m v i 2 + mgy i = ½ m v f 2 + mgy f where v i 2 = v xi 2 +v yi 2 + v zi 2 ½ m v 2 terms are defined to be kinetic energies (A scalar quantity of motion)

26. Energy l If only “conservative” forces are present, the total energy () of a system is conserved (sum of potential, U, and kinetic energies, K) of a system is conserved E mech = K + U l K and U may change, but E mech = K + U remains a fixed value. E mech = K + U = constant E mech is called “mechanical energy”

27. Example of a conservative system: The simple pendulum. l Suppose we release a mass m from rest a distance h 1 above its lowest possible point.  What is the maximum speed of the mass and where does this happen ?  To what height h 2 does it rise on the other side ? v h1h1 h2h2 m

28. Example: The simple pendulum. y y= 0 y=h 1  What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum.

29. Example: The simple pendulum. v h1h1 y y=h 1 y=0  What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum E = mgh 1 at top E = mgh 1 = ½ mv 2 at bottom of the swing

30. Example: The simple pendulum. y y=h 1 =h 2 y=0 To what height h2 does it rise on the other side? E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point. E = mgh 1 = mgh 2 or h 1 = h 2

31. Cart Exercise Revisited: How does this help? l 1 st Part: Find v at bottom of incline 5.0 m 30 ° 7.5 m N mg a i = g sin 30°  = 5 m/s 2 d = 5 m / sin 30° = ½ a i  t 2 10 m = 2.5 m/s 2  t 2 2s =  t v = a i  t = 10 m/s v x = v cos 30° = 8.7 m/s i j x y E mech is conserved K i + U i = K f + U f 0 + mgy i = ½ mv 2 + 0 (2gy) ½ = v = (100) ½ m/s v x = v cos 30° = 8.7 m/s One step, no FBG needed

32. Exercise : U, K, E & Path l A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless 1. Ball is dropped 2. Ball slides down a straight incline 3. Ball slides down a curved incline After traveling a vertical distance h, how do the speeds compare? (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell h 13 2