1. Chapter 9 Momentum

2. Impulse Momentum The impulse-momentum theorem Conservation of momentum Inelastic collisions Topics: Sample question: Male rams butt heads at high speeds in a ritual to assert their dominance. How can the force of this collision be minimized so as to avoid damage to their brains? Slide 9-1

3. Interactions Collision – a short duration interaction between two objects colliding Explosion – an interaction that forces two objects apart

4. Time Collision occur over a time or duration The duration of a collision depends on the material from which the objects are made The harder the objects the shorter the contact time Example – two steel balls – 1ms Example – tennis ball and racket -10ms

5. Impulse The force of the foot on the ball is an impulsive force. Slide 9-8

6. Impulse The amount by which the soccer ball is compressed is a measure of the magnitude of the force the foot exerts on the ball More compression greater force

7. Impulse

8. Impulsive force – A large force exerted during a small interval of time. Example – hammer on a nail Example – bat on a ball

9. Impulse The area under the curve of the force vs time graph is called the impulse of the force A harder kick or a kick of longer duration yields a larger impulse and a higher speed

10. Impulse

11. Impulse J = area under the force curve J = F ave Δt Can think of the collision in terms of the average force defined as the constant force that has the same duration and the same area under the curve as the real force

12. Units of Impulse Units are N s Equivalent to kg m/s Impulse has a sign + or - Impulse is a vector quantity pointing in the direction of the average force vector J = F ave Δt

13. How are impulse, mass, and velocity related

14. Derivation of momentum equation a = F ave /m a = Δv/ Δt =v f –v i / Δt a = F ave /m = v f –v i / Δt F ave Δt = m v f - m v i The product of an object’s mass and velocity is called momentum

15. Momentum Momentum is the product of an object’s mass and its velocity: Slide 9-10 p = mv 

16. Units Units of momentum are kg m/s These are also the units of impulse

17. The Impulse-Momentum Theorem Impulse causes a change in momentum: Slide 9-11 J =p f - p i = ∆p  

18. Impulse momentum Theorum An impulse delivered to an object causes the object’s momentum to change How can we slow down an object in the gentlest way possible? If the duration of the collision can be increased, the force of the impact will be decreased.

19. Conservation of Momentum

20. For the collision of the two balls Δp 1x = J 1x and Δp 2x = J 2x But since J 1x = -J 2x Δp 1x = -Δp 2x or Δp 1x + Δp 2x = 0 p 1x + p 2x = constant Then (p 1x ) f + (p 2x ) f = (p 1x ) i + (p 2x ) i

21. Conservation of Momentum The x component of total momentum is conserved (p x ) f = (p x ) i

22. Momentum of a System is Conserved Total change in the total momentum P of the system is zero Therefore the total momentum of the system remains constant

23. Isolated System Internal Forces – act only between particles within the system Isolated system is subject only to internal forces The total momentum of an isolated system is conserved

24. Non isolated system External Forces – act on the system from agents outside the system

25. Conservation of Momentum External forces can change the momentum of the system ΔP = Δp 1 + Δp 2 + Δp 3 Then by the impulse momentum theorem (F exton1 Δt) + (F exton2 Δt) + (F exton3 Δt) (F exton1 + F exton2 + F exton3 ) Δt F net Δt

26. Conservation of Momentum If F net =0 the total momentum of the system does not change An isolated system is one in which F net = 0 The total momentum of an isolated system is a constant. Interactions within the system do not change the system’s total momentum P f =P i

27. Conservation of Momentum The total momentum after an interaction is equal to the total momentum before the interaction

28. Example p 277 Two skaters Sandra and David stand facing each other on frictionless ice. Sandra has a mass of 45 Kg. David has a mass of 80 Kg. They then push off from each other. After the push Sandra moves at a speed of 2.2 m/s. What is David’s speed?

30. Example p 277 m S (v Sx ) f +m D (v Dx ) f = m S (v Sx ) i +m D (v Dx ) i = 0 Solving for (v Dx ) f David moves backward with a speed of 1.2 m/s

31. Inelastic Collisions For now, we’ll consider perfectly inelastic collisions: A perfectly elastic collision results whenever the two objects move off at a common final velocity. Slide 9-21

32. Elastic Collisions Two objects bounce apart after the collision Example p281 A 200 g air track glider and a 400g air track glider are pushed toward each other. The gliders have velcro tabs on the front so they will stick together when they collide. The 200 g glider is pushed with an initial speed of 3.0 m/s. The collision causes it to reverse direction at.50 m/s. What was the initial speed of the 400 g. glider?

34. Example p282 (m 1 +m 2 )(v x ) f = m 1 (v 1x ) i +m 2 (v 2x ) i

35. Momentum and Collisions in 2 dimensions Motion in a plane The total momentum P is a vector sum of the momenta p = mv of the individual particles. Momenta is conserved only if each component of P is conserved

36. Example p 283 Peregrine falcons grab their prey from above while both are in flight. A falcon flying at 18m/s, swoops down at a 45° angle from behind a pigeon flying horizontally at 9.o m/s. The falcon has a mass of.80 kg. The pigeon has a mass of.36 kg. What are the speed and direction of the falcon (now holding the pigeon) immediately after impact?

37. Example p 183 This is a perfectly inelastic collision. The total momentum of the falcon + pigeon system is conserved The x component of the total momentum before the collision must equal the x component of the total momentum after the collision The y component of the total momentum before the collision must equal the y component of the total momentum after the collision

38. Example p283

39. First find the x and y components of momentum before the collision For x (P x ) i = m F (v Fx ) i + m P (v px ) i =m F (-v F cosθ) +m P (-v P ) = (.80kg)(-18m/s)(cos 45°) + (.36kg)(-.90m/s) =-13.4 kg m/s For y (P y ) i = m F (v Fy ) i + m P (v Py ) i =m F (-v F sinθ) +0 = (.80kg)(-18m/s)(sin 45°) = -10.2 kg m/s

40. Example p 283 After the collision the birds move with common velocity at angle β from horizontal. The x component of the final momentum is (P x ) f = (m F +m P )v x ) f P xf = P xi because momentum is conserved

41. Example p283 Similarly P yf = P yi Find the angle β

42. Example p283 Use the Pythagorean theorem to find the final velocity The falcon with its meal is moving 37° below the horizontal at a speed of 15 m/s

43. Momentum and Collisions in 2 dimensions