2. Summary Lecture 8 Systems of Particles 9.2Centre of mass 9.3Newton 2 for system of particles 9.4-7Conservation of momentum 9.8-11Collisions 9.12Rocket propulsion Systems of Particles 9.2Centre of mass 9.3Newton 2 for system of particles 9.4-7Conservation of momentum 9.8-11Collisions 9.12Rocket propulsion Problems:Chap. 9: 1, 6, 10, 15, 20, 27, 40, 71, 73, 78 Problems:Chap. 9: 1, 6, 10, 15, 20, 27, 40, 71, 73, 78 Tomorrow 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time Tomorrow 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time Week March 20 – 24 20-minute test on material in lectures 1-7 Week March 20 – 24 20-minute test on material in lectures 1-7

4. So far we have considered the motion of POINT PARTICLES FINITE OBJECTS can move as a whole (translational motion) and also rotate about the “Centre of Mass” What happens depends on where we apply the force The “Centre of Mass” is that point where if we apply a force, the object will not rotate.

5. The motion of the Centre of Mass is a simple parabola. (just like a point particle) The motion of the entire object is complicated. This motion resolves to motion of the CM The Centre of Mass motion of points around the CM Lizzie Borden took an axe And gave her mother forty whacks. And when she saw what she had done, She gave her father forty-one

6. m1m1 m2m2 m1m1 m2m2 m1m1 m2m2 m1m1 m2m2 m1m1 m2m2 m1gm1gm2gm2g d1d1 d2d2 M = m 1 + m 2 The Centre of Mass CM M m 1 g x d 1 = m 2 g x d 2

7. Centre of Mass (1D) 0 x1x1 x2x2 x cm m1m1 M m2m2 M = m 1 + m 2 M x cm = m 1 x 1 + m 2 x 2 In general moment of M = moment of individual masses

8. Centre of Mass (3D) For a collection of masses in 3D cm r1r1 r3r3 r2r2

9. Centre of Mass (3D) For a collection of masses in 3D cm r cm = ix cm +jy cm So in a solid body we can find the CM by finding x cm and y cm

10. y x0 2 21 1 3 kg 8 kg 4 kg X cm = 16/15 = 1.07 m y cm = 20/15 = 1.33 m 15 kg r CM 1.07 1.33

11. becomes an integral: For odd shaped objects this probably needs to be determined experimentally For symmetric objects this can often be calculated 1.Look for a symmetry axis 2.Then carry out the integral to find the position of x cm along the axis. Finding the Centre of Mass For solid bodies the sum:

12. dx h R r dm =  r 2 dx but r = (R/h)x dm =  (R/h) 2 x 2 dx x Mass of cone M = 1/3  R 2 h x cm = ¾ h x cm Symmetry line x Solid cone

13. cm Sum of all EXTERNAL forces acting on system The total mass of the system The acceleration of the CM of the system For a system of particles, the dynamics of the Centre of Mass obeys Newton 2.

14. cm CM For a system of particles, the dynamics of the Centre of Mass obeys Newton 2. This also applies to a solid body, where the individual particles are rigidly connected. The dynamics of the Centre of Mass obeys Newton 2

15. You will recall that Where p=mv is the momentum of each particle For a system of particles P =  p = Mv cm Linear Momentum of system of particles This also applies to extended objects cm for system of particles

16. Conservation of Linear Momentum If F ext = 0 NO EXTERNAL forces act on the system P is a constant That is: P x, P y and P z remain constant if F ext-x, F ext-y and F ext-z are zero In an isolated system, momentum is conserved.

17. Exploding rocket Why? No external horizontal forces so horiz momentum unchanged C of M

18. m = 3.8 g, n =12 v = 1100 m s -1 Define system Initial momentum P i = n mv + M V i = n mv + 0 V M=12 kg + 12 m Final momentum p f = (M + nm) V = P i = n mv

19. m = 3.8 g, n =12 v = 1100 m s -1 Initial momentum P i = n mv + M V i = n mv + 0 V M=12 kg + 12 m Final momentum p f = (M + nm) V = P i = n mv KE initial ½ n mv 2 ½ x 12 x 0.0038 x (1100) 2 27588 J KE final ½ (M + 12m)V 2 ½ x (12.0456) x 4.2 2 163 J

21. What is a collision? Collisions An isolated event involving 2 or more objects No external forces Momentum is conserved Usually interact (often strongly) for short time Equal and opposite impulses are exerted on each other  p =  F dt

22. Elastic collisions Energy and momentum are conserved Inelastic collisions Only momentum is conserved Collisions But Energy is always conserved???

23. In 1 dimension

24. m 1 v 1i m 2 v 2i = 0 Before m 1 v 1f m 2 v 2f After Elastic Collision 1D We want to find V 1f and V 2f

25.  m 2 v 2f 2 = m 1 (v 1i - v 1f ) (v 1i + v 1f ) ……(3) Mom. Cons. m 1 v 1i = m 1 v 1f + m 2 v 2f ………………(1)  m 2 v 2f = m 1 (v 1i - v 1f )…………………(2) Energy Cons ½ m 1 v 1f 2 + ½ m 2 v 2f 2 = ½ m 1 v 1i 2  ½ m 2 v 2f 2 = ½ m 1 (v 1i 2 - v 1f 2 ) Mult. by 2 and factorise Divide equ. (3) by (2) V 1i is usually given, so to find v 2f we need to find an expression for v 1f. Get this from equ. (1). m 1 v 1f = m 1 v 1i - m 2 v 2f  Substitute this form of v 1f into equ 4  v 2f = v 1i + v 1i – m 2 /m 1 v 2f  v 2f (1 + m 2 /m 1 ) = 2v 1i  v 2f = v 1i + v 1f …………….…(4)

26. If m 1 >> m 2 v 2f  2v 1i If m 1 = m 2 If m 2 >>m 1 v 1f  0 v 2f  0 v 2f  v 1i v 1f  v 1i v 1f  -v 1i

28. m 1 v 1i m 2 v 2i =0 v cm CM What is V cm ? Mom of CM = mom of m 1 + mom of m 2 (m 1 + m 2 ) V cm = m 1 v 1i + m 2 v 2i Motion of the C of M

29. m 1 v 1i m 2 v 2i =0 v cm CM Motion of the C of M

32. Let’s observe the elastic collision from the view point of the centre of mass

35. In 1 dimension

36. m 1 v 1i m 2 v 2i =0 v cm CM What is V cm ? Mom of CM = mom of m 1 + mom of m 2 (m 1 + m 2 ) V cm = m 1 v 1i + m 2 v 2i

37. m 1 v 1i m 2 v 2i v cm CM

40. Let’s observe the elastic collision from the view point of the centre of mass

42. Elastic billiard balls comets  -particle scattering Collisions in 2 dimensions

43. Momentum is conserved Consider x-components m 1 v 1i = m 1 v 1f cos  1 + m 2 v 2f cos  2 Consider y-components 0= -m 1 v 1f sin  1 + m 2 v 2f sin  2 Since elastic collision energy is conserved 7 variables!3 equations Elastic collisions in 2-D m 1 v 1i before 22 11 m 2 v 2f m 1 v 1f after Impact parameter

44. Inelastic Almost any real collision! an example Automobile collision Collisions in 2 dimensions

45. m B = 550 kg v B = 78 kph  V f = pApA pBpB pfpf P fy = p f sin  P fx = p f cos   m A = 830 kg v a = 62 kph

46. Cons. Momentum ==> p A + p B = p f X component P A = P f cos  m A v A = (m A + m B ) v f cos  ………….(1) Y component P B = P f sin  m B v B = (m A + m B ) v f sin  ………….(2) pApA pBpB pfpf P fx = p f cos  P fy = p f sin   = mAvAmAvA Divide equ (2) by (1) ____________________ m A v A = (m A + m B ) v f cos  Gives  = 39.8 0

47. Cons. Momentum ==> p A + p B = p f X component P A = P f cos  m A v A = (m A + m B ) v f cos  ………….(1) Y component P B = P f sin  m B v B = (m A + m B ) v f sin  ………….(2) pApA pBpB pfpf P fx = p f cos  P fy = p f sin   = mAvAmAvA  = 39.8 0 Use equ 2 to find V f Gives V f = 48.6 kph

48. Can the investigators determine who was speeding? http://www.physics.ubc.ca/~outreach/phys420/p420_96/danny/danweb.htm

51. IN THE EARTH REF. FRAME Vel of gas rel me = vel of gas rel. rocket - vel of rocket rel me = U - v v v+  v mm U = Vel. of gas rel. to rocket Burns fuel at a rate Mom. of gas =  m(U - v) = -change in mom. of rocket (impulse) i.e. F dt =  m(v - U) = v dm - U dm

52. Note since m is not constant Now the force pushing the rocket is F = i.e. F dt =  m(v - U) = v dm - U dm Fdt = v dm + m dv

53. If I want to find out the TOTAL effect of throwing out gas, from when the mass was m i and velocity was v i, to the time when the mass is m f and the velocity v f, I must integrate. v f = U ln Thus

54. Fraction of mass burnt as fuel Speed in units of gas velocity 1 2.2.4.6.8 1 Constant mass (v = at) Reducing mass (m f = 0)

55. An example M i = 850 kg m f = 180 kg U = 2800 m s -1 dm/dt = 2.3 kg s -1 Thrust = dp/dt of gas = dm/dt U =2.3 x 2800 = 6400 N Initial accelerationF = ma ==> a = F/m = 6400/850 = 7.6 m s -2 Final vel.