Momentum

1. Momentum Chapter-6

2. Momentum • Symbol is lower case p • Momentum is a product of mass and velocity • p = mv

3. Units • p = mv, so the units are: • p = (kg)(m/s) kgm because velocity is a vector s momentum is a vector quantity Or, N-s

5. 82,288 kg x 0.02m/s = 1600 kgm/s

6. 100kg astronaut in a 60 kg suit moving At 10 m/s 1600 kg m/s

7. Conservation of momentum • Momentum of a system is conserved: • Before and after objects join • Before and after an explosion • Newton’s Third Law applies • FB on A = - FA on B

8. 1600 kgm/s astronaut grabs a 100 kg Sputnik, how fast do both go? 10 m/s

9. m1v1 = m2v21600 kgm/s = 260 kg v2v2 = 6.2 m/s

11. In a collision, the total momentum before doesn’t change: 2500 N-s + 800 N-s = 3300 N-s 1500 N-s + 1800 N-s = 3300 N-s

12. Explosions between vehicles

13. Recoil

14. Let’s say it’s a 5 kg rifle and a 10 g bullet fired at 1000m/s • m1v1 = m2v2 • (0.01 kg)(1000m/s) = 5kg v2 • 10 kgm/s = 5kg v2 • 2m/s = v2 of course you usually put your shoulder there

15. Converting to Newton-S • A Newton is a kgm/s2 • A Newton-second is a (kgm/s2)(s) = kgm/s • So Newton-seconds are equivalent to units of momentum

16. Deriving Impulse Theorem from Newton’s 2nd Law • F = ma = m Dv/Dt • F Dt = m Dv • So, the average force over the time interval is equal to the change in momentum • F Dt = p2 - p1

17. So it’s the same rifle on your shoulder, & the recoil lasts 0.01s • mDv = FDt • 10 kgm/s = F(0.01 s) • F = 1000NThat kind of hurts

18. Hitting the wall at the Indianapolis Speedway The average force you feel depends on rate of change in momentum Lets say you’re going 100km/hr, How fast do you decelerate to 0 km/hr as the front of your car crumples on the concrete? What force is felt by the 100kg driver?

19. (100 km/h)(1000 m/km)(1 h/3600 s) = 27.8 m/s Let’s say the hood is one meter long d = 1/2vt t = 0.072 s The impulse is F Dt = m Dv F (0.0726 s) = (100 kg) (27.8 m/s - 0 m/s) F = 38,600 N This is fatal

20. So we use straw bales around race tracks. … and some soft squishy spectators.

21. (100 km/h)(1000 m/km)(1 h/3600 s) = 27.8 m/s Let’s say the straw makes the car stop in 10m, d = 1/2vt t = 0.72 s The impulse is F Dt = m Dv F (0.72 s) = (100 kg) (27.8 m/s - 0 m/s) F = 3,860 N 3860 N / 980 N = about 4g (4x gravity) this is likely survivable

22. Billiards: the break

23. 1-D conservation of momentum

24. Very few shots line up perfectly

25. Momentum of the constituents before and after contact is equal

26. A stationary ball (red 3-ball) hit with a cue-ball. They share momentum after collision. Forward or back spin can affect the trajectory of the cue ball. Little to no spin is transferred to the target ball.

27. Conservation of Momentum in 2-D Cueball-1 60o Greenball Cueball-2 0.1 kg ball Rolls at 0.5 m/s

28. Conservation of Momentum in 2-D Cueball-1 60o Greenball Cueball-2 Cue ball starts with a momentum p=(0.1 kg)(0.5 m/s)= 0.05 kgm/s After collision (time 2): pcb2 = (sinq)(pcb1) pgb = (cosq)(pcb1) pcb2 = (sin 60o))(0.05kgm/s) pgb = (cos 60o))(0.05kgm/s) pcb2 = 0.0433 kgm/s pgb = 0.025 kgm/s vcb2 = 0. 433 m/s vgb = 0. 25 m/s

29. Conservation of Momentum in 2-D Cueball-1 60o Greenball Cueball-2 0.433 m/s 0.1 kg ball Rolls at 0.5 m/s 0.25 m/s

30. So, I looked this over… • No this is right…it seems counterintuitive but it checks • Momentum is conserved here because these are in different directions, their vector sum is the same as the original input momentum • And, if you check the sum total kinetic energy of the system KE = 1/2 mv2, that too is conserved.

32. Impulse in Rocket Flight • A rocket engine provides thrust force over a short time interval • Solid fuel rocket engines are rated in letters

37. So let’s say I use an A8-3 engine on a 45 gram rocket • F∆t = m∆v • 2.5 N-s = (.045kg) ∆v • ∆v = 55.6 m/s • How much acceleration? • F=ma • 8N = (.045 kg) a • a = 177.8 m/s2 -9.81 m/s2 • a = 168 m/s2

38. How far up does it get before it coasts and then in total before *? • There are 2 phases to the flight • v2 = vo2 + 2a(d-do) • (55.6 m/s)2 = 02 + 2(168m/s 2)(d-do) • d = 9.2 m (under acceleration) • Then its simple ballistic flight for 3 s • d = do +vot + 1/2at2 • d = 9.2m+(55.6m/s)(3s) + 1/2(-9.8m/s2)(3s)2 • d = 131.9 m total height attained

39. Multiple stage rockets drop excess mass during the flight. What about calculating the Height attained by a rocket that has multiple stages? These have three phases, Initial acceleration at mr+b With the the thrust of the initial engine C-6-0, no coasting time Then an acceleration and coast phase for the rocket mass without booster mr

40. Release of booster from second stage

41. So, currently under construction in my laboratory is this kit: We will mass and launch this on Friday, weather permitting We will calculate the booster, upper stage, and parachute deployment heights of this Two-stage rocket. We must mass the rocket parts and record the Specs of the Engines used We must have several students sight and record angles and horizontal observation Distances. Then share the data Monday.

42. Angular Momentum

45. The spinning “beacon” of a neutron star can make it a “pulsar” as observed from Earth

46. Radiotelescopes http://www.jb.man.ac.uk/~pulsar/Education/Sounds/sounds.html

47. So, lets say I have a massive star, 10x the mass of our sun. It runs out of hydrogen fuel collapses and explodes most of its mass into space in a supernova. The core of the star has a Mass 1.5 x that of our sun. 3 x 1030kg. Its initial radius is 7x108 m it collapses to 2x104 m as the matter is compressed into neutrons. What is the velocity of the surface of the neutron star?

48. Rotation speed of a star v = 2πr T v = 2π( 7 x 108m) 30d x 24h/d x 3600s/h Sunspots rotate in 30 d v = 1700 m/s

49. Conservation of angular momentum (w) w = mvr m1v1r1 = m2v2r2 (3 x 1030 kg)(1700m/s)(7 x 108m) = (3 x 1030 kg)v2 (2 x 104m) 6 x 107 m/s = v2 If the speed of light is 3 x 108 m/s its about 20% of light speed v = 2πr T 6 x 107 m/s = 2π (2 x 104m) T T = 0.002 sec…. http://www.jb.man.ac.uk/~pulsar/Education/Sounds/crab.au