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Motion II

Motion II. Momentum and Energy. Momentum. Obviously there is a big difference between a truck moving 100 mi/ hr and a baseball moving 100 mi/hr. We want a way to quantify this. Newton called it Quantity of Motion. We call it Momentum. Definition: Momentum = (mass)x(velocity) p = m v

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Motion II

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  1. Motion II Momentum and Energy

  2. Momentum • Obviously there is a big difference between a truck moving 100 mi/hr and a baseball moving 100 mi/hr. • We want a way to quantify this. • Newton called it Quantity of Motion. We call it Momentum.

  3. Definition: Momentum = (mass)x(velocity) p = mv • Momentum is a vector. • Combination of mass and velocity • Large momentum for massive objects moving slowly or small objects moving quickly

  4. Newton’s 2nd Law • Note: Sometimes we write t instead of just t. They both represent the time interval for whatever change occurs in the numerator.

  5. Newton’s 2nd Law • Since p = mv, we have p = (mv) • If m is constant (not true for rockets and ballons), p = mv and v/t = a

  6. Rearrange Newton’s 2nd Law • This is the impulse • Impulse is the total change in momentum

  7. Example • Baseball • p = pf – pi = mvf – (-mvi) = m(vf + vi) F = p/t Where t is the time the bat is in contact with the ball

  8. Realistic numbers • M = 0.15 kg • Vi = -40 m/s • Vf = 50 m/s • t = 0.0007 s • F = 19286 N • a = 128571 m/s2 • slow motion photography

  9. Ion Propulsion Engine: very weak force (0.1 N, about the weight of a paper clip) but for very long times (months) • Large momentum change • Very efficient

  10. Crumple zones on cars increase the time for a car to stop. This greatly reduces the average force during accident. • Air bags do the same thing

  11. A 70 kg person traveling at +15 m/s stops.What is the impulse? • -500 kgm/s • -1000 kgm/s • -1500 kgm/s • -2000 kgm/s

  12. Example • 70 kg person traveling at 15 m/s stops. • Dp = pf – pi = 0 kgm/s – 1050 kg·m/s = -1050 kg·m/s

  13. Example • 70 kg person traveling at 15 m/s stops. • Dp = pf – pi = 0 kgm/s – 1050 kg·m/s = -1050 kg·m/s • If the person is stopped by the windshield Dt = 0.01 s • What is the force?

  14. Example • 70 kg person traveling at 15 m/s stops. • Dp = pf – pi = 0 kgm/s – 1050 kg·m/s = -1050 kg·m/s • If the person is stopped by the windshield Dt = 0.01 s • F = -1050 kgm/s/0.01 s = -105000 N

  15. Example • 70 kg person traveling at 15 m/s stops. • Dp = pf – pi = 0 kgm/s – 1050 kg·m/s = -1050 kg·m/s • If the person is stopped by an airbag Dt = 0.40 s • What is the force?

  16. Example • 70 kg person traveling at 15 m/s stops. • Dp = pf – pi = 0 kgm/s – 1050 kg·m/s = -1050 kg·m/s • If the person is stopped by an airbag Dt = 0.40 s • F = -1050 kgm/s/0.40 s = -2625 N

  17. A woman has survived after falling from the 23rd floor of a hotel in the Argentine capital, Buenos Aires. Her fall was broken by a taxi, whose driver got out moments before the impact crushed the roof and shattered the windscreen. Eyewitness said the woman had climbed over a safety barrier and leapt from a restaurant at the top of the Hotel Crowne Plaza Panamericano. She was taken to intensive care for treatment for multiple injuries. The woman, who has not been named, is reported to be an Argentine in her 30s. The taxi driver, named by local media as Miguel, said he got out of his vehicle just before the impact after noticing a policeman looking up. "I got out of the car a second before. If I had not got out, I would have been killed," he told Radio 10. "I was only 10 metres from the impact. It made a terrible noise," he added.

  18. Estimate m = 50kg • H= 23 x 3.5m = 80.5m

  19. EstimateAcceleration BUT Thus

  20. Case 1: stops in deformation distance of body ~ 0.30 m (3 cm). • Case 2: stops in height of roof, ~ 0.50 m (50 cm).

  21. Conservation of Momentum • Newton’s 3rd Law requires F1 = -F2 • p1/t = -p2/t • Or • p1/t + p2/t = 0 • Or • p1 + p2 = 0 • NO NET CHANGE IN MOMENTUM

  22. Example: Cannon • 1000 kg cannon • 5 kg cannon ball Before it fires Pc + Pb = 0 Must also be true after it fires. If Vb= 400 m/s McVc + MbVb = 0 Vc = -MbVb/Mc = -(5kg)(400m/s)/(1000kg) = -2m/s

  23. Conservation of Momentum • Head on collision • Football • Train crash • Wagon rocket • Cannon Fail

  24. ENERGY & WORK • In the cannon example, both the cannon and the cannon ball have the same momentum (equal but opposite). • Question: Why does a cannon ball do so much more damage than the cannon?

  25. ENERGY & WORK • In the cannon example, both the cannon and the cannon ball have the same momentum (equal but opposite). • Question: Why does a cannon ball do so much more damage than the cannon? • Answer: ENERGY

  26. ENERGY & WORK • Essentially, energy is the ability to make something move. • What transfers or puts energy into objects? • A push

  27. WORK • We will begin by considering the “Physics” definition of work. • Work = (Force) x (displacement traveled in direction of force) • W=Fd • 1) if object does not move, NO WORK is done. • 2) if direction of motion is perpendicular to force, NO WORK is done

  28. Example • Push a block for 5 m using a 10 N force. • Work = (10 N) x (5 m) = 50 Nm = 50 kgm2/s2 = 50 J (Joules)

  29. I have not done any work if I push against a wall all day long but is does not move. • True • False

  30. Doing WORK on a body changes its ENERGY • ENERGY is the capacity to do work. • The release of energy does work, and doing work on something increases its energy E = W = F  d • Two basics type of energy: Kinetic and Potential.

  31. Kinetic energy: Energy of motion • Potential energy: Energy of situation • Position • Chemical structure • Electrical charge

  32. Kinetic Energy: Energy of Motion • Derivation: Push a ball of mass m with a constant force. KE = W = Fd = (ma)d But d = ½ at2 KE = (ma)(½ at2) = ½ m(at)2 But v = at KE = ½ mv2

  33. A 6000 kg truck travels at 10 m/s.What is its kinetic energy? • 3000 J • 30000 J • 300000 J • 3000000 J

  34. If the speed of the truck doubles, what happens to the energy? • Increases by ½x • Increases by 2x • Increases by 4x

  35. Potential Energy • Lift an object through a height, h. PE = W = Fd= (mg)h PE = mgh Note: We need to decide where to set h=0.

  36. Units of energy and work • W = Fd • KE = mv2/2 • PE = mgh

  37. Other types of energy • Electrical • Chemical • Thermal • Nuclear

  38. Conservation of Energy • We may convert energy from one type to another, but we can never destroy it. • We will look at some examples using only PE and KE. KE + PE = const. OR KEi + PEi = KEf + PEf

  39. Drop a rock of 100m cliff • Call bottom of cliff h = 0 • Initial KE = 0; Initial PE = mgh • Final PE = 0 • Use conservation of energy to find final velocity mgh = ½mv2 v2 = 2gh

  40. For our 100 m high cliff • Note that this is the same as we got from using kinematics with constant acceleration

  41. A 5 kg ball is dropped from the top of the Sears tower in downtown Chicago. Assume that the building is 300 m high and that the ball starts from rest at the top. • 5880 m/s • 2940 m/s • 76.7 m/s • 54.2 m/s

  42. A 5 kg ball is dropped from rest from the top (300 m high) of the Sears Tower. From conservation of energy, what is its speed at the bottom? • 5880 m/s • 2940 m/s • 76.7 m/s • 54.2 m/s

  43. Pendulum • At bottom, all energy is kinetic. • At top, all energy is potential. • In between, there is a mix of both potential and kinetic energy. • Bowling Ball Video

  44. Energy Content of a Big Mac • 540 Cal = 540000 cal • 1 Cal = 4186 Joules

  45. How high can we lift a 1kg object with the energy content of 1 Big Mac? • PE = mgh • h = PE/(mg) • PE = 2,260,440 J • m = 1 kg • g = 9.8 m/s2

  46. 2 Big Macs have enough energy content to lift a 1 kg object up to the orbital height of the International Space Station. • 150 Big Macs could lift a person to the same height. • It would take about 4,000,000 Big Macs to get the space shuttle to the ISS height. • Note: McDonald’s sells approximately 550,000,000 Big Macs per year.

  47. Revisit: Cannon Momentum • 1000kg cannon • 5kg cannon ball Before it fires Pc + Pb = 0 Must also be true after it fires. Assume Vb = 400 m/s McVc + MbVb =0 Vc= -MbVb/Mc = -(5kg)(400m/s)/(1000kg) = -2m/s

  48. Cannon: Energy KEb= ½ mv2 = 0.5 (5kg)(400m/s)2 = 400,000 kgm2/s2 = 400,000 J KEc= ½ mv2 = 0.5 (1000kg)(2m/s)2 = 2,000 kgm2/s2 = 2,000 J Note: Cannon Ball has 200 times more ENERGY

  49. When I drop a rock, it start off with PE which gets converted in to KE as it falls. At the end, it is not moving, so it has neither PE or KE. What happened to the energy? • Energy is not conserved in this case. • The energy is converted into Thermal energy • The energy is transferred to whatever it is dropped on.

  50. Important Equations

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