Impulse Contents: Impulse/WB Area Under Force Graph Ft = mv Rocket Science Rocket Demos

Change in momentum = Impulse Impulse = F t where F = Force t = time that the force is exerted (F or t = 0???) Example: What impulse is imparted by exerting a 12 N force for 4.0 s? Impulse = F t = (12 N)(4.0 s) Impulse = 48 Ns

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What is the impulse of a 6.12 N force acting for 2.3 seconds? 14 Ns Impulse = F t Impulse = (6.12 N )(2.3 s)=14 Ns W

A rocket engine is rated at 14 Ns of impulse, and burns for 1 A rocket engine is rated at 14 Ns of impulse, and burns for 1.7 seconds. What is the thrust of the engine? 8.2 N Impulse = F t 14 Ns = F(1.7 s) F = (14 Ns)/ (1.7 s) = 8.24 = 8.2 N W

Impulse is area under = (40 N)(6.0s) = 240 Ns Example - Impulse is area under F vs t graph t F 100 N 5.0 s Impulse is area under = (40 N)(6.0s) = 240 Ns

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What is the impulse? 100 N F t W 5.0 s Average F = 70 N, t = 8.0 s, Impulse = (70 N)(8.0 s) = 560 Ns W 560 Ns

What is the impulse? 100 N F t W 5.0 s Impulse = (40 N)(3.0 s) + (70 N)(5.0 s)= 470 Ns W 470 Ns

What is the impulse of the D12 engine? Impulse = (10. N)(1.73 s-0.11 s) + (28 N-10 N)(0.42 - 0.21) = 19.98 Ns (about 20 Ns) W 20 Ns

Force, time, mass, and velocity Impulse = F t = mv F = Force (N) t = Elapsed time (s) m = Mass (kg) v = Change in velocity (m/s) Example: A pitcher pitches a 0.145 kg baseball at 40. m/s, and the batter hits it directly back at 50. m/s to the outfield. What is the average force exerted by the bat if the collision lasted .013 s? F t = mv (F )(.013 s) = (0.145 kg)(-90 m/s) F = (0.145 kg)(-90 m/s)/(.013 s) F = -1003.8 N ≈ -1.0x103 N (units/sign)

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What force for 10. seconds makes a 2 What force for 10. seconds makes a 2.0 kg rocket speed up to 75 m/s from rest? 15 N (m)(v) = (F )( t) (2.0 kg)(75 m/s) = (F )(10 s) F = (2.0 kg)(75 m/s)/(10 s) = 15 N

A baseball bat exerts a force of 200. N on a. 50 kg ball for A baseball bat exerts a force of 200. N on a .50 kg ball for .10 seconds. What is the ball’s change in velocity? 40. m/s (m)(v) = (F )( t) (.50 kg)(v) = (200. N )(.10 s) v = (200. N )(.10 s)/(.50 kg) = 40. m/s

Jolene exerts a 50. N force for 3. 0 seconds on a stage set Jolene exerts a 50. N force for 3.0 seconds on a stage set. It speeds up from rest to .25 m/s. What is the mass of the set? 6.0 x 102 kg (m)(v) = (F )( t) (m)(.25 m/s) = (50. N )(3.0 s) m = (50. N )(3.0 s)/(.25 m/s) = 600 kg = 6.0 x 102 kg

One last thing: Deriving Newton’s second law: m v = F  t Finally: F = m v = ma = p t t

Rocket Propulsion Ft = mv So: F = engine thrust  t = time to burn fuel m = mass of fuel burned v = exhaust gas velocity Ft = mv

Rocket Propulsion Example 1: A rocket burns fuel at a rate of 1.2 kg/s, with an exhaust velocity of 1250 m/s. What thrust does it develop? Ft = mv v = 1250 m/s m = 1.2 kg t = 1 s (F)(1 s) = (1.2 kg)(1250 m/s), F = 1500 N

Rocket Propulsion Example 2: A model rocket engine develops 12 N of thrust with an exhaust velocity of 718 m/s. What is its fuel burn rate? Ft = mv F = 12 N v = 718 m/s m = ? kg t = 1 s (12 N)(1 s) = m(718 m/s), m = .0167 kg Burn rate = .0167 kg/s

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A certain rocket engine burns A certain rocket engine burns .0352 kg of fuel per second with an exhaust velocity of 725 m/s. What thrust does it generate? 25.5 N Ft = mv F(1 s) = (.0352 kg)(725 m/s), F = 25.5 N W

The Saturn V’s first stage engines generated 33. 82 MN of thrust (33 The Saturn V’s first stage engines generated 33.82 MN of thrust (33.82 x 106 N) with an exhaust velocity of 2254.7 m/s. What was its fuel burn rate? 15,000 kg/s Ft = mv (33.82 x 106 N)(1 s) = m(2254.7 m/s), m = 15,000 kg Burn Rate = 15,000 kg/s W

A D12 engine generates 11. 80 N of thrust burning fuel at a rate of A D12 engine generates 11.80 N of thrust burning fuel at a rate of .0143 kg/s. What is the exhaust velocity? 825.2 m/s Ft = mv (11.80 N)(1 s) = (.0143 kg)v, v = = 825.2 m/s W

Which Rocket Accelerates faster? B Demo – F14

Rocket Propulsion Solid Fuel Engine: © Microsoft Encarta

Rocket Propulsion Thrust of the Engine: F = mv t Big v means big F Solid Fuel Engine: Big v means big F Big v also means lots of pressure Danger Will Robinson! © Microsoft Encarta

© Estes

m = .5 kg, Ft = 20 Ns Ft = m v 20 Ns = .5kg v Lookout! m = .5 kg, Ft = 20 Ns Ft = m v 20 Ns = .5kg v v = ? (x 2.237 for mph) © Estes

Rocket Propulsion Liquid Fuel Engine: How do you keep it from tipping? Why the “steam” coming off? © Microsoft Encarta

Saturn V Rocket - Apollo missions Escape Tower Command Module L.E.M Third stage Second stage First stage Saturn V Animation (Flash in IE)

Ion Propulsion systems Low thrust/high Δv 20-50 km/s exhaust velocity Dawn

Gridded Xenon ion drive (Like Dawn has) where’s the power come from? Gridded Xenon ion drive (Like Dawn has)

How what is the “burn” rate for this engine? The Dawn spacecraft has a mass of about 770 kg, and an ion drive with an exhaust velocity of 35,000 m/s that generates about 92 millinewtons of thrust. (.092 N) How what is the “burn” rate for this engine? Ft = mv (.092 N)(1.0 s) = m(35,000 m/s) m = 2.62857E-06 kg/s How big a change in velocity can 50 kg of fuel give the spacecraft? The final mass will be 770 – 50 = 720 kg, on the average the spacecraft will have a mass of 745 kg So the delta v is approximately: (745 kg)(Δv) = (50 kg)(35,000 m/s) Δv ≈ 2350 m/s