Physics – Ch8,9 Ch8.1 Universal Gravitation All matter is attracted to all other matter. - More matter = larger force - Distance apart is important (Inverse square law)

Physics – Ch8,9 Ch8.1 Universal Gravitation All matter is attracted to all other matter. - More matter = larger force - Distance apart is important (Inverse square law) Newton’s Law of Universal Gravitation Universal gravitational constant Ex1) Compute the gravitational attraction between 2 60kg students sitting 0.5m apart.

Physics – Ch8,9 Ch8.1 Universal Gravitation All matter is attracted to all other matter. - More matter = larger force - Distance apart is important (Inverse square law) Newton’s Law of Universal Gravitation Universal gravitational constant Ex1) Compute the gravitational attraction between 2 60kg students sitting 0.5m apart.

Ex2) What is the gravitational attraction between a 75 kg person and the earth? Solve 2 ways. ME = 5.97x1024 kg RE= 6.4x106 m

Ex2) What is the gravitational attraction between a 75 kg person and the earth? Solve 2 ways. ME = 5.97x1024 kg RE= 6.4x106 m Fg = m∙g = (75kg)(9.8m/s2) = 735 N

Ex3) Two masses both of mass m are set a distance d apart, so that the force of attraction between them is F How does the force change if a) Both masses double b) Instead the distance between them doubles c) Instead both masses triple and the distance between them is cut in half Ch8 HW#1 1 – 5

of attraction between them is F d2 How does the force change if Ex3) Two masses both of mass m are set a distance d apart, so that the force of attraction between them is F F = Gmm d2 How does the force change if a) Both masses double F1 = G( 2∙m )( 2∙m ) = 4∙Gmm 4X bigger ( d )2 d 2 b) Instead the distance between them doubles F2 = G( m )( m ) = Gmm 4X smaller ( 2∙d )2 4d 2 c) Instead both masses triple and the distance between them is cut in half F3 = G( 3∙m )( 3∙m ) = 9∙Gmm 36X bigger ( ½ d )2 ¼ d 2 Ch8 HW#1 1 – 5

Ch8 HW#1 1 – 5 1. What is the gravitational attraction between 2 7.25kg bowling balls placed 2m apart? 2. What is the gravitational attraction between you (60kg) and a 0.1kg spoon, placed 0.25m away?

Ch8 HW#1 1 – 5 1. What is the gravitational attraction between 2 7.25kg bowling balls placed 2m apart? 2. What is the gravitational attraction between you (60kg) and a 0.1kg spoon, placed 0.25m away?

3. What is the gravitational attraction between the moon and the earth? 4. What is the gravitational attraction between the moon and the sun?

3. What is the gravitational attraction between the moon and the earth? 4. What is the gravitational attraction between the moon and the sun?

5. Two masses both of mass m are set a distance d apart, so that the force of attraction between them is F How does the force change if a) 1 mass doubles b) Both masses double and the distance between them doubles c) 1 mass tripled, 1 mass cut in half, and the distance quadrupled

Ch8.2 – Using Big G - There must be something that causes matter to seek smallest volume - Objects appear to be circling each other - Tycho Brahe and Johannes Kepler

Ch8.2 – Using Big G - There must be something that causes matter to seek smallest volume - Objects appear to be circling each other - Tycho Brahe and Johannes Kepler Kepler’s Laws 1.) Paths of all planets are ellipses, with Sun at one focus 2.) An imaginary line from the Sun sweeps out equal areas in equal time intervals 3.) The square of the ratio of the periods of 2 planets = the cube of the ratio of radii

Satellite Orbits

Satellite Orbits

Satellite Orbits

Satellite Orbits Earth drops off by 4.9 meters every 8000 meters (curvature). Orbital speed for a low satellite: 8000 m/s What type of force deals with circles? What force holds the satellite from flying off? Orbital Speed:

Satellite Orbits Earth drops off by 4.9 meters every 8000 meters (curvature). Orbital speed for a low satellite: 8000 m/s What type of force deals with circles? FC What force holds the satellite from flying off? FG FC = FG Orbital Speed: M is the mass of the object being circled. r is the radius of orbit from center. (both equations independent of mass of satellite.)

r Ex3) A satellite orbits Earth 225km above its surface. x What speed must it have? h rE r = rE + h = 6.38x106m + 225,000m = 6.605x106m r

Ex3) A satellite orbits Earth 225km above its surface. x a. What speed must it have? h r rE r = rE + h = 6.38x106m + 225,000m = 6.605x106m a.

Ex4) Satellite at 150 km above earth. Orbital speed: Ch8 HW#2 6 – 10

Ex4) Satellite at 150 km above earth Orbital speed: r = rE + h = 6.38x106m + 150,000m = 6.53x106m Einstein's theory of Gravity -Newton’s Laws are great approximation F = Gmm can get us to the moon r2 -Einstein: ( part of General Theory of Relativity ) Mass distorts space-time, kind of like a bowling ball distorts a bed. Masses are attracted to each other because of this. Black holes (predicted by Einstein ) – If an object is large enough, it’s g-force becomes so large that its escape speed becomes larger than the speed of light. Light gets pulled back in so it can’t be seen. Ch8 HW#2 6 – 10

6. What is the gravitational attraction between the earth and a 2000kg Ch8 HW#2 6 – 10 6. What is the gravitational attraction between the earth and a 2000kg satellite, sitting in a satellite factory, on the earth’s surface? (U can solve this the long or short way) 7. What is the gravitational attraction between a 2000kg satellite and the earth, when the satellite is 100 km above the earth? dE = ( rE + h ) = ( 6.38 x 106 + 100,000 ) = 6.48 x 106 m

6. What is the gravitational attraction between the earth and a 2000kg Ch8 HW#2 6 – 10 6. What is the gravitational attraction between the earth and a 2000kg satellite, sitting in a satellite factory, on the earth’s surface? (U can solve this the long or short way) | F = m.g | = (2000kg)(9.8m/s2) | = 19,600 N | 7. What is the gravitational attraction between a 2000kg satellite and the earth, when the satellite is 100 km above the earth? dE = ( rE + h ) = ( 6.38 x 106 + 100,000 ) = 6.48 x 106 m

7. What is the gravitational attraction between a 2000 kg satellite Ch8 HW#2 6 – 10 7. What is the gravitational attraction between a 2000 kg satellite and the earth, when the satellite is 100 km above the earth? dE = ( rE + h ) = ( 6.38 x 106 + 100,000 ) = 6.48 x 106 m 8. What speed does that satellite have to travel to maintain a circular orbit?

9. F = Gmm d2 a. Mass 1 doubles, mass 2 is ¼ the size, distance is ⅓ the size b. Both masses double, distance is 10x smaller c. Mass1: 4x bigger, mass2: 3x bigger, distance 8x greater

10. Speed of a satellite 1000 km above earth: r = rE + h = 6.38x106m + 1,000,000m = 7.38x106m

10. Speed of a satellite 1000 km above earth: r = rE + h = 6.38x106m + 1,000,000m = 7.38x106m

Ch9.1 – Momentum momentum = mass ∙ velocity Units: p = m∙v Ex1) What is the momentum of a 5 gram bullet traveling at 300m/s ?

Ch9.1 – Momentum momentum = mass ∙ velocity Units: p = m∙v Ex1) What is the momentum of a 5 gram bullet traveling at 300m/s ? p = m∙v = ( .005 kg )( 300 m/s ) = 1.5 kg∙m/s Ex2) What is the change in momentum of a 1000 kg car traveling at 50m/s that breaks to a stop?

Ch9.1 – Momentum momentum = mass ∙ velocity Units: p = m∙v Ex1) What is the momentum of a 5 gram bullet traveling at 300m/s ? p = m∙v = ( .005 kg )( 300 m/s ) = 1.5 kg∙m/s Ex2) What is the change in momentum of a 1000 kg car traveling at 50m/s that breaks to a stop? ∆p = pf – pi = mvf – mvi = (1000)(0) – (1000)(50) = – 50,000

Impulse: force x time (F∙t) Ex3) What kind of impulse is placed on a baseball that is hit with a bat with a force of 2500N over 50 ms (milliseconds)? Ex4) A 47g golf ball is blasted from rest to 70m/s when hit by a driver. If the impact lasts 1 ms, what impulse is acted on the ball? t = .001 sec vi = 0 vf = 70 m/s m = .047 kg

Impulse – force x time (F∙t) Ex3) What kind of impulse is placed on a baseball that is hit with a bat with a force of 2500N over 50 ms (milliseconds)? Impulse = F∙t = ( 2,500 N )( .05 sec ) = 125 N∙s Ex4) A 47g golf ball is blasted from rest to 70m/s when hit by a driver. If the impact lasts 1 ms, what impulse is acted on the ball? t = .001 sec vi = 0 vf = 70 m/s m = .047 kg

Impulse – force x time (F∙t) Ex3) What kind of impulse is placed on a baseball that is hit with a bat with a force of 2500N over 50 ms (milliseconds)? Impulse = F∙t = ( 2,500 N )( .05 sec ) = 125 N∙s Ex4) A 47g golf ball is blasted from rest to 70m/s when hit by a driver. If the impact lasts 1 ms, what impulse is acted on the ball? t = .001 sec vf = vi + at vi = 0 vf = 70 m/s a = vf – vi m = .047 kg t F = m∙a a = 70,000 m/s2 = (0.047kg)(70,000 m/s2) Alternate method: = 3290N F∙t = ∆p Imp = F∙t = mvf – mvi = 3290N.(0.001s) = 3.29 Ns

Impulse causes a change in momentum F∙t = ∆p Ex5) A 60kg person is sitting in a car traveling at 26m/s. In an accident, the person is brought to a stop by the seatbelt in 0.22 seconds. How much force acts on the person? Imp = ∆p F∙t = mvf – mvi Ch9 HW #1 1 – 6

Impulse causes a change in momentum F∙t = ∆p Ex5) A 60kg person is sitting in a car traveling at 26m/s. In an accident, the person is brought to a stop by the seatbelt in 0.22 seconds. How much force acts on the person? Imp = ∆p F∙t = mvf – mvi F = mvf – mvi = 0 – ( 60 )( 26 ) = -7,091 N t ( .22 ) Ch9 HW #1 1 – 6

1. Compact car of mass 725 kg moving at 28 m/s east. Find momentum. Ch9 HW#1 1 – 6 1. Compact car of mass 725 kg moving at 28 m/s east. Find momentum. A second car of mass 2,175 kg has same momentum. Find its velocity. 2. The driver of the 725kg compact car applies the brakes for 2 seconds. There was an average force of 5x103 N. a. What is the impulse? F∙t = b. Final velocity? F∙t = ∆p

F∙t = mvf – mvi –10,000 = 725(vf) – 725(28) vf = 14 m/s Ch9 HW#1 1 – 6 1. Compact car of mass 725 kg moving at 28 m/s east. Find momentum. A second car of mass 2,175 kg has same momentum. Find its velocity. 2. The driver of the 725kg compact car applies the brakes for 2 seconds. There was an average force of 5x103 N. a. What is the impulse? F∙t = 5,000N ∙ 2s = 10,000 N∙s b. Final velocity? F∙t = ∆p F∙t = mvf – mvi –10,000 = 725(vf) – 725(28) vf = 14 m/s

3. A driver accelerates a 240 kg snowmobile, speeding it up from 6 m/s to 28 m/s over 60 seconds. a. Find ∆p b. Impulse c. Force

3. A driver accelerates a 240 kg snowmobile, speeding it up from 6 m/s to 28 m/s over 60 seconds. a. Find ∆p b. Impulse F∙t = ∆p = 5,280 N∙s c. Force F = ∆p = 5,280 N∙s = 88 N t 60 s

4. A 75kg running back is running at 5 m/s North on the football field. a. What is his change in momentum if he slows to 2 m/s in 2 seconds? b. What average force does his muscles exert to slow him in the 2 sec?

4. A 75kg running back is running at 5 m/s North on the football field. a. What is his change in momentum if he slows to 2 m/s in 2 seconds? b. What average force do his muscles exert to slow him in the 2 sec? 5. A 75kg running back is running at 5 m/s North on the football field. a. What is his change in momentum if he is hit and stops in 0.2 sec? b. What average force does the opposing player exert on him? 6. A 75kg running back is running at 5 m/s North on the football field. a. What is his change in momentum if he turns and runs south at 5 m/s? b. What force to do this in 2 sec?

pi = pf Ch9.2 – Conservation of Momentum The total momentum = the total momentum before a collision after the collision Two types: Elastic collision (objects don’t stick together) Inelastic collision (objects stick together)

pi = pf Ch9.2 – Conservation of Momentum The total momentum = the total momentum before a collision after the collision Two types: Elastic collision (objects don’t stick together) m1v1i + m2v2i = m1v1f + m2v2f Inelastic collision (objects stick together) m1v1i + m2v2i = ( m1 + m2 )vf

Elastic collision m1v1i + m2v2i = m1v1f + m2v2f (Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf (Stick together) Ex1) A 0.510kg lab cart moving at 1.5m/s collides with and sticks to a 0.605kg lab cart at rest. What speed do they roll away at?  2 2 1 1

Elastic collision m1v1i + m2v2i = m1v1f + m2v2f (Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf (Stick together) Ex1) A 0.510kg lab cart moving at 1.5m/s collides with and sticks to a 0.605kg lab cart at rest. What speed do they roll away at?  m1v1i + m2v2i = ( m1 + m2 )vf (.510)(1.5) + 0 = (1.115)vf vf = .686 m/s 2 2 1 1

Elastic collision m1v1i + m2v2i = m1v1f + m2v2f (Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf (Stick together) Ex1) A 2,275kg car going 28m/s rear-ends a 875kg compact car going 16m/s in the same direction. The two cars stick together. If friction is negligible, how fast does the wreckage move after the collision? 

Elastic collision m1v1i + m2v2i = m1v1f + m2v2f (Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf (Stick together) Ex1) A 2,275kg car going 28m/s rear-ends a 875kg compact car going 16m/s in the same direction. The two cars stick together. If friction is negligible, how fast does the wreckage move after the collision? m1v1i + m2v2i = (m1 + m2)vf (2275)(+28) + (875)(+16) = (3150)vf 63700 + 14000 = 3150 vf 77700 = 3150 vf vf = 24.7 m/s Ch9 HW#2 7 – 11

Lab9.1 – Conservation of Momentum in Inelastic Collisions - due tomorrow - Ch9 HW#2 due at beginning of period

Ch9 HW#2 7 – 11 7. 2 freight cars, each with a mass of 3.0x105 kg collide. One was initially moving at 2.2 m/s, the other was at rest. They stick together. Final speed? 8. A 0.105kg hockey puck moving at 24 m/s, is caught by a 75kg goalie at rest. At what speed does the goalie slide on the ice? 9. A 35.0g bullet strikes a 5.0kg stationary wooden block and embeds itself in the block. They fly off together at 8 m/s, what was the initial speed of the bullet?

Ch9 HW#2 7 – 11 7. 2 freight cars, each with a mass of 3.0x105 kg collide. One was initially moving at 2.2 m/s, the other was at rest. They stick together. Final speed? m1v1i + m2v2i = ( m1 + m2 )vf ( 3x105 )( 2.2 ) + 0 = ( 6x105 )vf 8. A 0.105kg hockey puck moving at 24 m/s, is caught by a 75kg goalie at rest. At what speed does the goalie slide on the ice? ( 0.105 )( 24 ) + 0 = ( 0.105 + 75 )vf 9. A 35.0g bullet strikes a 5.0kg stationary wooden block and embeds itself in the block. They fly off together at 8 m/s, what was the initial speed of the bullet? m1v1i + m2v2i = ( m1 + m2 )vf ( 0.035 )v1i + 0 = ( 0.035 + 5 )( 8.6 )

Ch9 HW#2 7 – 11 10. A 0.705kg lab cart is moving at 1 m/s collides with and sticks to a 0.500kg cart initially at rest. What speed do they roll away at? 11. A 0.500kg lab cart had an unknown mass placed in it, and is pushed at 0.50 m/s. It collides with a 0.505kg cart at rest. They stick together and their final velocity is measure to be 0.30 m/s. What is the unknown mass? m1v1i + m2v2i = ( m1 + m2 )vf (0.500 + m)(0.50) + 0 = ([0.500 + m] + .505)(0.30)

Ch9 HW#2 7 – 11 10. A 0.705kg lab cart is moving at 1 m/s collides with and sticks to a 0.500kg cart initially at rest. What speed do they roll away at? m1v1i + m2v2i = ( m1 + m2 )vf 11. A 0.500kg lab cart had an unknown mass placed in it, and is pushed at 0.50 m/s. It collides with a 0.505kg cart at rest. They stick together and their final velocity is measure to be 0.30 m/s. What is the unknown mass? (0.500 + m)(0.50) + 0 = ([0.500 + m] + .505)(0.30) 0.25 + 0.50m = (1.005 + .505m)(0.30) 0.25 + 0.50m = 0.3015 + 0.1515m 0.349m = 0.0515

Ch9.2B – Conservation of Momentum cont Elastic collision m1v1i + m2v2i = m1v1f + m2v2f (Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf (Stick together) Ex1) A 0.510kg lab cart moving at 1.5m/s collides with and a 0.605kg lab cart at rest. After the collision, the 0.605kg cart is moving at 1.26 m/s. What speed does the 0.510 kg cart roll at?  2 2 1 1

(.510)(1.5) + 0 = (.510)v1f + (.605)(1.26) v1f = 0 m/s Ch9.2B – Conservation of Momentum cont Elastic collision m1v1i + m2v2i = m1v1f + m2v2f (Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf (Stick together) Ex1) A 0.510kg lab cart moving at 1.5m/s collides with and a 0.605kg lab cart at rest. After the collision, the 0.605kg cart is moving at 1.26 m/s. What speed does the 0.510 kg cart roll at?  m1v1i + m2v2i = m1v1f + m2v2f (.510)(1.5) + 0 = (.510)v1f + (.605)(1.26) v1f = 0 m/s 2 2 1 1

Elastic collision m1v1i + m2v2i = m1v1f + m2v2f (Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf (Stick together) Ex2) A 0.510kg lab cart moving at 1.5m/s collides with and a 0.705kg lab cart at rest. After the collision, the 0.705kg cart is moving at 1.10 m/s. What speed does the 0.510 kg cart roll at?  2 2 1 1

(.510)(1.5) + 0 = (.510)v1f + (.705)(1.10) v1f = – 0.02 m/s Elastic collision m1v1i + m2v2i = m1v1f + m2v2f (Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf (Stick together) Ex2) A 0.510kg lab cart moving at 1.5m/s collides with and a 0.705kg lab cart at rest. After the collision, the 0.705kg cart is moving at 1.10 m/s. What speed does the 0.510 kg cart roll at?  m1v1i + m2v2i = m1v1f + m2v2f (.510)(1.5) + 0 = (.510)v1f + (.705)(1.10) v1f = – 0.02 m/s 2 2 1 1

Elastic collision m1v1i + m2v2i = m1v1f + m2v2f (Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf (Stick together) Ex3) A 0.5 kg cue ball is traveling at 5 m/s when it strikes a stationary 0.4 kg 8-ball. After the collision, the 8-ball is traveling at 8 m/s straight ahead. What is the speed and direction of the cue ball?

( 0.5 kg ) ( 0.4 kg ) ( 0.5 kg ) ( 0.4 kg ) O→ O = O O→ Elastic collision m1v1i + m2v2i = m1v1f + m2v2f (Dont stick together) Inelastic collision m1v1i + m2v2i = ( m1 + m2 )vf (Stick together) Ex3) A 0.5 kg cue ball is traveling at 5 m/s when it strikes a stationary 0.4 kg 8-ball. After the collision, the 8-ball is traveling at 8 m/s straight ahead. What is the speed and direction of the cue ball? ( 0.5 kg ) ( 0.4 kg ) ( 0.5 kg ) ( 0.4 kg ) O→ O = O O→ v1i = 5 m/s v2i = 0 v1f = ? v2f = 8 m/s

( 0.5 kg ) ( 0.4 kg ) ( 0.5 kg ) ( 0.4 kg ) O→ O = O O→ Ex3) A 0.5 kg cue ball is traveling at 5 m/s when it strikes a stationary 0.4 kg 8-ball. After the collision, the 8-ball is traveling at 8 m/s straight ahead. What is the speed and direction of the cue ball? ( 0.5 kg ) ( 0.4 kg ) ( 0.5 kg ) ( 0.4 kg ) O→ O = O O→ v1i = 5 m/s v2i = 0 v1f = ? v2f = 8 m/s m1v1i + m2v2i = m1v1f + m2v2f ( 0.5 )( 5 ) + 0 = ( 0.5 )v1f + ( 0.4 )( 8 ) 2.5 = 0.5v1f + 3.2 -0.7 = 0.5 vf vf = -1.4 m/s Ch9 HW#3 12 – 15

Lab9.2 – Conservation of Momentum in Elastic Collisions - due tomorrow

at 2.5 m/s. What is the final speed of the 0.710 lab cart? Ch9 HW#3 12 – 15 12. A 0.710kg lab cart travelling at 2 m/s collides with a 0.500kg lab cart initially at rest. The collision is elastic. The 0.500kg cart moves away at 2.5 m/s. What is the final speed of the 0.710 lab cart? 2 2 1 1

m1v1i + m2v2i = m1v1f + m2v2f m1v1i + m2v2i = m1v1f + m2v2f Ch9 HW#3 12 – 15 12. A 0.710kg lab cart travelling at 2 m/s collides with a 0.500kg lab cart initially at rest. The collision is elastic. The 0.500kg cart moves away at 2.5 m/s. What is the final speed of the 0.710 lab cart? m1v1i + m2v2i = m1v1f + m2v2f 13. A 0.505kg lab cart travelling at 2.5 m/s collides with a 0.610kg cart initially at rest. The collision is elastic. The 0.610kg cart moves away at 2.2 m/s. What is the final speed of the 0.505 lab cart? m1v1i + m2v2i = m1v1f + m2v2f 2 2 1 1 2 2 1 1

14. Ball 1, with a mass of 0.355 kg, rolls along a frictionless table with a velocity of 0.095 m/s. It collides with ball 2, with a mass of 0.710 kg and a speed of .045 m/s in the same direction. After the collision, ball 1 continues in the same direction at 0.035m/s. What is the speed of ball 2? O→ + O → = O → + O 15. A 0.50kg ball traveling at 6.0 m/s collides with a 1.00kg ball moving in the opposite direction at 12.0 m/s. The 0.50kg ball bounces backwards at 14 m/s after the collision. Final speed of 1.00kg ball? O→ + ←O = ←O O

14. Ball 1, with a mass of 0.355 kg, rolls along a frictionless table with a velocity of 0.095 m/s. It collides with ball 2, with a mass of 0.710 kg and a speed of .045 m/s in the same direction. After the collision, ball 1 continues in the same direction at 0.035m/s. What is the speed of ball 2? O→ + O → = O → + O m1v1i + m2v2i = m1v1f + m2v2f 15. A 0.50kg ball traveling at 6.0 m/s collides with a 1.00kg ball moving in the opposite direction at 12.0 m/s. The 0.50kg ball bounces backwards at 14 m/s after the collision. Final speed of 1.00kg ball? O→ + ←O = ←O O ( + ) ( – ) ( – ) ( ? )

Ch9.3 – Explosions Ex1) A .44 caliber handgun has a mass of 1.276 kg. It fires a 10g lead bullet with a muzzle velocity of 366 m/s. What is the recoil velocity of the gun? What is the momentum of the gun after firing Before After

p = m.v = (1.276kg)(2.9m/s) = 3.66 kg.m/s Ch9.3 – Explosions Ex1) A .44 caliber handgun has a mass of 1.276 kg. It fires a 10g lead bullet with a muzzle velocity of 366 m/s. What is the recoil velocity of the gun? What is the momentum of the gun after firing Before After m1v1i + m2v2i = m1v1f + m2v2f 0 + 0 = (1.276kg)(v) + (0.010kg)(366m/s) v = 2.9m/s p = m.v = (1.276kg)(2.9m/s) = 3.66 kg.m/s

Ex2) An astronaut floating in space wants to get back to the space station. (He’s trying to get back to the space station restaurant. Apparently, it has great food, but no atmosphere!!!) He has a jetpack on, fires the thrusters, and 35 g of hot gasses are expelled at 875 m/s. If the astronaut has a mass of 84 kg, how fast does he move? In what direction should he direct the exhaust gases? Before After

m1v1i + m2v2i = m1v1f + m2v2f 0 + 0 = (0.035kg)(875m/s) + (84kg).v2 Ex2) An astronaut floating in space wants to get back to the space station. (He’s trying to get back to the space station restaurant. Apparently, it has great food, but no atmosphere!!!) He has a jetpack on, fires the thrusters, and 35 g of hot gasses are expelled at 875 m/s. If the astronaut has a mass of 84 kg, how fast does he move? In what direction should he direct the exhaust gases? Before After m1v1i + m2v2i = m1v1f + m2v2f 0 + 0 = (0.035kg)(875m/s) + (84kg).v2 v2 = 0.37 m/s

HW# 17. Velcro holds two carts together HW# 17. Velcro holds two carts together. After the spring is released, it pushes the carts apart, giving the 1.5 kg cart a speed of 3 m/s to the left. What is the velocity of the 4.5 kg cart? Before After

Ch9 HW#4 16 – 19 17. Velcro holds two carts together. After the spring is released, it pushes the carts apart, giving the 1.5 kg cart a speed of 3 m/s to the left. What is the velocity of the 4.5 kg cart? Before After m1v1i + m2v2i = m1v1f + m2v2f 0 + 0 = (1.5)(3) + (4.5)(v2f) v2f = 1 m/s

16. A 4.0 kg model rocket is launched, shooting 50 g of burned fuel Ch9 HW#4 16 – 19 16. A 4.0 kg model rocket is launched, shooting 50 g of burned fuel out its exhaust at 625 m/s. What is the speed of the rocket? 18. A 5 kg rifle fires a 50 g bullet at 100 m/s. What is its recoil velocity?

what speed does it move backward at? Ch9 HW#4 16 – 19 18. A 5 kg rifle fires a 50 g bullet at 100 m/s. What is its recoil velocity? m1v1i + m2v2i = m1v1f + m2v2f 19. 2 campers dock a canoe. One camper has a mass of 80 kg and steps off the canoe at 4.0 m/s. If the rest of the canoe totals 115 kg, what speed does it move backward at?

Lab9.3 – Explosions - due tomorrow - Ch9 HW#3 due at beginning of period - Want to get ahead? Work on Ch8,9 Rev tonite.

Ch8,9 Review 1. Assume you have mass of 50kg and the Earth has mass of 5.97x1024kg. The radius of the Earth is 6.38x106 m. What is the force of gravitational attraction between you and earth? How does this compare to your weight? Fg = m∙g = 490N 2. 2 masses m1 and m2 at a distance d apart exert a force F on each other. How does the force change if both masses triple and the distance between them is cut in half? 3. What is the orbital speed of a satellite circling the earth at a distance of 1x1010m from the center of the earth?

1. Assume you have mass of 50kg and the Earth has mass of 5.97x1024 kg. The radius of the Earth is 6.38x106 m. What is the force of gravitational attraction between you and earth? How does this compare to your weight? Fg = m∙g Fd = Gmm = ( 6.67x10-11 )( 5.97x1024 )( 50 ) = 490 N d2 ( 6.38x106 )2 = 489 N 2. 2 masses m1 and m2 at a distance d apart exert a force F on each other. How does the force change if both masses triple and the distance between them is cut in half? F = G( 3∙m )( 3∙m ) = 9∙Gmm 36X bigger ( ½ d )2 ¼ d 2 3. What is the orbital speed of a satellite circling the earth at a distance of 1x1010m from the center of the earth?

4. A car moving at 10 m/s crashes into a barrier and stops in 0.50s. There is a 20kg child in the car. Assume the child’s velocity is changed by the same amount in the same time period. a. What is the change in momentum of the child? ∆p = pf – pi b. What is the impulse needed to stop the child? Impulse = c. What is the average force exerted on the child?

∆p = pf – pi = mvf – mvi = (20)(0) – (20)(10) = -200 kgm/s 4. A car moving at 10 m/s crashes into a barrier and stops in 0.50s. There is a 20kg child in the car. Assume the child’s velocity is changed by the same amount in the same time period. a. What is the change in momentum of the child? ∆p = pf – pi = mvf – mvi = (20)(0) – (20)(10) = -200 kgm/s b. What is the impulse needed to stop the child? Impulse = ∆p = 200 Ns c. What is the average force exerted on the child? F.t = ∆p F(0.50s) = -200 Ns F = -4000N

5. Marble A with a mass of 5.0g moves at a speed of 20 cm/s. It collides with marble B, mass of 10.0g, moving at 10 cm/s in the same direction. After the collision, marble A continues with a speed of 8 cm/s in the same direction. What is the speed of marble B? 20 cm/s 10 cm/s 8 cm/s = 10 10 5 5

(0.005)(20) + (0.010)(10) = (0.005)(8) + (0.010)(v2f) vf = 16 cm/s 5. Marble A with a mass of 5.0g moves at a speed of 20 cm/s. It collides with marble B, mass of 10.0g, moving at 10 cm/s in the same direction. After the collision, marble A continues with a speed of 8 cm/s in the same direction. What is the speed of marble B? 20 cm/s 10 cm/s 8 cm/s = m1v1i + m2v2i = m1v1f + m2v2f (0.005)(20) + (0.010)(10) = (0.005)(8) + (0.010)(v2f) vf = 16 cm/s 10 10 5 5

6. A 2575kg van runs into the back of a 825kg car at rest. They move off together at 8.5m/s. Assuming friction is negligble, find the initial speed of the van. ? m/s 0m/s 8.5m/s + =

m1v1i + m2v2i = ( m1 + m2 )vf (2575kg)(v1i) + 0 = (2575+825)(8.5m/s) 6. A 2575kg van runs into the back of a 825kg car at rest. They move off together at 8.5m/s. Assuming friction is neglible, find the initial speed of the van. ? m/s 0m/s 8.5m/s + = m1v1i + m2v2i = ( m1 + m2 )vf (2575kg)(v1i) + 0 = (2575+825)(8.5m/s) v1i = 11.2 m/s

7. Marble A with a mass of 5.0g moves at a speed of 20 cm/s east. It collides with marble B, mass of 10.0g, moving at 10 cm/s west. After the collision, marble A rebounds with a speed of 18 cm/s west. What is the speed and direction of marble B? 20 cm/s 10 cm/s 18 cm/s = 10 10 5 5

(0.005)(+20) + (0.010)(–10) = (0.005)(–8) + (0.010)(v2f) 7. Marble A with a mass of 5.0g moves at a speed of 20 cm/s east. It collides with marble B, mass of 10.0g, moving at 10 cm/s west. After the collision, marble A rebounds with a speed of 18 cm/s west. What is the speed and direction of marble B? 20 cm/s 10 cm/s 18 cm/s = m1v1i + m2v2i = m1v1f + m2v2f (0.005)(+20) + (0.010)(–10) = (0.005)(–8) + (0.010)(v2f) v2f = +9 cm/s (east) 10 10 5 5

Ex1) Galileo discovered the moons of Jupiter Ex1) Galileo discovered the moons of Jupiter. He could measure their orbital sizes by using the diameter of Jupiter as a unit of measure. Io had a period of 1.8 days and was 4.2 units from the center of Jupiter. Callisto has a period of 16.7 days. Predict Callista’s distance from Jupiter’s center. 1.82 = 4.23 3.24∙rC3 = 20,639 16.72 rC3 rC3 = 6,370 rC = 18.6 units