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 = c. Force F = ∆p = 5,280 N∙s = 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?

Ex1) 2 cars enter an icy intersection and skid into each other, Ch9.4 2-Dimensional Collisions Ex1) 2 cars enter an icy intersection and skid into each other, the 2500kg sedan heading south at 20m/s and the 1450kg coupe heading east at 30m/s. The 2 vehicles entangle. What speed and direction do they slide off at?

Ex2) A 0.1kg ball slides at 1.4 m/s across a table and collides with another 0.1kg stationary ball. The second ball moves off at 30°, with a speed of 0.3 m/s, as shown. What is the speed and direction of the 1st ball after the collision? 0.3 m/s 1.4 m/s 30° Ch9 HW#5 20,21 1 2

20. 2 cars enter an icy intersection and skid into each other, Ch9 HW#5 20,21 20. 2 cars enter an icy intersection and skid into each other, the 1000kg car #1 heading north at 10m/s and the 1500kg car #2 heading east at 15m/s. The 2 vehicles entangle. What speed and direction do they slide off at?

21. A 0.500kg hockey puck slides at 10 m/s across the ice and collides with another 0.500kg stationary puck. The second puck moves off at 10°, with a speed of 5 m/s, as shown. What is the speed and direction of the 1st ball after the collision? 5 m/s 10 m/s 15° 1 2

Ch9 Review 1. 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 1. 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

2. 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 2. 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

3. 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) 3. 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

4. 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) 4. 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