3. + What do you think? Imagine an automobile collision in which an older model car from the 1960s collides with a car at rest while traveling at 15 mph. Now imagine the same collision with a 2007 model car. In both cases, the car and passengers are stopped abruptly. List the features in the newer car that are designed to protect the passenger and the features designed to minimize damage to the car. How are these features similar?
4. + What do you think? What are some common uses of the term momentum? Write a sentence or two using the term momentum. Do any of the examples provided reference the velocity of an object? Do any of the examples reference the mass of an object?
5. + Momentum Momentum (p) is proportional to both mass and velocity. A vectorquantity, so direction matters!! Things moving right are positive Tings moving left are negative SI Units: kg • m/s
6. + Example A 2250kg pick-up truck has a velocity of 25m/s to the east. What is the momentum of the truck? m= 2250kg v=25m/s east (+) P= mv P= (2250)(25) P= 56,250 kg • m/s
7. + Momentum and Newton’s 2nd Law Provethat the two equations shown below are equivalent. F = ma and F = p/ t Newton actually wrote his 2nd Law as F = p/ t. Force depends on how rapidly the momentum changes.
8. + Impulse and Momentum The quantity F t is called impulse. SI units: N•m or kg•m/s Impulse equals change in momentum. Another version of Newton’s 2nd Law Changes in momentum depend on both the force and the amount of time over which the force is applied.
9. + Changing momentum Greater changes in momentum ( p) require more force (F) or more time ( t) . A loadedtruck requires more time to stop. Greater pfor truck with more mass Same stopping force
10. + Movie file:///Volumes/Physics_mac/inquiry_ppts/f iles/ch06/70598.html
11. + Example A 1400kg car moving westward with a velocity of 15m/s collides with a utility pole and is brought to rest in 0.3sec. Find the force exerted on the car during collision. Given: m=1400kg t= 0.30sec F=?? vi= 15m/s west (-) vf= 0m/s
12. + Example Ft= mvf- mvi mv f mvi F t (1400)(0) (1400)( 15) F 0.30 F=70,000N to the east
13. + Example A 2240kg car traveling west slows down uniformly from 20m/s to 5m/s. how long does it take the car to slow down if the force on the car is 8410N to the east? How far does the car travel during the time it slows down?
14. + Example mv f mvi Given: t F m= 2240kg (2240)( 5) (2240)( 20) v i= 20m/s west (-) t 8410 vf= 5m/s west (-) F= 8410N east (+) t= 4 sec t= ?? d=??
15. + Example d= ½ (vi + vf)t d= ½ (-20-5) 4 d= -50 m or 50 m west
16. + Stopping Time F t = p = mv When stopping, p is the same for rapid or gradual stops. Increasing the time ( t) decreases the force (F). What examples demonstrate this relationship? Air bags, padded dashboards, trampolines, etc Decreasing the time ( t) increases the force (F). What examples demonstrate this relationship? Hammers and baseball bats are made of hard material to reduce the time of impact.
17. +6.2 Conservation of MomentumPg. 205-211
18. + What do you think? • Two skaters have equal mass and are at rest. They are pushing away from each other as shown. • Compare the forces on the two girls. • Compare their velocities after the push. • How would your answers change if the girl on the right had a greater mass than her friend? • How would your answers change if the girl on the right was moving toward her friend before they started pushing apart?
19. + Momentum During Collisions When the bumper cars collide, F1 = -F2 so F1 t = -F2 t, and therefore p1 = - p2. Thechange in momentum for one object is equal and opposite to the change in momentum for the other object. Total momentum is neither gained nor lost during collisions.
20. + Conservation of Momentum Total momentum remains constant during collisions. The momentum lost by one object equals the momentum gained by the other object. Conservation of momentum simplifies problem solving.
21. + Example A 76kg boater, initially at rest, in a stationary 45kg boat steps out of the boat onto the dock. If the boater moves out of the boat with a velocity of 2.5m/s to the right, what is the final velocity of the boat?
22. + Given m 1= 76kg m 2= 45kg v1i= 0m/s v2i= 0m/s v1f= 2.5 m/s v2f= ?? m1v1i + m2v2i= m1v1f + m2v2f (76)(0)+(45)(0)=(76)(2.5)+(45)(v2f) 0= 190 + 45v2f -190 = 45v2f v2f= -4.2 m/s to the right but notice the negative v2f= 4.2 m/s to the left
23. + Now what do you think? • Two skaters have equal mass and are at rest. They are pushing away from each other as shown. • Compare the forces on the two girls. • Compare their velocities after the push. • How would your answers change if the girl on the right had a greater mass than her friend? • How would your answers change if the girl on the right was moving toward her friend before they started pushing apart?
24. +6.3 Elastic and Inelastic CollisionsPg. 212-220
25. + What do you think? • Collisions are 1. A baseball and a bat sometimes described 2. A baseball and a glove as elastic or inelastic. To the right is a list of 3. Two football players colliding objects. Rank 4. Two billiard balls them from most elastic 5. Two balls of modeling clay to most inelastic. 6. Two hard rubber toy balls • What factors did you consider when ranking 7. An automobile collision these collisions?
26. + Collisions Perfectly inelastic collisions A collision where 2 objects stick together after colliding Perfectlyinelastic collisions are analyzed in terms of momentum Dueto the fact that the objects become one object after the collision
27. + Perfectly Inelastic Collisions Two objects collide and stick together. Two football players A meteorite striking the earth Momentum is conserved. Masses combine.
28. + Perfectly Inelastic Collisions v1i is + (m1 is moving to the right) V2i is – (m2 is moving to the left)
29. + Example A 1850kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975kg. Te 2 cars become entangled as a result of the collision. If the compact car was moving at a velocity of 22m/s to the north before the collision, what is the velocity of the mass after the accident?
30. + Example m 2= 975kg m 1= 1850kg v2i= 22m/s north (+) v1i= 0m/s vf= ?? m1v1i + m2v2i= (m1+ m2) vf (1850)(0) (975)(22) vf 1850 975 vf= 7.59 m/s north
31. + Classroom Practice Problem Gerard is a quarterback and Tyler is a defensive lineman. Gerard’s mass is 75.0 kg and he is at rest. Tyler has a mass of 112 kg, and he is moving at 8.25 m/s when he tackles Gerard by holding on while they fly through the air. With what speed will the two players move together after the collision? Answer: 4.94 m/s
32. + Classroom Practice Problems An 2.0 x 105 kg train car moving east at 21 m/s collides with a 4.0 x 105 kg fully-loaded train car initially at rest. The two cars stick together. Find the velocity of the two cars after the collision. Answer: 7.0 m/s to the east
33. + Inelastic Collisions Kinetic energy is less after the collision. It is converted into other forms of energy. Internal energy - the temperature is increased. Sound energy - the air is forced to vibrate. Some kinetic energy may remain after the collision, or it may all be lost.
34. + Example 2 clay balls collide in a perfect inelastic collision. The 1st ball has a mass of 0.500kg and an initial velocity of 4.0m/s to the right. The 2nd ball has a mass of 0.250 kg. and an initial velocity of 3.0m/s to the left. What is the decrease in kinetic energy during the collision?
35. + Example m 1= 0.500 kg KE= v1= 4m/s right (+) KE= KEf – KEi m 2= 0.250kg KEi= ½ m1v1i + m2v2i v2= 3m/s left (-) KEf= ½ (m1+m2)vf2
36. + Example m1v1i + m2v2i= (m1+ m2) vf m1v1i m2v2i (0.5)(4) (0.25)( 3) vf (m1 m2 ) (0.5 0.25) vf= 1.67 m/s KEi= ½ (0.5)(42) + ½ (0.25)(-32)= 5.125J KEf= ½ (0.5 + 0.25) 1.672 = 1.05J KE= 1.05 – 5.125 = -4.08J Means that KE is lost
37. + Elastic Collisions Objects collide and return to their original shape. Kinetic energy remains the same after the collision. Perfectly elastic collisions satisfy both conservation laws shown below.
38. + Elastic Collisions Two billiard balls collide head on, as shown. Which of the following possible final velocities satisfies the law of conservation of momentum? vf,A = 2.0 m/s, vf,B = 2.0 m/s m = 0.35 kg m = 0.35 kg vf,A = 0 m/s, vf,B = 4.0 m/s vf,A = 1.5 m/s, vf,B= 2.5 m/s Answer: all three v = 4.0 m/sv= 0 m/s
39. + Elastic Collisions Two billiard balls collide head on, as shown. Which of the following possible final velocities satisfies the law of conservation of kinetic energy? vf,A = 2.0 m/s, vf,B = 2.0 m/s m = 0.35 kg m = 0.35 kg vf,A = 0 m/s, vf,B = 4.0 m/s vf,A = 1.5 m/s, vf,B= 2.5 m/s Answer: only vf,A = 0 m/s, vf,B = 4.0 m/s v = 4.0 m/s v= 0 m/s
40. + Example A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head on collision with a 0.030 kg shooter marble moving to the left at 0.180m/s. After the collision, the smaller marble moves to the left at 0.315m/s. Assume neither marble rotates at all and both marbles are on a frictionless surface. What is the velocity of the 0.030kg marble after the collision?
41. + Example m 1= 0.015kg m 2= 0.030kg v1i= 0.225m/s right (+) v2i= 0.18m/s left (-) v1f= 0.315m/s left (-) vf= ?? m1v1i + m2v2i= m1v1f + m2v2f (0.015)(0.225) + (0.030)(-0.18) = (0.015)(-0.315) + (0.030)(vf) 0.003375 + -0.0054 = -0.004725 + 0.030 vf -0.002025= - 0.004725 + 0.030 vf 0.0027 = 0.030vf vf= 0.09 m/s right
42. + Types of Collisions Click below to watch the Visual Concept. file:///Volumes/Physics_mac/inquiry_ppts/f iles/ch06/70600.html
43. + Types of Collisions
44. + Now what do you think? • To the right is a list of 1. A baseball and a bat colliding objects. Rank 2. A baseball and a glove them from most elastic to most inelastic. 3. Two football players 4. Two billiard balls • What factors did you consider when ranking 5. Two balls of modeling clay these collisions? 6. Two hard rubber toy balls 7. An automobile collision
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