1 / 20

Warm-up U7-#1 1/13/14

Warm-up U7-#1 1/13/14. m=3200kg v i =16m/s v f = 0m/s t=25.6s F=?. 1) What is the force felt on a 3200kg truck moving at 16 m/s which runs into a haystack bringing the truck to a stop in 25.6 seconds. Ft = m v F (25.6 s ) = 3200 kg (0-16 m/s )

amalie
Download Presentation

Warm-up U7-#1 1/13/14

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Warm-up U7-#1 1/13/14 m=3200kg vi =16m/s vf = 0m/s t=25.6s F=? 1) What is the force felt on a 3200kg truck moving at 16 m/s which runs into a haystack bringing the truck to a stop in 25.6 seconds. Ft = mv F (25.6s) = 3200kg (0-16m/s) F(25.6s) = - 51200 kgm/s  F = - 2,000 N B

  2. Law of Conservation of Momentum- Momentum is neither gained nor lost in the absence of an external force momentum before = momentum after pbefore = pafter

  3. Collision #1 – Object at rest or Explosions after before • BEFORE – object at rest  momentum zero. • AFTER – cannon and ball go in opposite directions. Momentums cancel  total momentum = zero. pcannonbefore + pballbefore = 0 pcannonafter + pcballafter = 0

  4. ex 1) A 2 kg rifle shots a 0.001kg bullet at 200 m/s. What will be the recoil velocity of the rifle? G: mrifle = 2 kg mbullet = 0.001kg vbullet = 200m/s pbefore = pafter pbefore= 0 U: vrifle = ? Eq: pafter= 0 pafter = (mv)bullet + (mv)rifle Sub: 0=(0.001kg*200m/s)+(2kg*vrifle) -2kg*vrifle = 0.2kgm/s “” Solve:vrifle = -0.1 m/s “-”

  5. Collision #2 - Elastic Collisions • Elastic collision- When objects collide without being permanently deformed and without generating heat. Objects do not stick together! • Kinetic Energy is conserved in elastic collisions. (m1v1 + m2v2)Before = (m1v1 + m2v2)after

  6. Ex. G: A 1000 kg car traveling at 20.0 m/s hits a 3000 kg truck at rest. If the truck is traveling 10 m/s forward after the elastic collision, what is the cars final velocity? (mcvc + mtvt)Before = (mcvc + mtvt)after 1000kg*20m/s+0=1000kg*vcar+3000kg*10m/s 20,000kgm/s - 30,000kgm/s = 1000kg * vcar -10,000kgm/s = 1000kg * vcar “” vcar = -10 m/s Before: mcar=1000kg mtruck=3000kg vcar=20 m/s vtruck=0 After: vcar = ? vtruck = 10m/s “-”

  7. Collision #3 - Inelastic Collisions • Inelastic collision- collision where the objects become distorted or generate heat. • Objects stick together so the after the collision there is only one object. (m1v1 + m2v2)before = (m1+m2)(vf)after

  8. Example 3: Sam, who is 85kg, jumps into a 300 kg rowboat initially at rest. His initial velocity was 5 m/s forward. What is the velocity of Sam in the boat after he lands? Before: msam=85kg mrow=300kg vsam=5m/s vrow = 0 (msvs + mrbvrb)before = (ms+mrb)(vf) (85kg*5m/s + 0) = (300kg+85kg)vf 425kgm/s = 385kg * vf vf = 1.1 m/s “” After: vf= ?

  9. Warm-up 1/14/14 Mike is traveling forward at 20.0 m/s in his 1000 kg car and hits Justice’s 1,500 kg car going slower at 8 m/s in the same direction. Justice’s car is traveling 15 m/s forward after the elastic collision, what is the final velocity of Mike’s car? GIVEN: EQUATION: SOLVE: m1= 1000 kg v1f= 9.5 m/s m1v1i + m2v2i= m1v1f+ m2v2f m2= 1500 kg SUBSTITUTE: v1i= 20 m/s (1000)(20) + (1500)(8) = (1000)(v1f) + (1500)(15) v2f= 15 m/s v2i= 8 m/s 20,000 + 12,000 = (1000)(v1f) + 22,500 UNKNOWN: 20,000 + 12,000 – 22,500 = (1000)(v1f) v1f= ? m/s 9500 = (1000)(v1f) 9500/1000 = v1f

  10. Problem1 A 40 kg child runs across a store at 4.0 m/s and jumps onto a 15 kg shopping cart initially at rest. At what speed will the shopping cart and the child move together across the store assuming negligible friction? What type of collision is this? Inelastic SOLVE: GIVEN: EQUATION: m1= 40 kg m1v1i + m2v2i= (m1 + m2)vf Vf = 2.9 m/s m2= 15 kg SUBSTITUTE: v1i= 4.0 m/s v2i= 0 m/s (40)(4.0) + (15)(0) = (40 + 15)vf 160 + 0 = (55)vf UNKNOWN: 160 = (55)vf vf= ? m/s 160/55 = vf

  11. Problem 2 Tanner throws a 0.20 kg football and knocks over a 0.90 kg vase at rest. (bad Tanner!) After the collision the football bounces straight back with a speed of 3.9 m/s while the vase is moving at 2.6 m/s in the opposite direction. How fast did Tanner throw the football? EQUATION: m1v1i + m2v2i= m1v1f+ m2v2f GIVEN: m1= 0.20 kg SUBSTITUTE: m2= 0.90 kg (0.20)(v1i) + (0.90)(0) = (0.20)(-3.9) + (0.90)(2.6) v1f= -3.9 m/s (0.20)(v1i) + 0 = (-0.78) + (2.34) v2f= 2.6 m/s v2i= 0 m/s (0.20)(v1i) = 1.56 v1i= 1.56/0.20 UNKNOWN: v1i= ? m/s SOLVE: v1i= 7.8 m/s

  12. Problem 3 After missing an easy lay up, Whitney tosses a 0.75 kg basketball at a 1.2 kg water jug initially at rest on the sidelines. The ball is thrown to the right at 8.5 m/s and continues to move to the right at 3.0 m/s after the collision. What is the velocity of the jug after the collision? EQUATION: GIVEN: m1v1i + m2v2i= m1v1f+ m2v2f m1= 0.75 kg SUBSTITUTE: m2= 1.2 kg (0.75)(8.5) + (1.2)(0) = (0.75)(3.0) + (1.2)(v2f) v1i= 8.5 m/s v1f= 3.0 m/s 6.375 + 0 = 2.25 + (1.2)(v2f) v2i= 0 m/s 6.375 – 2.25 = (1.2)(v2f) 4.125 = (1.2)(v2f) UNKNOWN: 4.125/1.2 = v2f v2f= ? m/s SOLVE: v2f= 3.4 m/s

  13. Problem 4 What is the final velocity of a 85 kg halfback rushing to the right at 10 m/s that hits a 130 kg linebacker running to the left at 8 m/s. After the elastic collision, the linebacker has slowed to 2 m/s. SOLVE: GIVEN: EQUATION: m1= 85 kg m1v1i + m2v2i= (m1 + m2)vf Vf = 2.9 m/s m2= 15 kg SUBSTITUTE: v1i= 4.0 m/s v2i= 0 m/s (40)(4.0) + (15)(0) = (40 + 15)vf UNKNOWN: vf= ? m/s

  14. In an experiment, a toy wooden car with a mass of 300g, initially at rest, is struck in the rear by a 30g dart traveling at 15 m/s as shown. With what speed does the car with the dart stuck in it move after the collision?

  15. A 50 kg astronaut traveling at 8 m/s to the left catches a 10 kg meteor traveling at 20 m/s to the left. What is the final velocity of the astronaut holding the meteor?

  16. 2) In an experiment, a toy wooden car with a mass of 300g, initially at rest, is struck in the rear by a 30g dart traveling at 15 m/s as shown. With what speed does the car with the dart stuck in it move after the collision? V= 15 m/s V= 0 m/s V= ? 300g 30g 300g 30g

  17. 3) A 50 kg astronaut traveling at 8 m/s to the left catches a 10 kg meteor traveling at 20 m/s to the left. What is the final velocity of the astronaut holding the meteor?

  18. Warm-up 1/14/14 Megan is traveling forward at 20.0 m/s in her 1000 kg car and hits Samantha’s 1,500 kg car going slower at 8 m/s in the same direction. Sam’s car is traveling 15 m/s forward after the elastic collision, what is the final velocity of Megan’s car? GIVEN: EQUATION: SOLVE: m1= 1000 kg v1f= 9.5 m/s m1v1i + m2v2i= m1v1f+ m2v2f m2= 1500 kg SUBSTITUTE: v1i= 20 m/s (1000)(20) + (1500)(8) = (1000)(v1f) + (1500)(15) v2f= 15 m/s v2i= 8 m/s 20,000 + 12,000 = (1000)(v1f) + 22,500 UNKNOWN: 20,000 + 12,000 – 22,500 = (1000)(v1f) v1f= ? m/s 9500 = (1000)(v1f) 9500/1000 = v1f

  19. Warm-up 1/14/14 Mike is traveling forward at 20.0 m/s in his 1000 kg car and hits Joseph’s 1,500 kg car going slower at 8 m/s in the same direction. Joseph’s car is traveling 15 m/s forward after the elastic collision, what is the final velocity of Mike’s car? GIVEN: EQUATION: SOLVE: m1= 1000 kg v1f= 9.5 m/s m1v1i + m2v2i= m1v1f+ m2v2f m2= 1500 kg SUBSTITUTE: v1i= 20 m/s (1000)(20) + (1500)(8) = (1000)(v1f) + (1500)(15) v2f= 15 m/s v2i= 8 m/s 20,000 + 12,000 = (1000)(v1f) + 22,500 UNKNOWN: 20,000 + 12,000 – 22,500 = (1000)(v1f) v1f= ? m/s 9500 = (1000)(v1f) 9500/1000 = v1f

More Related