1 / 80

Linear Momentum

Linear Momentum. Momentum (vector quantity). p = mass X velocity p = m v Unit = kg-m/s. Momentum: Example 1. What is the momentum of a 100.0 kg football player running at 6.0 m/s? p = mv = (100.0 kg)(6.0 m/s) = 600 kg-m/s. Momentum: Example 2.

fremont
Download Presentation

Linear Momentum

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. Linear Momentum

  2. Momentum (vector quantity) • p = mass X velocity • p = mv • Unit = kg-m/s

  3. Momentum: Example 1 What is the momentum of a 100.0 kg football player running at 6.0 m/s? p = mv = (100.0 kg)(6.0 m/s) = 600 kg-m/s

  4. Momentum: Example 2 If Mr. West is running at 2 m/s and has a momentum of 80 kg m/s, what is his mass?

  5. The Law of Conservation of Momentum m1v1 + m2v2 = m1v1’ + m2v2’ momentumbefore = momentumafter

  6. A 0.015 kg bullet is fired with a velocity of 200 m/s from a 6 kg rifle. What is the recoil velocity of the rifle (consider it’s direction)?

  7. The Rhino has a mass of 900 kg and hits the Hulk running at 2 m/s. After the collision, the Hulk is moving at 1.3 m/s and Rhino stops. What is the Hulk’s mass?

  8. A 10,000 kg railroad car moving at 24.0 m/s strikes an identical car that is stationary. They lock together. What will be there common speed afterwards?

  9. m1v1 + m2v2 = m1v1’ + m2v2’ m1v1 + m2v2 = (m1 + m2)vf (10,000 kg)(24.0 m/s) + 0 = (20,000 kg)(vf) vf = +12.0 m/s (Notice that the speed was cut exactly in half. Why?)

  10. Two boats meet in the middle of a lake. Boat A has a mass of 150 kg, and boat B a mass of 250 kg. The person in Boat A pushes against Boat B and has a speed of -1.4 m/s. • What is the initial momentum of Boat A? • Calculate the velocity of Boat B.

  11. Rocket Ship • Before moving, p = 0 • Which moves more, the rocket or the gas?

  12. A 0.145 kg ball is thrown at +40.2 m/s. The bat has a mass of 0.840 kg. How fast must you swing the bat to return the ball at 20.1 m/s? Assume the bat stops when you hit the ball. (ANS: -10.4 m/s) • How fast must you swing the bat to return the ball at 40.2 m/s? (ANS: -13.9 m/s)

  13. Types of Collisions • Elastic • KE and momentum are conserved • Inelastic • Momentum is conserved • KE is not conserved • Lost (heat, sound) • Added (chemical reaction) • Perfectly inelastic • Two objects stick together after the collision

  14. Elastic Collisions • Both KE and momentum are conserved m1v1i + m2v2i = m1v1f + m2v2f m1v1i - m1v1f = m2v2f - m2v2i m1(v1i - v1f) = m2(v2f - v2i)

  15. ½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2 m1v1i2 - m1v1f2 = m2v2f2 - m2v2i2 m1(v1i2 - v1f2) = m2(v2f2 - v2i2) Remember that (a2 – b2) = (a + b)(a – b) m1(v1i+ v1f) (v1i - v1f) = m2 (v2i+ v2f)(v2f- v2i) Divide the yellow box equations: m1(v1i+ v1f) (v1i - v1f) = m2 (v2i+ v2f)(v2f- v2i) m1(v1i - v1f) m2(v2f - v2i) v1i+ v1f = v2i+ v2f

  16. Elastic Collisions: Example 1 A cue ball moving at 2.0 m/s strikes the red ball. What is the speed of both balls after the elastic collision if they have equal mass?

  17. mvci + mvri = mvcf + mvrf all masses are equal mvci + 0 = mvcf + mvrf The red ball is still vci = vcf + vrf Factored out the m 2.0 m/s = vcf + vrf Need another equation vci+ vcf = vri+ vrf 2.0 m/s + vcf = 0 + vrf` Substitute vrf` = 2.0 m/s + vcf 2.0 m/s = vcf + 2.0 m/s + vcf 0 = 2vcf or vcf = 0 2.0 m/s = vcf + vrf = 0 + vrf vrf = 2.0 m/s

  18. Elastic Collisions: Example 2 Two pool balls of equal mass collide. One is moving to the right at 20 cm/s, and the other to the left at 30 cm/s. Calculate their velocities after they collide elastically. +20 cm/s -30 cm/s

  19. mvwi + mvbi = mvwf + mvbf vwi + vbi = vwf + vbf 20 cm/s + (-30 cm/s) = vwf + vbf -10 cm/s = vwf + vbf Need another equation vwi+ vwf = vbi+ vbf KE is conserved vwf = vbi+ vbf - vwi vwf = -30 cm/s+ vbf – 20 cm/s vwf = -50 cm/s+ vbfNow substitute -10 cm/s = -50 cm/s+ vbf + vbf 40 cm/s = 2vbf vbf = +20 cm/s

  20. vwf = -50 cm/s+ vbf vwf = -50 cm/s+ (+20 cm/s) = -30 cm/s +20 cm/s -30 cm/s -30 cm/s +20 cm/s Note how they exchanged velocities (because they have equal mass)

  21. Elastic Collisions: Example 3 A proton of mass 1.01 amu moving at 3.60 X 104 m/s elastically collides head on with a still Helium nucleus (4.00 amu). What are the velocities of the particles after the collision? vpf = -2.15 X 104 m/s vhf = 1.45 X 104 m/s

  22. vpf = vpi(mp – mh) = 3.60 X 104 m/s(1.01 amu-4,00 amu) (mp + mh) (1.01 amu + 4.00 amu) vpf = -2.15 X 104 m/s vpi+ vpf = vhf vhf = 3.60 X 104 m/s + (-2.15 X 104m/s) vhf = 1.45 X 104 m/s

  23. Perfectly Inelastic Collision: Ex 1 An 1800-kg Cadillac is stopped at a traffic light. It is struck in the rear by a 900-kg Cube moving at 20.0 m/s. The cars become entangled. Calculate their velocity after the collision. KE is not conserved m1v1 + m2v2 = m1v1’ + m2v2’ m1v1 + m2v2 = (m1 + m2)vf 0 + (900 kg)(20.0 m/s) = (1800kg + 900kg)vf vf= +6.67 m/s

  24. How much KE was lost as a result of the collision? KEi = ½ (900 kg)(20 m/s)2 = 180,000 J KEf = ½ (2700 kg)(6.67 m/s)2 = 60,000 J DKE = 60,000 J – 180,000 J = -120,000J

  25. Perfectly Inelastic Collision: Ex 2 Two balls of mud collide head on and stick together. The first ball of mud had a mass of 0.500 kg and was moving at +4.00 m/s. The second had a mass of 0.250 kg and was moving at –3.00 m/s. Find the velocity of the ball after the collision. m1v1 + m2v2 = m1v1’ + m2v2’ m1v1 + m2v2 = (m1 + m2)vf (0.500 kg)(4.00 m/s)+(0.250 kg)(-3.00 m/s) = (0.750 kg)(vf) vf = +1.67 m/s

  26. Perfectly Inelastic Collision: Ex 3 A 95.0 kg running back running at 3.50 m/s collides with a 120.0 kg linebacker running in the opposite direction at 4.00 m/s. They grapple and stick together • Calculate the velocity after the collision. (-0.69 m/s) • Calculate the kinetic energy of the system before the collision. (1542 J) • Calculate the kinetic energy after the collision. (51 J) • What percent of the kinetic energy was lost in the collision? (97%)

  27. Ballistic Pendulum Used to determine velocities Perfectly Inelastic collisions (KE converts to PE)

  28. Perfectly Inelastic Collision: Ex 3 A 5.00 gram bullet is fired into a 1.00 kg block of wood. The wood-bullet system rises 5 cm. Calculate the initial velocity of the bullet.

  29. m1v1 + m2v2 = m1v1’ + m2v2’ m1v1 + m2v2 = (m1 + m2)vf 0 +(0.005 kg)(vi) = (1.00 kg + 0.005 kg)(vf) (0.005 kg)(vi) = (1.005 kg)(vf) We need a second equation (two unknowns) ½ mv21 + mgy1 = ½ mv22 + mgy2 ½(1.005 kg)(vf2) +0 =0 + (1.005 kg)(9.8 m/s2)(0.05 m) vf = 0.990 m/s (0.005 kg)(vi) = (1.005 kg)(0.990 m/s) vi = 199 m/s Moment of impact

  30. A 7.00 g bullet is fired into a 0.950 kg ballistic pendulum. The bob rises to a height of 22.0 cm. Calculate the initial speed of the bullet. (284 m/s)

  31. An 8.00 g bullet is fired at 350 m/s into a ballistic pendulum bob of 2.00 kg. Calculate how high the coupled bullet and bob will rise.

  32. A 10.0 g bullet is fired into a 1.20 kg ballistic pendulum at a speed of 320 m/s. The string on the pendulum is 150 cm. • Calculate the height that the pendulum rose. (35.1 cm) • Calculate the angle the pendulum rises. (40.5o)

  33. Collisions in 2 or 3 Dimensions (Glancing collisions) • A moving marble collides with a stationary marble. • Which of the following situations is/are not possible after:

  34. If both masses are equal, then the angles of deflection add to 90o.

  35. Collisions in 2 or 3 Dimensions Problem Solving: • Resolve all vectors into their x and y components • Spix = Spfx Spiy = Spfy

  36. 2-D Collisions: Example 1 At an intersection, a 1500-kg car travels east at 25 m/s. It collides with a 2500-kg van traveling north at 20 m/s. If the vehicles stick together afterwards, calculate the magnitude and velocity of the cars after the collision.

  37. Let’s first work with the initial components Spix = (1500 kg)(25 m/s) Spix = 37,500 kg-m/s Spiy = (2500 kg)(20 m/s) Spiy = 50,000 kg-m/s After the collision: v vy = v sinq q vx = v cosq

  38. After the collision: Spix = Spfx 37,500 kg-m/s = mvx 37,500 kg-m/s = (1500 kg + 2500 kg) vx 37,500 kg-m/s = (4000 kg) vx vx = 9.375 m/s Spiy = Spfy 50,000 kg-m/s = mvy 50,000 kg-m/s = (1500 kg + 2500 kg)vy 50,000 kg-m/s = (4000 kg) vy vy= 12.5 m/s

  39. Now solve for v v2 = vx2+vy2 v = 16 m/s Divide the two equations tanq = vy = 12.5 vx 9.38 q = 53o

  40. 2-D Collisions: Example 1 A pool ball moving at +3.00 m/s strikes another pool ball (same mass) that is initially at rest. After the collision, the two balls move off at 45.0o angles. Calculate the speeds of the two balls. vred q = +450 q = -450 vgreen

  41. x coordinate mgvg + mrvr = mgvg + mrvr 3m + 0 = mgvg + mrvr 3 = vg + vr 3 = vgcos(45o) + vrcos(-45o) 3 = 0.707vg + 0.707vr 4.24 = vg + vr

  42. y coordinate 0 = mgvg + mrvr 0 = mgvg + mrvr 0 = vg + vr 0 = vgsin(45o) + vrsin(-45o) 0 = 0.707vg - 0.707vr vg = vr Two equations, two unknowns 4.24 = vg + vr vg = vr 2.12 m/s

  43. Two zombie heads of unequal mass sit on an icerink. The first head (m1 = 4.00 kg) is propelled toward the stationary second head (m2 = 2.00 kg) at a velocity of v1= 4.00 m/s. After the collision, both heads move off at 35o to the x-axis. Calculate their final speeds. 2.44 m/s, 4.88 m/s

  44. A 1000 kg car travelling south at 15.0 m/s is hit by a 1500 kg van travelling west at 20.0 m/s. They stick together. • Calculate the final speed of the cars. (13.4 m/s) • Calculate the direction that the cars will travel in. (26.6o South of West) • Can you express your answer in “i and j” notation?

  45. Two different pucks are placed on an air hockey table. A 15.0 g red puck is pushed to the right at 1.00 m/s. A 20.0 g green puck is pushed to the left at 1.20 m/s. After the collision, the green puck travels at 1.10 m/s at an angle of 40.0o south of the horizontal. • Calculate the x and y components of the green puck’s velocity after the collision. (-0.843 m/s, -0.707 m/s) • Calculate the x and y components of the green puck’s momentum after the collision. (-0.0169 kg m/s, -0.0141 kg m/s) • Find the speed and direction of the red puck. (1.08 m/s, 60.7o north of east) • Can you express your answer to (c) in “i and j” notation? ((0.527 i + 0.940j) m/s)

  46. Center of Mass Center of mass – one point on a particle that follows the same path.

  47. Center of Mass • Stand perpendicular to the wall. Both feet must be against the baseboard. Now lift your outer leg. • Stand with your back against the wall. Be sure your heels touch the baseboard. Now try to pick up an eraser without bending your knees.

  48. Center of Mass 3. Get down on your elbows and knees on the floor. Place your elbows right in front of your knees and then move your hands behind you back. Try to move an eraser with your nose.

  49. General Motion • Translational Motion • all points of an object follow the same path • Sliding a book across a table • Rotational Motion • General Motion – combination of translational and rotation motion

More Related