1 / 21

Chapter 7

Chapter 7. Linear Momentum. Why study momentum ? In an isolated system, net force is zero,  momentum is a conserved quantity Applications: collision problems  mass or velocity determination discovery of missing objects or sub-atomic particles (neutrino 中微子 ).

bona
Download Presentation

Chapter 7

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. Chapter 7 Linear Momentum

  2. Why study momentum ? • In an isolated system, net force is zero, • momentum is a conserved quantity • Applications: • collision problems  mass or velocity • determination • discovery of missing objects or • sub-atomic particles (neutrino中微子) http://www.hk-phy.org/articles/neutrino/neutrino.html

  3. Linear Momentum - Like energy, it is conserved Define momentum Consider the rate of change of momentum 0 Mass of the system does not change in general Newton’s Second Law Newton actually stated his Second Law of motion as  Net External force equals the change in momentum of a system divided by the time over which it changes.

  4. q q Collision with a stationary object Axis-aligning bounding box  means that the sides of the box are horizontal and vertical If the stationary boundary isvertical, vf = [-vix, viy] If the stationary boundary ishorizontal, vf = [vix, -viy] For incoming velocity vi = [vix, viy] The angle that the ball comes in at must equal the angle at which it leaves, that is the angle of incidence (incoming) must equal the angle of reflection (outgoing). Example • Suppose you are coding a simple Pong game, and you want to model thecollision of the ball with the paddle. If the ball is approaching the paddle with an incoming velocity of [40, 75] when it hits, what should be its resulting velocity be? Answer The final velocity is [-40, 75]

  5. Example 7.1 • Calculate the momentum of a 110Kg football player running at 8 m/s. • Compare the player’s momentum with that of a hard-thrown 0.41Kg football that has a speed of 25 m/s. • m=110Kg, v=8 m/s, mfb=0.41 Kg, vfb=25 m/s • p/pfb = ? = 85.9 • Example 7.2 • What is the average force does exert on a 0.14Kg baseball by a bat, given that the ball’s initial velocity is 45 m/s and that its final velocity, after a 1.3 ms impact, is 65 m/s in exactly the opposite direction? • m=0.14 Kg, v0=45 m/s, vf=-65 m/s, in 1.3 ms • F = ?  Dp/Dt = (pf– pi)/Dt = m(vf-v0)/0.0013 = 11800 N  ~2600 lb

  6. 7.2 Impluse Effect of a force on an object depends on how long it acts. Dp = (net F) Dt The change in momentum equals the average net external force multiplied by the time this force acts. (net F) Dt is called the impulse, I Impulse is the same as change in momentum, I=Dp notice I // Dp 體操 : for a given Dp change, net F ↓ if Dt ↑ less impact A piano hammer striking a string would generate a force similar to Factual but its impluse might be the same as that of Feff.

  7. Example 7.3 Calculate the final speed of a 110Kg football player running at 8 m/s who collides head on with a padded goalpost and experiences a backward force of 17600 N for 0.055 s. M=110 Kg, vi=8 m/s, net F=17600 N, Dt=0.0550 s  vf = ? Use I = (net F) Dt = M(vf– vi) -17600*0.0550 = 110(vf– 8) vf = -0.800 m/s The minus sign indicates the player bounces backward.

  8. 7.3 Conservation of momentum Under what condition is momentum conserved ? Net external force = 0  consider a larger system • p1 + p2 = constant • Ptotal = p1 + p2 = p1’ + p2’ • Conservation of Momentum (isolated system, net F = 0) • Ptotal = constant • Ptotal = P’total • Consider the impulse • Dp1 = F1Dt • Dp2 = F2Dt • Newton’s Third Law F2 = -F1 • Dp2 = (-F1)Dt = -Dp1 • Dp1 + Dp2 =0

  9. - The three dimension in nature is independent - Momentum can be conserved along one direction and not another. - Momentum is conserved along the X-direction, but not in Y-direction

  10. 7.4 Elastic collisions in one dimension The two collided bodies moving the along the same direction Elastic collision –both momentum and internal kinetic energy conserved Very nearly elastic collision because some KE  heat, sound

  11. Example 7.4An elastic collision • Calculate the velocities of two masses following an elastic collision, • given that mA = 0.500 Kg, mB = 3.5 Kg, vA = 4.00 m/s, vB = 0. • What are the final velocities of mA and mB ?  vA’ = ? , vB’ = ? • Since vB=0,  pA = pA’ + pB’ • mA vA = mAvA’ + mBvB’ • ½ mAvA2 = ½ (mAvA’2 + mBvB’2) • Solve the conservation of momentum for vB’ first, and substitute into • conservation of internal KE to eliminate vB’, leaving only vA’ unknown. • There are two solutions for vA’, vA’= 4.00 m/s or -3.00 m/s • First solution  same as initial condition  discarded 4.00 m/s • vB’ = 1.00 m/s • A small mass m, collide with a larger mass M, the larger mass is knocked forward with a lower speed (here mB = 7mA).

  12. 7.5 Inelastic collision in one dimension Inelastic collision  internal kinetic energy is not conserved, some internal KE may be converted into heat or sound energy Two equal masses head toward one another at equal speeds and stick together. KEint = mv2, KE’int = 0  internal KE not conserved The two masses come to rest after collision  momentum conserved Perfectly inelastic, KE’int = 0

  13. Example 7.5 Inelastic collision of puck and goalie • Find the recoil velocity of a 70 Kg hockey goalie, originally at rest, who catches a 0.15 Kg hockey puck slapped at him at a velocity of 35 m/s. • How much KE is lost in the collision? Assume friction between the ice and the puck-goalie system is negligible. • m1=0.15 Kg, v1=35 m/s, m2=70.0 Kg, v2 = 0 • v1’=v2’=v’= ? • How much KE is lost in collision ? • (a) m1 v1 + m2 v2 = (m1 + m2) v’ solve for v’ • v’ = 0.0748 m/s • KE’int– KEint = ½ (m1+m2)v’2–½ m1v12 • = -91.7 J

  14. Example 7.6 A collision that releases stored energy is inelastic Two carts collide inelastically, and a spring releases its PE and converts it into internal KE. Mass of the cart on the left is m1= 0.35Kg, v1= 2.00m/s. Cart on the right has m2= 0.5Kg, v2= -0.5m/s. After collision, the first cart has a recoil velocity v1’= -4.00m/s. (a) What is the final vel. of the first cart, v2’=? (b) how much energy is released by the spring (assume all internal PE is converted into internal KE) (a) m1v1 + m2v2 = m1v1’ + m2v2’ Solve the conservation of momentum for v2’ first, v2’=3.7m/s (b) KEint = ½ (m1v12 + m2 v22)= 0.763 J KE’int = ½ (m1 v1’2 + m2 v2’2) = 6.22 J KE’int– KEint = 5.46 J (internal KE increased energy released by the spring

  15. 7.6 Collision of point masses in two dimensions • Complication, if • object rotation • Point masses • Neglect rotation Many scattering experiments have a target mass that is stationary in laboratory X component : m1v1 = m1v1’cosq1 + m2v2’cosq2 Y component : 0 = m1v1’sinq1 + m2v2’sinq2

  16. Example 7.7 Determine the final velocity of an unseen mass from the scattering of another mass A 0.25Kg mass is slid on a frictionless surface into a dark room, where it strikes an object with m2=0.4, v2=0. The 0.25Kg mass emerges from the room at an angle of 45 with its incoming direction, The speed of the 0.25Kg mass was v1=2 m/s, and it is 1.5 m/s after the collision. Calculate the magnitude and direction of the velocity (v’2 and h) of the 0.4Kg mass after the collision. X : m1v1 = m1v1’cos45 + m2v2’cosq2 Y : 0 = m1v1’sin45 + m2v2’sinq2 m1= 0.250Kg, m2=0.400Kg v1= 2.00m/s, v2=0 v1’=1.5 m/s solve for sinq2 and cosq2 tanq2 q2 = -48.5o Given v2’ one can determine mass of the unseen mass m2

  17. Elastic Collisions of Two Equal Masses • See p.185 • a special case m1 = m2 like the case with billiard balls, or the case • with some subatomic collisions • assume v2 =0, and elastic collisions  internal KE conserve • ½ mv12 = ½ m (v1’2 + v2’2) • X and Y components of momentum conserve • X component : m1v1 = m1v1’cosq1 + m2v2’cosq2 • Y component : 0 = m1v1’sinq1 + m2v2’sinq2 •  (X component)2 : v12= (v1’cosq1 + v2’cosq2)2 • (Y component)2 : 0 = (v1’sinq1 + v2’sinq2)2 ½ mv12 = ½ m (v1’2 + v2’2) + m v1’v2’ cos(q1 + q2)  m v1’v2’ cos(q1 + q2)=0 (1) v1’= 0  head on collision, incoming mass stop (2) v2’ = 0  no collision, incoming mass unaffected (3) cos(q1 + q2) = 0  angle of separation is 90o after collision

  18. Particle Accerlator CERN - European Organization for Nuclear Research 27 Km in diameter

  19. World Class Particle accelerator • CERN • http://cernenviro.web.cern.ch/CERNenviro/web/main/main.php • Large Hadron Collider (LHC) • http://livefromcern.web.cern.ch/livefromcern/antimatter/history/historypictures/LHC-drawing-half.jpg • FermiLab • http://www.cs.cmu.edu/~zollmann/pics/2004_10_chicago/slides/fermilab.html • Stanford Linear Accelerator Center (SLAC) • http://www.pbs.org/wgbh/nova/einstein/toda-ocon-01.html

  20. A submicroscopic particle scatters straight backward from a target particle. In experiments seeking evidence for quarks, electrons were observed t occasionally scatter straight backward from a proton.

  21. 7.7 Rockect Propulsion

More Related