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Chapter 9 Linear Momentum and Collisions EXAMPLES

Chapter 9 Linear Momentum and Collisions EXAMPLES. Example 9.1 The Archer. The archer of mass 60kg is standing on a frictionless surface (ice). He fires a 0.50kg arrow horizontally at 50 m/s . With what velocity does the archer move across the ice after firing the arrow? Can we use:

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Chapter 9 Linear Momentum and Collisions EXAMPLES

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  1. Chapter 9Linear Momentum and Collisions EXAMPLES

  2. Example 9.1 The Archer • The archer of mass 60kg is standing on a frictionless surface (ice). He fires a 0.50kg arrow horizontally at 50 m/s. With what velocity does the archer move across the ice after firing the arrow? • Can we use: • Newton’s Second Law ? NO No information about F or a • Energy? NO No information about work or energy • Momentum? YES • The System will be: the archer with bow (particle 1) and the arrow (particle 2)

  3. Example 9.1 The Archer, final • ΣFx = 0, so it is isolated in terms of momentum in the x-direction • Total momentum before releasing the arrow is 0: p1i + p2i = 0 • The total momentum after releasing the arrow is p1f + p2f = 0 m1v1f + m2v2f = 0  v1f = – (m2/m1 )v2f v1f = – (0.50kg/60kg)(50.0î)m/s  v1f = –0.42 î m/s • The archer will move in the opposite direction of the arrow after the release • Agrees with Newton’s Third Law • Because the archer is much more massive than the arrow, his acceleration and velocity will be much smaller than those of the arrow

  4. Example 9.2 Railroad Cars Collide(Perfectly Inelastic Collision) • Initial Momentum = Final Momentum (One-Dimension) If: m1v1+m2v2 = (m1 + m2)v’ Given: v2 = 0, v’1= v’2 = v’ = ? v’ =m1v1/(m1 + m2) = 240,000/20,000m/s v’ =12 m/s m1 = 10,000kg m2 = 10,000kg m1 + m2 = 20,000 kg

  5. Example 9.3 How Good Are the Bumpers? • Mass of the car is 1500 kg. The collision lasts 0.105s. • Find: p =I and the average force (Favg) exerted on the car.

  6. Example 9.4 Tennis Ball Hits the Wall • p =I ? Given to the ball. If m = 0.060 kg and v = 8.00 m/s p =I : change in momentum  wall. • Momentum ∕∕ to wall doesn’t change. • Impulsewill be  wall. • Take + direction toward wall, p =I = mv = m (vf – vi)  p =I = m(–vsin45–vsin45) p =I = –2mvsin45 = –2.1 N.s • Impulse on wall is in opposite direction: 2.1 N.s vf = –vsin45 vi = vsin45

  7. Example 9.5 Explosion as a Collision • Initial Momentum = Final Momentum (One-Dimension) m1v1+m2v2 = m1v’1+ m2v’2 • Initially: v = 0 Explodes! • Finally: mv = 0 = m2v’2 + m1v’1 Given: m1 , m2, v’2, you may compute v’1  v’1= – (m2/m1)v’2 m v = 0

  8. Example 9.6 Rifle Recoil • Momentum Before = Momentum After m1v1+m2v2 = m1v’1+ m2v’2 • Given:mB= 0.02 kg, mR= 5.00 kg, v’B = 620 m/s 0 = mBv’B + mRv’R v’R=–mBv’B/mR =– (0.02)(620)/5.00 m/s = – 2.48 m/s (to the left, of course!)

  9. Example 9.7 Ballistic Pendulum • Perfectly inelastic collision – the bullet is embedded in the block of wood • Momentum equation will have two unknowns • Use conservation of energy from the pendulum to find the velocity just after the collision • Then you can find the speed of the bullet

  10. Example 9.7 Ballistic Pendulum, final • Before: • Momentum Conservation: • After: • Conservation of energy: Solving for vB: Replacing vB into 1st equation and solving for v1A:

  11. Example 9.8 Collision at an Intersection • Mass of the car mc= 1500kg Mass of the van mv = 2500kg • Find vf if this is a perfectly inelastic collision (they stick together). • Before collision: • The car’smomentum is: Σpxi= mcvc Σpxi= (1500)(25) = 3.75x104 kg·m/s • The van’smomentum is: Σpyi= mvvv Σpyi= (2500)(20) = 5.00x104 kg·m/s • After collision: both have the same x- and y-components: Σpxf = (mc + mv )vf cos Σpyf = (mc + mv )vf sin

  12. Example 9.8 Collision at an Intersection, final • Because the total momentum is both directions is conserved: Σpxf = Σpxi 3.75x104 kg·m/s = (mc + mv )vf cos = 4000vf cos (1) Σpyf = Σpyi 5.00x104 kg·m/s = (mc + mv )vf sin = 4000vf sin (2) • Dividing Eqn (2) by (1) 5.00/3.75 =1.33 = tan   = 53.1° • Substituting in Eqn (2) or (1)  5.00x104 kg·m/s = 4000vf sin53.1°  vf =5.00x104/(4000sin53.1° )  vf =15.6m/s

  13. Example 9.9 Center of Mass (Simple Case) • Both masses are on the x-axis • The center of mass (CM) is on the x-axis • One dimension xCM = (m1x1 + m2x2)/M M = m1+m2 xCM≡(m1x1 + m2x2)/(m1+m2) • The center of mass is closer to the particle with the larger mass • If: x1 = 0, x2 = d & m2 = 2m1 xCM≡(0 + 2m1d)/(m1+2m1)  xCM≡2m1d/3m1  xCM = 2d/3

  14. Example 9.10 Three Guys on a Raft A group of extended bodies, each with a known CM and equivalent mass m. Find the CM of the group. xCM = (Σmixi)/Σmi xCM = (mx1+ mx2+ mx3)/(m+m+m)  xCM = m(x1+ x2+ x3)/3m = (x1+ x2+ x3)/3  xCM = (1.00m + 5.00m + 6.00m)/3 = 4.00m

  15. Example 9.11 Center of Mass of a Rod • Find the CM position of a rod of mass M and length L. The location is on the x-axis (A). Assuming the road has a uniform mass per unit length λ = M/L (Linear mass density) • From Eqn 9.32 • But:λ = M/L

  16. Example 9.11 Center of Mass of a Rod, final. (B). Assuming now that the linear mass density of the road is no uniform: λ = x • The CM will be: • But mass of the rod and  are related by: The CM will be:

  17. Material for the Final Exam • Examples to Read!!! • Example 9.2 (page 239) • Example 9.5 (page 247) • Example 9.10 (page 256) • Material from the book to Study!!! • Objective Questions: 7-8-13 • Conceptual Questions: 3-5-6 • Problems: 1-9-11-15-25-26-27-37-40-65

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