Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 9B - Conservation of Momentum A PowerPoint Presentation by Paul E. Tippens, Emeritus Professor Southern Polytechnic State University A PowerPoint.

Similar presentations


Presentation on theme: "Chapter 9B - Conservation of Momentum A PowerPoint Presentation by Paul E. Tippens, Emeritus Professor Southern Polytechnic State University A PowerPoint."— Presentation transcript:

1

2 Chapter 9B - Conservation of Momentum A PowerPoint Presentation by Paul E. Tippens, Emeritus Professor Southern Polytechnic State University A PowerPoint Presentation by Paul E. Tippens, Emeritus Professor Southern Polytechnic State University © 2007

3 Momentum is conserved in this rocket launch. The velocity of the rocket and its payload is determined by the mass and velocity of the expelled gases. Photo: NASA NASA

4 Objectives: After completing this module, you should be able to: State the law of conservation of momentum and apply it to the solution of problems.State the law of conservation of momentum and apply it to the solution of problems. Distinguish by definition and example between elastic and inelastic collisions.Distinguish by definition and example between elastic and inelastic collisions. Predict the velocities of two colliding bodies when given the coefficients of restitution, masses, and initial velocities.Predict the velocities of two colliding bodies when given the coefficients of restitution, masses, and initial velocities.

5 A Collision of Two Masses When two masses m 1 and m 2 collide, we will use the symbol u to describe velocities before collision. The symbol v will describe velocities after collision. Before m1m1 u1u1 m2m2 u2u2 m1m1 v1v1 m2m2 v2v2After

6 A Collision of Two Blocks m1m1 Bm2m2 “u” = Before“ v ” = After m1m1 u1u1 m2m2 u2u2 Before m2m2 v2v2 m1m1 v1v1After Collision

7 Conservation of Energy m1m1 m2m2 u1u1 u2u2 before after lost The kinetic energy before colliding is equal to the kinetic energy after colliding plus the energy lost in the collision.

8 Example 1. A 2-kg mass moving at 4 m/s collides with a 1-kg mass initially at rest. After the collision, the 2-kg mass moves at 1 m/s and the 1-kg mass moves at 3 m/s. What energy was lost in the collision? It’s important to draw and label a sketch with appropriate symbols and given information. m2m2 u 2 = 0 m1m1 u 1 = 4 m/s m 1 = 2 kgm 1 = 1 kg BEFORE m2m2 v 2 = 2 m/s m1m1 v 1 = 1 m/s m 1 = 2 kgm 1 = 1 kg AFTER

9 Example 1 (Continued). What energy was lost in the collision? Energy is conserved. m2m2 u 2 = 0 m1m1 u 1 = 4 m/s m 1 = 2 kg m 1 = 1 kg m2m2 v 2 = 2 m/s m1m1 v 1 = 1 m/s m 1 = 2 kg m 1 = 1 kg BEFORE: AFTER Energy Conservation: K(Before) = K(After) + Loss Loss = 16 J – 3 J Energy Loss = 15 J

10 Impulse and Momentum A B uAuA uBuB A B vAvA vBvB B -FA t-FA t-FA t-FA t F B  t Opposite but Equal F  t F  t = mv f – mv o F B  t = -F A  t Impulse =  p m B v B - m B u B = -(m A v A - m A u A ) m A v A + m B v B = m A u A + m B u B Simplifying:

11 Conservation of Momentum A B uAuA uBuB A B vAvA vBvB B -FAt-FAt-FAt-FAt F B  t The total momentum AFTER a collision is equal to the total momentum BEFORE. Recall that the total energy is also conserved: K A0 + K B0 = K Af + K Bf + Loss Kinetic Energy: K = ½mv 2 m A v A + m B v B = m A u A + m B u B

12 Example 2: A 2-kg block A and a 1-kg block B are pushed together against a spring and tied with a cord. When the cord breaks, the 1-kg block moves to the right at 8 m/s. What is the velocity of the 2 kg block? A B The initial velocities are zero, so that the total momentum before release is zero. m A v A + m B v B = m A u A + m B u B 00 m A v A = - m B v B v A = - mBvBmAmBvBmA

13 Example 2 (Continued) m A v A + m B v B = m A u A + m B u B 0 0 m A v A = - m B v B v A = - mBvBmAmBvBmA A B 2 kg 1 kg A B 8 m/s v A2 v A = - (1 kg)(8 m/s) (2 kg) v A = - 4 m/s

14 Example 2 (Cont.): Ignoring friction, how much energy was released by the spring? A B 2 kg 1 kg A B 8 m/s 4 m/s ½kx 2 = ½ m A v A + ½m B v B 2 Cons. of E: ½kx 2 = ½ m A v A + ½m B v B 2 2 ½kx 2 = ½(2 kg)(4 m/s) 2 + ½(1 kg)(8 m/s) 2 ½kx 2 = 16 J + 32 J = 48 J ½kx 2 = 48 J

15 Elastic or Inelastic? An elastic collision loses no energy. The deform- ation on collision is fully restored. In an inelastic collision, energy is lost and the deformation may be permanent. (Click it.)

16 Completely Inelastic Collisions Collisions where two objects stick together and have a common velocity after impact. BeforeAfter

17 Example 3: A 60-kg football player stands on a frictionless lake of ice. He catches a 2-kg football and then moves at 40 cm/s. What was the initial velocity of the football? Given: u B = 0; m A = 2 kg; m B = 60 kg; v A = v B = v C v C = 0.4 m/s A B m A v A + m B v B = m A u A + m B u B Momentum: 0 (m A + m B )v C = m A u A (2 kg + 60 kg)(0.4 m/s) = (2 kg) u A Inelastic collision: u A = 12.4 m/s

18 Example 3 (Cont.): How much energy was lost in catching the football? 0 ½(2 kg)(12.4 m/s) 2 = ½(62 kg)(0.4 m/s) 2 + Loss 154 J = 4.96 J + Loss Loss = 149 J 97% of the energy is lost in the collision!!

19 General: Completely Inelastic Collisions where two objects stick together and have a common velocity v C after impact. Conservation of Momentum: Conservation of Energy:

20 Example 4. An 87-kg skater B collides with a 22-kg skater A initially at rest on ice. They move together after the collision at 2.4 m/s. Find the velocity of the skater B before the collision. A B u B = ? u A = 0 Common speed after colliding: 2.4 m/s. 22 kg 87 kg v B = v A = v C = 2.4 m/s (87 kg) u B = (87 kg + 22 kg)(2.4 m/s) (87 kg)u B =262 kg m/s u B = 3.01 m/s

21 Example 5: A 50 g bullet strikes a 1-kg block, passes all the way through, then lodges into the 2 kg block. Afterward, the 1 kg block moves at 1 m/s and the 2 kg block moves at 2 m/s. What was the entrance velocity of the bullet? 2 kg1 kg 1 m/s 2 m/s 1 kg 2 kg u A = ?

22 2 kg1 kg 1 m/s 2 m/s 1 kg 2 kg Find entrance velocity of bullet: m A = 0.05 kg; u A = ? u A 2.05 kg (0.05 kg) u A =(1 kg)(1 m/s)+(2.05 kg)(2 m/s) m A u A + m B u B + m C u C = m B v B + (m A +m C ) v AC Momentum After = Momentum Before 50 g A C B 00 u A (0.05 kg) u A =(5.1 kg m/s) u A = 102 m/s

23 Completely Elastic Collisions Collisions where two objects collide in such a way that zero energy is lost in the process. APPROXIMATIONS!

24 Velocity in Elastic Collisions ABAB uBuB uAuA vAvA vBvB 1. Zero energy lost. 2. Masses do not change. 3. Momentum conserved. (Relative  v After) = - (Relative  v Before) Equal but opposite impulses (F  t) means that: For elastic collisions: v A - v B = - (u A - u B )

25 Example 6: A 2-kg ball moving to the right at 1 m/s strikes a 4-kg ball moving left at 3 m/s. What are the velocities after impact, assuming complete elasticity? AB AB 3 m/s 1 m/s vAvAvAvA vBvBvBvB 1 kg 2 kg v A - v B = - (u A - u B ) v A - v B = u B - u A v A - v B v A - v B = (-3 m/s) - (1 m/s) From conservation of energy (relative v): v A - v B = - 4 m/s

26 Example 6 (Continued) AB AB 3 m/s 1 m/s vAvAvAvA vBvBvBvB 1 kg 2 kg m A v A + m B v B = m A u A + m B u B Energy: v A - v B = - 4 m/s v A v B (1 kg) v A +(2 kg) v B =(1 kg)(1 m/s)+(2 kg)(-3 m/s) v A + 2v B v A + 2v B = -5 m/s Momentum also conserved: v A - v B = - 4 m/s Two independent equations to solve:

27 Example 6 (Continued) AB AB 3 m/s 1 m/s vAvAvAvA vBvBvBvB 1 kg 2 kg v A + 2 v B = -5 m/s v A - v B = - 4 m/s Subtract: v B2 = - 1 m/s 0 + 3v B2 = - 1 m/s v B = - 0.333 m/s Substitution: v A - v B = - 4 m/s v A2 - (-0.333 m/s) = - 4 m/s v A = -3.67 m/s

28 Example 7. A 0.150 kg bullet is fired at 715 m/s into a 2- kg wooden block at rest. The velocity of block afterward is 40 m/s. The bullet passes through the block and emerges with what velocity? BA u B = 0 (0.150 kg) v A + (2 kg)(40 m/s) = (0.150 kg)(715 m/s) 0.150 v A + (80 m/s) = (107 m/s) 0.150 v A = 27.2 m/s) v A = 181 m/s

29 Example 8a: Inelastic collision: Find v C. AB 5 kg 7.5 kg u B =0 2 m/s AB Common v C after vCvCvCvC After hit: v B = v A = v C (5 kg)(2 m/s) = (5 kg + 7.5 kg) v C 12.5 v C =10 m/s v C = 0.800 m/s In an completely inelastic collision, the two balls stick together and move as one after colliding.

30 Example 8. (b) Elastic collision: Find v A2 and v B2 AB 5 kg 7.5 kg v B1 =0 2 m/s Conservation of Momentum: (5 kg)(2 m/s) = (5 kg)v A2 + (7.5 kg) v B AB vAvAvAvA vBvBvBvB 5 v A + 7.5 v B = 10 m/s For Elastic Collisions: Continued...

31 Example 8b (Cont). Elastic collision: Find v A & v B A B 5 kg 7.5 kg v B =0 2 m/s A B vAvAvAvA vBvBvBvB Solve simultaneously: 5 v A + 7.5 v B = 10 m/s -5 v A + 5 v B = +10 m/s x (-5) 12.5 v B = 20 m/s v A - 1.60 m/s = -2 m/s v A = -0.400 m/s v B = 1.60 m/s

32 General: Completely Elastic Collisions where zero energy is lost during a collision (an ideal case). Conservation of Momentum: Conservation of Energy:

33 Example 9: A 50 g bullet lodges into a 2-kg block of clay hung by a string. The bullet and clay rise together to a height of 12 cm. What was the velocity of the 50-g mass just before entering? uAuAuAuA B A B A 12 cm The ballistic pendulum!

34 Example (Continued): B A 12 cm 50 g uAuA 2.05 kg 2 kg Collision and Momentum: m A u A +0= (m A +m B ) v C (0.05 kg) u A = (2.05 kg) v C To find v A we need v C. energy After collision, energy is conserved for masses. v C = 2gh

35 Example (Continued): B A 12 cm 50 g uAuA 2.05 kg 2 kg m A u A +0= (m A +m B )v C (0.05 kg) u A = (2.05 kg)(1.53 m/s) v C = 2gh = 2(9.8)(0.12) After Collision: v C = 1.53 m/s u A = 62.9 m/s Momentum Also Conserved:

36 Summary of Formulas: Conservation of Momentum: Conservation of Energy: For elastic only:

37 CONCLUSION: Chapter 9B Conservation of Momentum


Download ppt "Chapter 9B - Conservation of Momentum A PowerPoint Presentation by Paul E. Tippens, Emeritus Professor Southern Polytechnic State University A PowerPoint."

Similar presentations


Ads by Google