Momentum

1. Momentum Linear Momentum Impulse Collisions One and Two Dimensional Collisions

2. Linear Momentum Momentum is a measure of how hard it is to stop or turn a moving object. p = mv (single particle) P = Σpi (system of particles)

3. Conservation of Linear Momentum The linear momentum of a system is conserved – the total momentum of the system remains constant Ptot = Σp = p1 + p2 + p3 = constant p1i + p2i = p1f + p2f Σpix = Σpfx, Σpiy = Σpfy,Σpiz = Σpfz “Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant.”

4. mp= 1.672 x 10-27 kg me= 9.109 x 10-31 kg ve= 5.507 x 105m/s • Practice Problem 1: • How fast must an electron move to have the same momentum as a proton moving at 300 m/s?

5. ptot= 750 kgm/s @ 53.1° N of E • Practice Problem 2: • A 90-kg tackle runs north at 5.0 m/s and a 75-kg quarterback runs east at 8.0 m/s. What is the momentum of the system of football players?

6. Impulse • An impulse is described as a change in momentum: • I = Dp • Impulse is also the integral of force over a period of time: • I = Fdtor I = FDt • The change in momentum of a particle is equal to the impulse acting on it

7. F=ma!!! F = ma = m (Δv/Δt) m(Δv) = FΔt Δp = FΔt Practice Problem 3: Show that Impulse actually comes from Newton’s 2nd Law

8. Impulsive forces are generally of high magnitude and short duration. Force time

9. ptot= 2.12 kgm/s @ 62.8° above -x Practice Problem 4: • A 150-g baseball moving at 40 m/s 15o below the horizontal is struck by a bat. It leaves the bat at 55 m/s 35o above the horizontal. What is the impulse exerted by the bat on the ball? • If the collision took 2.3 ms, what was the average force?

10. The log will move at .53 m/s in the opposite direction of the lumberjack Practice Problem 5: An 85-kg lumberjack stands at one end of a floating 400-kg log that is at rest relative to the shore of a lake. If the lumberjack jogs to the other end of the log at 2.5 m/srelative to the shore, what happens to the log while he is moving?

11. a) v2f= -6 m/s b) Us = 4.5 J • Practice Problem 6: • Two blocks of mass 0.5 kg and 1.5 kg are placed on a horizontal, frictionless surface. A light spring is compressed between them. A cord initially holding the blocks together is burned; after this, the block of mass 1.5 kg moves to the right with a speed of 2.0 m/s. • What is the speed and direction of the other block? • What was the original elastic energy in the spring?

12. Vcf = -80 m/s Practice Problem 7:(1-D conservation of momentum) What is the recoil velocity of a 120-kg cannon that fires a 30-kg cannonball at 320 m/s?

13. Collisions Elastic Inelastic Perfectly Inelastic

14. Collisions: • In all collisions, momentum is conserved. • Elastic Collisions: No deformation occurs • Kinetic energy is also conserved • Inelastic Collisions: Deformation occurs • Kinetic energy is lost • Perfectly Inelastic Collisions • Objects stick together, kinetic energy is lost • Explosions • Reverse of perfectly inelastic collision, kinetic energy is gained

15. Elastic Collisions • Objects collide and return to their original shape • Kinetic energy remains the same after the collision • Perfectly elastic collisions satisfy both conservation laws shown below

16. Inelastic Collisions • Momentum is Conserved: • m1iv1i +m2iv2i = m1fv1f +m2fv2f • Kinetic energy is less after the collision • It is converted into other forms of energy • Internal energy - the temperature is increased • Sound energy - the air is forced to vibrate • Some kinetic energy may remain after the collision, or it may all be lost

17. Perfectly Inelastic Collisions • Two objects collide and stick together • Two football players • A meteorite striking the earth • Momentum is conserved • Masses combine

18. vf = ½ v1i This plot illustrates the tremendous complexity of the proton-Proton collisions that occur at the Large Hadron Collider. • Practice Problem 8: (1-D elastic collision) • A proton, moving with a velocity of vi, collides elastically with another proton initially at rest. If the two protons have equal speeds after the collision, find • the speed of each in terms of vi and ... • the direction of the velocity of each

19. Vf = 1.125 m/s m1 = 1.5 kg V1i = 1.5 m/s m2 = .5 kg V2i = 0 m/s • Practice Problem 9:(1-D perfectly inelastic collision) • A 1.5 kg cart traveling at 1.5 m/s collides with a stationary 0.5 kg cart and sticks to it. At what speed are the carts moving after the collision?

20. V1f = .75 m/s V2f = 2.25 m/s m1 = 1.5 kg V1i = 1.5 m/s m2 = .5 kg V2i = 0 m/s • Practice Problem 10:(1-D elastic collision) • A 1.5 kg cart traveling at 1.5 m/s collides elastically with a stationary 0.5 kg cart. At what speed are each of the carts moving after the collision?

21. 2-D collisions • Use conservation of momentum independently for x and ydimensions • You must resolve your momentum vectors into x and y components when working the problem: • m1v1ix + m2v2ix = m1v1fx + m2v2fx • m1v1iy + m2v2iy = m1v1fy + m2v2fy

22. V1f = .58 m/s @ 17.25° below -x Practice Problem 11:(2-D collision) A pool player hits a cue ball in the x-direction at 0.80 m/s. The cue ball knocks into the 8-ball, which moves at a speed of 0.30 m/s at an angle of 35o angle above the x-axis. Determine the velocity and angle of deflection of the cue ball. (assume m1 = m2)