Momentum Linear Momentum Impulse Collisions Physics C Energy Momentum 11/18/2018 Linear Momentum Impulse Collisions One and Two Dimensional Collisions Bertrand

Physics C Energy 11/18/2018 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) Bertrand

Conservation of Linear Momentum Physics C Energy 11/18/2018 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.” Bertrand

Physics C Energy 11/18/2018 Practice Problem 1: How fast must an electron move to have the same momentum as a proton moving at 300 m/s? mp= 1.672 x 10-27 kg me= 9.109 x 10-31 kg ve= 5.507 x 105 m/s Bertrand

Physics C Energy 11/18/2018 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? ptot= 750 kgm/s @ 53.1° N of E Bertrand

Impulse An impulse is described as a change in momentum: I = Dp Physics C Energy 11/18/2018 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 =  Fdt or I = FDt The change in momentum of a particle is equal to the impulse acting on it Bertrand

F=ma!!! Practice Problem 3: Physics C Energy 11/18/2018 Practice Problem 3: Show that Impulse actually comes from Newton’s 2nd Law F=ma!!! F = ma = m (Δv/Δt) m(Δv) = FΔt Δp = FΔt Bertrand

Impulsive forces are generally of high magnitude and short duration. Physics C Energy 11/18/2018 Impulsive forces are generally of high magnitude and short duration. time Force Bertrand

ptot= 2.12 kgm/s @ 62.8° above -x Physics C Energy 11/18/2018 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? ptot= 2.12 kgm/s @ 62.8° above -x Bertrand

Physics C Energy 11/18/2018 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/s relative to the shore, what happens to the log while he is moving? The log will move at .53 m/s in the opposite direction of the lumberjack Bertrand

Physics C Energy 11/18/2018 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? a) v2f= -6 m/s b) Us = 4.5 J Bertrand

Practice Problem 7: (1-D conservation of momentum) Physics C Energy 11/18/2018 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? Vcf = -80 m/s Bertrand

Physics C Energy Elastic Inelastic Perfectly Inelastic 11/18/2018 Collisions Elastic Inelastic Perfectly Inelastic Bertrand

Collisions: In all collisions, momentum is conserved. Physics C Energy 11/18/2018 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 Bertrand

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 Billiard balls colliding are nearly perfectly elastic. Ideal gases undergo perfectly elastic collisions between molecules and between the walls of the container.

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 If the objects are still moving after the collision, then there is still some KE. If both objects have stopped, such as might occur in some head-on collisions, then all of the KE is converted into other forms of energy.

Perfectly Inelastic Collisions Two objects collide and stick together Two football players A meteorite striking the earth Momentum is conserved Masses combine Ask students to suggest additional examples of perfectly inelastic collisions.

Practice Problem 8: (1-D elastic collision) Physics C Energy 11/18/2018 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 vf = ½ v1i This plot illustrates the tremendous complexity of the proton-Proton collisions that occur at the Large Hadron Collider. Bertrand

Practice Problem 9: (1-D perfectly inelastic collision) Physics C Energy 11/18/2018 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? Vf = 1.125 m/s m1 = 1.5 kg V1i = 1.5 m/s m2 = .5 kg V2i = 0 m/s Bertrand

Practice Problem 10: (1-D elastic collision) Physics C Energy 11/18/2018 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? 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 Bertrand

Physics C Energy 11/18/2018 2-D collisions Use conservation of momentum independently for x and y dimensions You must resolve your momentum vectors into x and y components when working the problem: m1v1ix + m2v2ix = m1v1fx + m2v2fx m1v1iy + m2v2iy = m1v1fy + m2v2fy Bertrand

Practice Problem 11: (2-D collision) Physics C Energy 11/18/2018 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) V1f = .58 m/s @ 17.25° below -x Bertrand