MOMENTUM & COLLISIONS

MOMENTUM & COLLISIONS

Linear Momentum • Moving objects have momentum • Vector quantity • Points in the same direction as the velocity vector • Momentum: • Equals the product of an objects mass and velocity • Proportional to mass and velocity p = mv p = momentum (kg * m/s) m = mass (kg) v = velocity (m/s)

TAXI PROBLEM What is the taxi cab’s momentum? * Mass of the taxi = 53 kg * Velocity of the taxi = 1.2 m/s Answer: p = mv p = (53 kg)(1.2 m/s) p = 63.6 kg * m/s to the left p = 63.6 kg * m/s v = 1.2 m/s

Momentum & Newton’s 2nd Law • Newton’s 2nd Law • ΣF = ma = m(Δv/Δt) • ΣF = m(Δv/Δt) • Momentum • p = mv  m = p/v • ΣF = (p/v)(Δv/Δt)  ΣF = Δp/Δt

Impulse: change In momentum • If the momentum of an object changes, either mass, velocity, or both change • If mass remains the same  than velocity changes  acceleration occurs • What produces an acceleration? • FORCE • Greater the force acting on the object  greater its change in velocity  greater its change in momentum

Impulse: change in momentum • How long the force acts is also important… • Stalled car • Apply a force over a brief amount of time  produce a change in momentum • Apply the same force over an extended period of time  produce a greater change in the car’s momentum • A force suspended for a long time produces more change in momentum than does the same force applied briefly • Both force and time are important in changing momentum

Impulse: change in momentum • IMPULSE (J) = Δp = pf – pi = mvf – mvi • J = Favg Δt • FavgΔt = mΔv • Favg = mΔv/Δt • Impulse (J) = Change in momentum • Impulse is also the product of the average force and the time during which the force is applied. • Vector quantity • Units: kg * m/s

Impulse problem • A long jumper's speed just before landing is 7.8 m/s. What is the impulse of her landing? (mass = 68 kg) • J = Δp • J = pf - pi • J = mvf – mvi • J = 0 - (68kg)(7.8m/s) • J = -530 kg * m/s • *Negative sign indicates that the direction of the impulse is opposite to her direction of motion

Impulsive force • Baseball player swings a bat and hits the ball, the duration of the collision can be as short as 1/1000th of a second and the force averages in the thousands of newtons • The brief but large force the bat exerts on the ball = Impulsive force

Kinetic Books • View Kinetic books section 8.4- Physics at play: Hitting a baseball BASEBALL PROBLEM The ball arrives at 40 m/s and leaves at 49 m/s in the opposite direction. The contact time is 5.0×10−4 s. What is the average force on the ball? J = Δp = Favg Δt = mΔv Favg Δt = mΔv FavgΔt = mΔv Favg = mΔv/Δt Favg = (0.14kg)(49 – (-40)m/s)/5.0×10−4 s Favg = 2.5×104 N

Impulse: change in momentum • Case 1: Increasing momentum • To increase the momentum of an object  apply the greatest force possible for as long as possible • Golfer teeing off and a baseball player trying for a home run • Swing as hard as possible (large force) • Follow through with their swing (increase in time)

Impulse: change in momentum • Case 2: Decreasing momentum • You are in a car that is out of control  Do you want to hit a cement wall or haystack? • In either case, your momentum is decreased by the same impulse… • But, the same impulse does not mean the same amount of force or the same amount of time  rather it means the same PRODUCT of force and time

Impulse: change momentum • Case 2 continued: Decreasing momentum • Hit the haystack  Extend the impact time • Change in momentum occurs over a long time  Small impact force • mv = Ft • Hit the cement wall • Change in momentum occurs over a short time  Large impact force • mv = Ft

Changing Momentum: Scenario 1 If you want to decrease a large momentum, you can have the force applied for a longer time ** If the change in momentum occurs over a long time  Force of impact is small • Examples: • Air bags in cars. • Crash test video FDt

Changing Momentum: Scenario 2 If the change in momentum occurs over a short time, the force of impact is large. • Karate link • Boxing video FDt

Impulse: change in momentum • QUESTION • When a glass falls, will the impulse be less if it lands on a carpet than if it lands on a hard floor? • NO Impulse is the same for either surface because the change in momentum is the same • Carpet: More time is available for the change in momentum  smaller force for the impulse • Hard floor: Less time is available for the change in momentum (due to less “give”)  larger force for the impulse

Conservation of momentum • Conservation of momentum: • Occurs when there are no net external force(s) acting on the system • Result  Total momentum of an isolated system is constant • Momentum before = Momentum after • Playing pool example: • Kinetic books 8.6

Conservation of momentum • Momentum • p = mv • Conservation of momentum • Momentum before = Momentum after • pi1 + pi2 +…+ pin = pf1 + pf2 +…+ pfn • pi1, pi2, …, pin = initial momenta • pf1, pf2, …, pfn = final momenta • m1vi1 + m2vi2 = m1vf1 + m2vf2 • m1, m2 = masses of objects • vi1, vi2 = initial velocities • vf1, vf2 = final velocities

Conservation of momentum A 55.0 kg astronaut is stationary in the spaceship’s reference frame. She wants to move at 0.500 m/s to the left. She is holding a 4.00 kg bag of dehydrated astronaut chow. At what velocity must she throw the bag to achieve her desired velocity? (Assume the positive direction is to the right.)

solution • VARIABLES: • Mass of astronaut ma = 55 kg • Mass of bag mb = 4 kg • Initial velocity of astronaut via = 0 m/s • Initial velocity of bag vib =0 m/s • Final velocity of astronaut vfa = -0.5 m/s • Final velocity of bag vfb = ? • EQUATION: • m1vi1 + m2vi2 = m1vf1 + m2vf2 • mavia + mbvib = mavfa + mbvfb • 0 = mavfa + mbvfb • Vfb = - (mavfa / mb) • Vfb = - ((55kg)(-0.5m/s))/(4kg) = 6.875 m/s

collisions • Collision of objects  Demonstrates the conservation of momentum • Whenever objects collide in the absence of external forces net momentumbefore collision = net momentumaftercollision

collisions • Momentum is conserved in ALL TYPES of collisions • Elastic Collisions • Objects collide without being permanently deformed and without generating heat • Inelastic Collisions • Colliding objects become distorted (tangled or coupled together) and generate heat

collisions • Problem • Consider a 6-kg fish that swims toward and swallows a 2-kg fish that is at rest. If the larger fish swims at 1 m/s, what is its velocity immediately after lunch? net momentumbefore collision = net momentumaftercollision (net mv)before = (net mv)after (6kg)(1m/s) +(2kg)(0) = (6kg + 2kg)(vafter) vafter = ¾ m/s

collisions • Problem • Consider a 6-kg fish that swims toward and swallows a 2-kg fish that is moving towards the larger fish at 2 m/s. If the larger fish swims at 1 m/s, what is its velocity immediately after lunch? net momentumbefore collision = net momentumaftercollision (net mv)before = (net mv)after (6kg)(1m/s) +(2kg)(-2m/s) = (6kg + 2kg)(vafter) vafter = 1/4 m/s

collisions • Perfectly Elastic collisions • Not common in the everyday world • Some heat is generated during collisions • Drop a ball and after it bounces from the floor, both the ball and the floor are a bit warmer • At the microscopic level  perfectly elastic collisions are common • Electrically charged particles bounce off one another without generating heat

Examples of Perfectly ELASTIC Collisions • Electron scattering • Hard spheres (Pool balls)

collisions • Elastic collision • Kinetic energy is conserved • KE before = KE after • KE = 1/2mv2 • Momentum is conserved in any collision  Elastic or inelastic

ELASTIC collisions in 1-dimension • Conservation of Kinetic Energy: • Conservation of Momentum: • Rearrange both equations and divide:

Elastic collisions • Final velocities in Head-On Two-Body Elastic Collisions (v2i = 0 m/s)

Examples of Perfectly INELASTIC Collisions • Catching a baseball: Video • Football tackle • Cars colliding and sticking • Bat eating an insect

collisions • Inelastic collision • Kinetic energy is NOT conserved • KE before ≠ KE after • Momentum is conserved in any collision  Elastic or inelastic

Perfectly INELASTIC collisionsin 1-dimension • Final velocities are the same

Problem A 5879-lb (2665 kg) Cadillac Escalade going 35 mph smashes into a 2342-lb (1061 kg) Honda Civic also moving at 35 mph (15.64 m/s) in the opposite direction. The cars collide and stick. a) What is the final velocity of the two vehicles? • m1v1i + m2v2i = (m1 +m2)vf • (2665kg)(15.64m/s) + (1061kg)(-15.64m/s) = (2665 + 1061kg)vf • vf = 6.73 m/s = 15.1 mph

Collisions • Momentum is always conserved in a collision • Collision video • Classification of collisions: • ELASTIC • Both energy & momentum are conserved • INELASTIC • Momentum conserved, not energy • Perfectly inelastic -> objects stick • Lost energy goes to heat

Center of mass • “Average” location of mass • An object can be treated as though all its mass were located at this point • For a symmetric object made from a uniformly distributed material, the center of mass is the same as its geometric center

Center of mass • Equation: xcm = m1x1 + m2x2 + …mnxn / m1 + m2 + …mn • xCM = x position of center of mass • mi = mass of object i • xi = x position of object i

Center of mass • View section 8.20 in Kinetic books • * Specifically example 1- Center of mass problem

Center of Mass • Video • Balancing Activity: video demo

Don’t use following slides???

Conservation of momentum • Key Facts: • Newton’s 2nd Law (F = ma) • To accelerate an object  Net force must be applied • To change the momentum of an object  exert an impulse on it • The momentum of a system cannot change unless it is acted on by external forces

Conservation of momentum • Law of Conservation of Momentum • In the absence of an external force, the momentum of a system remains unchanged • Examples in which the net momentum is the same before and after the event: • Radioactive decay • Cars colliding • Stars exploding

Conservation of Momentum mv(initial) = mv(final) An astronaut of mass 80 kg pushes away from a space station by throwing a 0.75-kg wrench which moves with a velocity of 24 m/s relative to the original frame of the astronaut. What is the astronaut’s recoil speed? (0.75kg)(24m/s) = 80kg(v) v = 0.225 m/s

Conservation of momentum • Question • Newton’s 2nd law states that if no net force is exerted on a system, no acceleration occurs. Does it follow that no change in momentum occurs? • Yes, because no acceleration (a = Δv/t)  means no change in velocity  and no change in momentum (p = mΔv) • Also, no net force means  no net impulse (J = Ft)  J = Δp  no change in momentum

Conservation of momentum • Question • Newton’s 3rd law states that the force a rifle exerts on a bullet is equal and opposite to the force the bullet exerts on the rifle. Does is follow that the impulse the rifle exerts on the bullet is equal and opposite to the impulse the bullet exerts on the rifle? • Yes, because the rifle acts on the bullet and bullet reacts on the rifle during the same time interval • Since time is equal and force is equal and opposite for both  Impulse, Ft, is also equal and opposite for both (Impulse – vector quantity and can be canceled)

Conservation of momentum • The law of conservation of momentum can be derived from Newton’s 2nd and 3rd laws • Newton’s 2nd law  F = ma • Newton’s 3rd law  Forces are equal but opposite * Refer to Kinetic Books- 8.7 For step-by-step derivation

collisions • Collisions: • Momentum- Useful concept when applied to collisions • In a collision, two or more objects exert forces on each other for a brief instant of time, and these forces are significantly greater than any other forces they may experience during the collision

Problem A proton (mp=1.67x10-27 kg) elastically collides with a target proton which then moves straight forward. If the initial velocity of the projectile proton is 3.0x106 m/s, and the target proton bounces forward, what are a) The final velocity of the projectile proton? b) The final velocity of the target proton? 0.0 m/s 3.0 x 106 m/s

Elastic collision in 1-dimension • Final equations for head-on elastic collision: • Relative velocity changes sign • Equivalent to Conservation of Energy

Problem An proton (mp=1.67x10-27 kg) elastically collides with a target deuteron (mD=2mp) which then moves straight forward. If the initial velocity of the projectile proton is 3.0x106 m/s, and the target deuteron bounces forward, what are a) The final velocity of the projectile proton? b) The final velocity of the target deuteron? vp = -1.0 x 106 m/s vd = 2.0 x 106 m/s Head-on collisions with heavier objects always lead to reflections

MOMENTUM & COLLISIONS