Linear Momentum = mass in motion A measure of how hard it is to stop an object. It is like a quantity of motion. How is it different from inertia?
Momentum (p) depends on: mass & velocity of object. p = mv m in kg v in m/s Units are … kg mno name. s
Momentum is a Vector Quantity Same direction as velocity All Energy KE too is a scalar
Ex 1. A 2250 kg pickup truck has v = 25 m/s east. What is the truck’s momentum? p = mv= (2250 kg)(25 m/s) = 5.6 x 10 4 kg m s
Change in momentum - accl occurs any time an object changes velocity (speed or direction).
Momentum Change & Newton’s 2 nd Law F = ma F = m( v/ t) F t =m vm (v f - v i ) for const mass. F t = p Impulse. p = Change in momentum
Equations of Momentum Change J =F t = p Impulse = change momentum. p f – p i. p = mv f – mv i for velocity change with constant mass can factor out mass you can write, m (v f - v i ) or m v.
Force is required to change velocity or momentum of a body in motion. Force must be in contact for some time.
Increased force & contact time on object give greatest impulse p = m v.
Hit a homerun needs large impulse. The more contact time, the less force needed to give same impulse p.
Impulse (J) is the momentum change. It has the same units. kg mor Ns s It is like force but includes a contact time component!
Ex 2. How long does it take an upward 100N force acting on a 50 kg rocket to increase its speed from 100 to 150 m/s?
F = 100 N v = 50 m/s m = 50 kg Ft = m v t = m v F 50 kg(50 m/s) 100 kg m/s 2 = 25 s.
Concept: A pitcher throws a fastball to a catcher. Who exerts a larger force on the ball? Explain.
Concept: Explain, in terms of impulse and momentum, how airbags help avoid injury in a car crash.
Examples of Impulse/ Change in Momentum Baseball batter swinging through ball. Applying brakes of car over time to stop.
Ex 3. How long does it take a 250 N force to increase to speed of a 100 kg rocket from 10 m/s to 200 m/s?
Ft = m vt = m v F F = 250 N m= 100 kg v =190 m/s t = 100kg(190m/s) 250 kg m/s 2. = 76 s.
Ex 4. The speed of a 1200 kg car increases from 5 to 29 m/s in 12 s. What force accelerated the car?
Ex 5: A 0.4 kg ball is thrown against a wall with a velocity of 15 m/s. If it rebounds with a velocity of 12 m/s: a) what was its v? b) What was its p?
v = v f – v i. -12 m/s – (15 m/s) = - 27 m/s. p = m v = 0.4kg(27m/s) =10.8 kg m/s
Running with momentum. 15 min. https://www.youtube.com/watch?v=jLIyDf kQcskhttps://www.youtube.com/watch?v=jLIyDf kQcsk Relaxing with impulse.13 minutes. https://www.youtube.com/watch?v=0nOHL Thv2mwhttps://www.youtube.com/watch?v=0nOHL Thv2mw
Understanding Car Crashes 22 min start 8:53 Hewitt Momentum 4:20 https://www.youtube.com/watch?v=2Fwhj UuzUDghttps://www.youtube.com/watch?v=2Fwhj UuzUDg http://www.youtube.com/watch?v=yUpiV2I_IRI
Hwk read text 208 – 211 do pg 214 #1- 4 concepts do p 211 #1 - 4. Impulse prbs. Also worksheet “Impulse Momentum”
Which are units of Impulse? Nm N/s Ns N/m A ball mass 0.10 kg is dropped from 12-m. Its momentum just as it strikes the ground is: 1.5 kgm/s 1.8 kgm/s 2.4 kgm/s 4.8 kgm/s
A 0.060-kg tennis ball, initially moving at 12 m/s, is struck by a racquet causing it to move in the opposite direction at a speed of 18 m/s. What is the impulse exerted by the racquet on the ball? 0.36 kgm/s 0.72 kgm/s 1.1 kgm/s 1.8 kgm/s
Recoil illustrates conservation of momentum where initial and final momentum = 0. 0 = p 1 + p 2.
1. The cannon is 100kg and the cannonball is 5 kg. If the ball leaves the cannon with a speed of 100 m/s, find the recoil velocity of the cannon.
Before FiringAfter Firing m 1 v 1 + m 2 v 2 = m 1f v 1f + m 2f v 2f 0 = (100kg)v cf + (5kg)(100m/s) -500 kgm/s = (100 kg) v cf - 5 m/s = v cf recoil velocity of cannon
Extra Example – not on sheet A 63-kg astronaut is in spacewalk when the tether breaks. The astronaut throws a 10-kg oxygen tank directly away from the spaceship at 12 m/s. Assuming the astronaut was initially at rest, what is his final speed after throwing the tank? 1.9 m/s
Hwk. Read text p 215-218. Do pg 221 #2, and pg 233 #17, 19, 20, 24, 25.
Let’s say a 4 kg fish swimming at 5 m/s, eats a 1 kg fish. What is their final velocity? Stick em together problems
Bg fishsm.fish Bg fish sm.fish m 1 v 1 + m 2 v 2 = m 1f v 1f + m 2f v 2f (4kg)(5m/s)+(1 kg)0 =(4kg)v 1 +(1kg)v 2 But the final velocities are equal so factor out the v f : 20 kg m/s = v f (4+1kg) v f = (20 kg m/s) / (5kg) = 4m/s
Fish lunch Hewitt 4:00 https://www.youtube.com/watch?v= MK0B5hEU7OI
Find the final velocity of the cart and brick together 2. A 2 kg brick is dropped on a 3 kg cart moving at 5.0 m/s.
cartbrickcartbrick m 1 v 1 + m 2 v 2 = m 1f v 1f + m 2f v 2f (3kg)(5.0m/s) + 0 = (3kg)v 1 + (2kg)v 2 150 kg m/s = v (3kg + 2 kg) (150 kg m/s )/5 kg = 3.0 m/s
Elastic & Inelastic Collisions Totally Elastic: no KE lost at all (to heat, light, sound etc.) Usu. Involves objects that don’t make contact or bounce off. Totally Inelastic: involves greatest loss of KE. Usu damage done. Most extreme case – objects stick together.
Do Now: On July 4 th my family likes to shoot off fireworks. One rocket was shot straight up, climbed to a height 18-m and exploded into hundreds of pieces in all directions at its highest point. Thinking about conservation laws, think about the rocket at its highest point just before & just after it explodes: How does the rocket’s momentum compare before & after the explosion? How does its KE compare compare before & after the explosion?
Inelastic Collisions Stick em together KE “lost” converted Elastic Collisions – no KE lost. Bounce off each other. Pg 233 #17, 19, 20, recoil prbs24, 25. Sentences, equations, show work w/units.