Take out tables. On a separate sheet: Make a list of every equation we’ve already used in this class that has the velocity term in it.

Here is your choice: a. I toss a bullet at you. b. I shoot a bullet at you from a gun. Which is more dangerous to you? Why?

Linear Momentum & Impulse

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. kQcskhttps:// kQcsk Relaxing with impulse.13 minutes. Thv2mwhttps:// Thv2mw

Understanding Car Crashes 22 min start 8:53 Hewitt Momentum 4:20 UuzUDghttps:// UuzUDg

Hwk read text 208 – 211 do pg 214 #1- 4 concepts do p 211 # 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 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

Graphs

Constant force f - t graph:  p /Impulse is area under curve F  t. Force N

Non-Constant Force Force vs. time graph. The area under the curve = impulse or  p change in momentum. What is the impulse during the 9 seconds of contact? 225 Ns

Consv Momentum Demos.

Conservation of Momentum If no external force acts on a closed system, the total momentum remains unchanged even if objects interact.

What is a system? Two or more objects that interact in motion. One may transfer part or all of its momentum to the other(s). Common examples: collisions, explosions.

One Ball transfers all its momentum.

The astronaut transfers part of his momentum to the second astronaut.

Conservation of Momentum Calc’s Total momentum before = total after interactions. Collisions. Explosions Pushing apart.

 P before =  p after m 1 v 1 + m 2 v 2 = m 1f v 1f + m 2f v 2f v 1 and v 2 velocities for objects one and two. m 1 and m 2 masses of objects To Calculate:

Recoil From Explosions

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 Do pg 221 #2, and pg 233 #17, 19, 20, 24, 25.

Recoil Hewitt 6:25

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 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 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.

Which is totally elastic? Inelastic?

Inelastic Collision m c = 1000 kg m t = 3000 kg v c = 20 m/s v t =0 p c = p t =

m 1 v 1 + m 2 v 2 = m 1f v 1f + m 2f v 2f ( 1000kg)(20m/s) + 0 = (1000)v + (3000)v ( kg m/s) = (1000kg kg)v ( kg m/s) = (4000 kg)v ( kg m/s) = v 4000 kg v = 5 m/s

Elastic Collision m c = 1000 kg m t = 3000 kg v c = 20 m/s v t =0 Find final velocity of the car if truck has final velocity of 10 m/s.

m 1 v 1 + m 2 v 2 = m 1f v 1f + m 2f v 2f (1000kg)(20m/s) + 0 = (1000kg)v c +(3000kg)(10m/s) 20,000 kg m/s = (1000kg)v c kg m/s 20,000 kg m/s – 30,000 kg m/s = v c (1000kg) - 10 m/s = v c

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.

In class pg 221 #1 write out and hand in to be graded. and do pg 219 #1 – 4 calcs

Film Car Crashes or Running with momentum.