Physics    

Ch 6 Section 1 Momentum & Impulse

TSWBAT

•      Compare the momentum of different moving objects.

 

•      Compare the momentum of the same object moving with different velocities.

 

•      Identify examples of change in the momentum of an object.

 

•      Describe changes in momentum in terms of force and  time.

 

Momentum               

•   Momentum is the property of an object that relates its mass and velocity.

–  We know that the more massive something is the greater its impact.

•  A bowling ball dropped on your foot has a greater effect than a beach ball would!

 

Momentum       

•   Momentum is the property of an object that relates its mass and velocity.

–  We know the faster something is moving the greater its impact.

•  A bullet moving at 100 m/s does more damage than one thrown at 10m/s

 

Momentum Equation

D p = m D v 

Dp stands for change in momentum

Dv is change in velocity

Yes, direction matters!

M is mass

 

 

Impulse Equation

•   Related to Newton’s 2nd Law

–  F = ma and a = D v/ D t           so F = m D v/ D t

•   This can be rewritten

–  F D t = m D v

•   F D t is known as impulse

•   Momentum = impulse

So, why do I care about that?

•   If you have the misfortune to be in a car accident, you would hope that everything possible has been done to minimize the force on your body.

This is done by:

 

•   reducing the velocity of the car (speed limits!)

•   increasing the time it takes to stop the motion.

How do we increase the time?

•     Objects that stretch or crumple help increase the time it takes for your body to come to a stop, thus minimizing the force & hopefully injuries

–  seatbelts stretch

–  airbags deflate

–  padded dashboards & seats give

–  fenders crumple

 

Importance

•     If I want to learn about the force that an object will exert when it collides with another object, I can use this relationship.

•     The force is directly proportional to the mass and velocity and inversely proportional to the time

Application

•     The egg drop project

•     Please read the directions thoroughly.

•     Pick your partners with care.

•     Note the due date and rain date(s).

•     Dress appropriately (it will probably turn cold again come egg dropping time)!

Impulse – Momentum Theorem

Sample problem

•     A 0.5 kg cart is moving at 2 m/s.  It hits a bumper that stops it in 0.1 s.

•     a) What is the cart’s momentum?

•     b) What is the force exerted on the cart to stop it?

•     c) How far does it travel in the time it takes to stop?

Solution

» Solution A - momentum

•   D p = m D v = 0.5 kg x -2m/s

•   D p = -1 kg m/s

»  Solution B - impulse

•     F = D p/ D t  =                              .     (-1 kg m/s) / (0.1 s)

•     F = -10 kg m/s2 = -10 N

Solution

» Solution C – stopping distance

•   D x = ½ (vi + vf) D t

•           = ½ (2m/s +0m/s) 0.1s

•   D x = 0.1 m

Classwork/Homework

•     Practice A – Momentum

–  P. 199 #1-3

•     Practice B – Impulse

–  P. 201 #1-4

•     Practice C – Stopping Distance

–  P. 203 #1-3

Physics    

Ch 6 Section 2

Conservation of Momentum

TSWBAT

•      Describe the interaction between two objects in terms of the change in momentum of each object.

 

•      Compare the total momentum of two objects before and after they interact.

 

•      State the law of conservation of momentum.

 

•      Predict the final velocities of objects after collisions, given the initial velocities, force, and time.

Conservation of Momentum

•     When no net external forces are acting on a system of objects the total vector momentum of the system remains constant.

•     For us, this means that the momentum before a collision equals the momentum after collision

The Law of
Conservation of Momentum:

The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between the objects.

 

m1v1,i + m2v2,i +…     = m1v1,f + m2v2,f+…

total initial momentum = total final momentum

 

 

Momentum is Conserved

Sample Problem

A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right,what is the final velocity of the boat?

 

Solution

Given:

     m1 = 76 kg m2 = 45 kg

     v1,i = 0        m/s  v2,i = 0 m/s

     v1,f = 2.5 m/s to the right

    Unknown:

     v2,f = ?

 

Solution

m1v1,i + m2v2,i = m1v1,f + m2v2,f

 

•      (76kg)(0 m/s)+(45kg)(0m/s) =

                   (76kg)(+2.5m/s)+(45kg)(v2,f)

•      0 kg.m/s = 190 kg.m/s +(45kg)(v2,f)

•      - (45kg)(v2,f) = 190 kg.m/s

•      v2,f = (190 kg.m/s)/(-45kg)

•      v2,f = -4.22 m/s or 4.22 m/s to the left

 

 

 

Classwork/Homework

•      Classwork

–  P. 209 Practice D #1-4

–  P. 204 Section 6.1 review #1-5

•      Homework

–  P. 211 Section 6.2 review #1-4

–  Continue work on Egg Drop Project

 

 

Physics    

Ch 6 Section 3

Collisions

TSWBAT

•      Identify different types of collisions.

 

•      Determine the changes in kinetic energy during perfectly inelastic collisions.

 

•      Compare conservation of momentum and conservation of kinetic energy in perfectly inelastic and elastic collisions.

 

•      Find the final velocity of an object in perfectly inelastic and elastic collisions.

 

Types of Collisions

4 types of Collisions

•     Perfectly inelastic

–  the objects stick together after collision

–  kinetic energy is not conserved

•     inelastic

–  the object deform (change shape - squish!)

–  kinetic energy is not conserved

4 types of Collisions (continued)

•     elastic

–  the objects bounce off each other

–  kinetic energy is conserved

•     super elastic

–  the objects bounce apart gaining EK as chemical or elastic EP is converted into EK(an explosion)

–  kinetic energy is not conserved

Please Note:

•     Total Energy is always conserved.

•     Total energy is kinetic, potential and thermal (friction, changing objects shape and temperature, etc.)

•     So its OK that kinetic energy is not conserved in most collisions.

Perfectly Inelastic Collisions

Equations

•     Using these equations and the masses & velocities of the objects before the collision, you can solve for velocity after the collision!

•     Perfectly Inelastic equation

•      m1v1i+ m2v2i= (m1+m2)vf

Elastic equation

•     to solve you must also be given information on the motion of one object after the collision OR use conservation of kinetic energy.

•      m1v1i+ m2v2i= m1v1f + m2v2f

•     Conservation of Kinetic Energy

•     ½m1v1i2 + ½m2v2i2 = ½m1v1f2 + ½m2v2f2

 

Sample problem

•      2 carts undergo a perfectly inelastic collision.  Cart 1 has a mass of 0.500 kg and moves at 4.00 m/s to the right. Cart 2 has a mass of 0.250 kg and moves at 3.00 m/s to the left.

•      What is the velocity of the carts after the collision?

•      What is the loss of Kinetic energy?

 

Practice

•     Homework

–  Practice E p. 214 #1,3,5

–  Practice F p.216 #1,3

–  Practice G p. 219 #1,3