Unit 8: Linear Momentum

Impulsive Force Model 1

Unit 8: Linear Momentum

Work on Unit 8 HW asapDemos: Collision carts and track

U7b: All students get perfect Score - answers online

What is implied?

Impulsive Force Particle Page 2

Agenda Review Dynamics for exam Operational Definitions

momentum units

Momentum Motion Map Elastic and Inelastic Collisions Conservation of Momentum

Mid Term Review


Scratch off Practice The University of New England was founded

in… A) 1953 B) 1978 C) 1996 D) 1939 E) 1831

Add 147 to the above answer to determine when UNE was first established.

A) 1978 B) 1953 C) 1996 D) 1831 E) 1939


Operational DefinitionsLinear momentum:

p mvUnits (kg.m/s)

Elastic Collision ~ “No-Stick”Inelastic Collision ~ “Stick”


Clicker ?

1 2 3 4

0% 0%0%0%

1. Ekf = Eki, pf ≠ pi

2. Ekf = Eki, pf = pi

3. Ekf ≠ Eki, pf ≠ pi

4. Ekf ≠ Eki, pf = pi

During elastic collisions…

60

+x

t=0

t=0

1 2 3 4

3

4

1


Clicker ?

1 2 3 4

0% 0%0%0%

1. Ekf = Eki, pf ≠ pi

2. Ekf = Eki, pf = pi

3. Ekf ≠ Eki, pf ≠ pi

4. Ekf ≠ Eki, pf = pi

During inelastic collisions…

60

+x

t=0

t=0

1 2 3 4

1


Elastic Collision Kinetic energy is conserved (Eki = Ekf) Linear momentum is conserved (pf=pi=p)

Ekf (J) pf (kg.m/s)

Eki (J)

1

1pi

(kg.m/s)

1

1

Ekf(J) = 0.95 Eki(J) pf(kg.m/s) = 0.95 p

i(kg.m/s)

0 0


Inelastic Collision Kinetic energy is NOT conserved (Eki ≠

Ekf)

Linear momentum is ALWAYS conserved (pf=pi=p)Ekf (J) pf (kg.m/s)

Eki (J)

1

1pi

(kg.m/s)

1

1

Ekf(J) = 0.5 Eki(J) pf(kg.m/s) = 0.95 pi(kg.m/s)

0 0


Elastic Collision Motion Map

Insert masses at positions proportional to those in the collision. No motion => no change in position. Time increases from the top down.

+x

t=0

t=0

1 2 3 4

3

4

1


Elastic Collision MM1 Insert velocity vectors at each block’s

position. Identify the interaction point with a dashed circle. Define initial and final momenta for each object’s mass and velocity.

p2i=0

t=0

t=0

1 2

1

3 4

3

4

p2fp1i

p1f=0


Elastic Collision MM2 Insert total initial and final momentum

as vector sums of either the initial or final momenta.

p2i=0

+ = + =

t=0

t=0

1 2

1

3 4

3

4

p1i + p2i = pi = p

p1f + p2f = pf = p

p2fp1i

p1f=0


Elastic Collision MM3 Predictive Power: Since pf = p2f = pi = p1i

=> m2v2f = m1v1i

since m1=m2 => v2f = v1i

+ = + =

p2i=0

t=0

t=0

1 2

1

3 4

3

4

p2fp1i

p1f=0

p1i + p2i = pi = p

p1f + p2f = pf = p


Alternative Representations

Bar graph to monitor before, after, and total momentum.

p2p1

pBefor

ep2p1

pAfter

Area of velocity vs. mass plot (same area).

m2

m1

v

Before

m2

m1

v

After

p

p


Inelastic Collision MM1 Insert velocity vectors at each block’s

position. Identify the interaction point with a dashed circle. Define initial and final momenta for each particle based on masses and velocities.

p1i p1&2f

t=0

t=0

1 2 3 4

p2i=0 1


Inelastic Collision MM2 Insert total initial and final momentum

as vector sums of either the initial or final momenta.

+ = =p1&2f = pi =

p

p1i p1&2f

t=0

t=0

1 2 3 4

p2i=0 1

p1i + p2i = pi = p


Inelastic Collision MM3 Predictive Power: Since pf = p1&2f = pi = p1i

=> (m1+m2)v1&2f = m1v1i

since m1=m2 => v1&2f = (1/2)v1i

+ = =

t=0

t=0

1 2 3 4

p1i p1&2f

p1&2f = pi = p

p1i + p2i = pi = p



Bar graph to monitor before, after, and total momentum.

Area of velocity vs. mass plot (same area!).

m2

m1

v

Before

m1+m

2

vAfter

p

p

p2p1

pBefor

ep1&2

pAfter


Work out problem Create an inelastic MM for m1=m striking

m2=3m.

Predict v1&2f if v1i = +4m/s, v2i = 0

t=02

t=0

p1i


Inelastic Collision MM5 Predictive Power: Since pf = p1&2f = pi = p1i

=> (m1+m2)v1&2f = m1v1i

m1=m, m2=3m => v1&2f=(1/4)v1i =(1/4)(+4m/s) =+1m/s

p1i

p2i=0

+ = =

p1&2f

t=0

t=0

1 2

1

3 4

p1&2f = pi = p

p1i + p2i = pi = p



Bar Graph to represent initial, final, and total momentum

p2p1

pBefor

ep1&2

pAfter

Area of velocity vs. mass plot (same area!!!).

m2

m1

v

Before

m1+m

2

vAfter

p

p


Clicker ?

1 2 3 4

0% 0%0%0%

1. 2m rebounds, m moves slowly2. 2m stops, m moves at medium pace 3. 2m continues slow, m moves quickly4. None of the above

If mass “2m” collides ELASTICALLY with mass “m” then…

60

p1i p2i=0

t=0

1 2


Elastic Collision MM4 Create an elastic MM for m1=2m striking

m2=m. Predict v1f and v2f if v1i = +3m/s, v2i=0. Very challenging because we have two

unknowns so that we need two equations. Solved generically. Details are provide

below but will not be discussed.


Elastic Collision Prediction Because momentum is conserved, pf=pi:

m1v1f + m2v2f = m1v1i

Because the collision is elastic, Ekf = Eki

(1/2)m1v1f2 + (1/2)m2v2f

2 = (1/2)m1v1i2

Two equations with two unknowns: (v1f and v2f given m1 = 2m, m2 = m, v1i =

+3m/s). To simplify let v1i = v, v2f = v2, v1f = v1 then:

m1v1+m2v2=m1v Solve for v1:

v1 = (m1v-m2v2)/m1


Crunch Time 1 Square v1 and multiply by m1…

v12= [(m1v-m2v2)/m1]2 =

v2-2vv2m2/m1+v22m2

2/m12

=> m1v12=m1v2-2vv2m2+v2

2m22/m1

Dropping (1/2) in Ek:

m1v12 + m2v2

2 = m1v2 Substitute m1v1

2 into Ek …

m1v2-2vv2m2+v22m2

2/m1 + m2v22 = m1v2

Cancel m1v2 on both sides and divide by v2…

-2vm2 + v2m22/m1 + m2v2 = 0


Crunch Time 2 Divide by m2 and isolate v2…

v2(m2/m1 + 1) = 2v Solve for v2…

v2 = 2v/(1+m2/m1) = v2f

Solve for v1…

v1 = v-m2v2/m1 = v - 2m2v/m1(1+m2/m1)

= v(m1+m2-2m2)/(m1+m2) = v(m1-m2)/(m1+m2)

v1 = v(1-m2/m1)/(1+m2/m1) = v1f


m1 > m2 => m1 Plows on Since m1=2m, m2=m and v1i=3m/s v1f = v(1-m2/m1)/(1+m2/m1) = +1m/s v2f = 2v/(1+m2/m1) = +4m/s

p1fp1i

p2i=0

+ = + =p1i + p2i = pi

t=0

t=0

1 2

1

3 43 4

p1f + p2f = pf

p2f

4



Bar graphs to represent initial, final and total momentum

p2p1

p

Before

p2

p

After

p1

Area of velocity vs. mass plot (same area!!!).

m2

m1

v

Before

vAfter

p

p

m2

m1


Clicker ?

1 2 3 4

0% 0%0%0%

1. m rebounds, 2m moves slowly2. m stops, 2m moves at medium pace 3. m continues slow, 2m moves quickly4. None of the above

If mass “m” collides ELASTICALLY with mass “2m” then…

60

p1i p2i=0

t=0

1 2


m1 < m2 => m1 bounces back!

If m1=m, m2=2m and v1i=3m/s

v1f = v(1-m2/m1)/(1+m2/m1) = -1m/s

v2f = 2v/(1+m2/m1) = +2m/s

p1f

p1i

p2i=0

+ = + =p1i + p2i = pi

t=0

t=0

2

1

3 4

p1f + p2f = pf

p2f4 3

2

Before

After

p2p1 p2p1


Unit 8a: Momentum ctd.

HW U8a due after ThanksgivingDemos: Collision carts and two air

pucks


Inelastic Collision Motion Map

Insert masses and positions proportional to those in the collision. No motion => no change in position. Time increases from the top down.

+x

t=0

t=0

1 2 3 4

1


Agenda Refined Operational Definitions:

Elastic Collision Inelastic Collision

Explosions Two objects in motion and colliding Two-Dimensional Collisions


Operational DefinitionsLinear momentum:

p mvUnits (kg.m/s)

Elastic Collision: Eki = Ekf

Inelastic Collision: Eki ≠ Ekf


What is your opinion?

1 2

56%

44%

1. True2. False

Explosions are examples of elastic collisions because kinetic energy before and after the explosion is conserved.


Explosions 1 MMM an explosion of an object into two

pieces, one with mass m, the other with mass 2m. What is the pi? Therefore what must be pf? Is this by definition an elastic or inelastic event?t=

0

p2p1

Before

After

p2p1


Explosions 2 Since Eki = 0 and Ekf≠0 => Inelastic “event”

Predictive Power: Since pf = p2f + pif = pi = 0

=> +m2v2f - m1v1f = 0

since m2=2m1 => v2f = (1/2)v1f

+ =

+ =

p1i + p2i = pi

p1f + p2f = pf

p2i=0

3 2 3

p2fp1f

p1i=0

t=01

44

p2p1

pBefore

pAfter

p2p1


2 Moving Objects 1 MMM the situation when two carts of

equal mass m collide elastically, but in opposite directions and different velocities. Look at the individual and total momenta we observe that momentum is transferred.

t=0

12t=01


2 Moving Objects 1 Prediction: Since momenta are transferred:

p1i = p2f and p2i = p1f

And since m1=m2=m v2f = v1i and v1f = v2i

p1i + p2i = pi

t=0

12 t=0

1p2i

p1i

p1f + p2f = pf

4 2 4p1f

p2f

33

Rigorous solution requires conservation of

kinetic energy, lots of algebra.

p2p1

pBefor

e pAfter

p2p1


Two Dimensional Collisions Motivation: A 1000 kg eastbound car

traveling 30m/s runs a red light and center-punches a 10,000 kg truck traveling southbound at 4 m/s. The two collide inelastically.

Draw a top view of the incident and qualitatively estimate where the two vehicles end up, assuming no interactions with any other obstacles.


Clicker ?

1 2 3 4

3% 7%76%14%

1. A2. B3. C4. D

Choose the correct path, either A, B, C, or D AND explain why the paths you did not choose are incorrect.

A

B

C

D

N

W E

S


Because p = constant… … only “C” can be correct since it is the

only one with momenta both in eastern and southern directions.

But what is pf and ? A

B

C

D

N

W E

S


Use Vector Tool: p=0 pf is the total momentum after the

collision. What is it comprised of? The momenta of the car (pc) and truck

(pt). What are these in terms of pf?

pf

N

W E

S


Seek Resultant Vector The components of total momentum! pcar = mV(east) = 30,000kgm/s (east) ptruck = Mv(south) = 40,000kgm/s(south) pf = ?, = ? pf = 50,000 kgm/s = tan-1(4/3) = 53° What is vf =? vf = pf/(m+M) = 4.5m/s

pf

pcar

ptruck


Arbitrary Impact Gets messy, though understandable

through tip to tail vector addition. For example, two billiard balls. Easy to solve if + = 90°

p1i

pi pf

p1f p2f


Arbitrary Impact p1f = pfcos and p2f = pfcos COM: pf = p1f + p2f = pi

p1fy = p1fsin = -p2fy = -p2fsin The vertical components of the final

momenta must cancel.

p1i

pi pf

p1f p2f


Clicker ?

1 2 3

89% 7%4%

1. A2. B3. C

Two identical billiard balls collide; what will be their post-collision paths?

A CB


Unit 8b: Impulse ForceHW U8b due after Thanksgiving

Demos: Force sensor and motion detector with collision cart, eggs and blanket, small and big balls.


Agenda Newton’s Third Law Impulsive Force and change of velocity Examples


Clicker ?

1 2 3 4

0% 0%0%0%

1. VW Bug2. Mac Truck3. Both feel same

force4. No Clue

Which feels the greater force, a VW Bug colliding at 60 mph with the Mac Truck or the Mac Truck colliding at 30 mph with the VW Bug?

60Pow

!


Clicker ?

1 2 3 4

0% 0%0%0%

1. VW Bug2. Mac Truck3. Both interact

over same t4. Texting buddies

Which vehicle interacts over a longer time during the collision?

60Pow

!


Recall Newton’s Third Law Recall, the forces between two colliding

carts, regardless of mass, are equal and opposite.

Examine the units of F vs. t, break them down into their simplest form.

t (s)

Pow!

F (N)


Clicker ?

1 2 3 4 5

4%6%

39%39%

13%

1. 32 Ns2. 16 kgm/s3. 3.2 kgm/s4. 1.6 Ns5. 0.8 kgm/s

Estimate the numerical value of the area of the F vs. t curve below.

F (N)

t (s)

4

0 0.8


Area under F vs. t Plot N.s = kg.m.s/s2 = kg.m/s Same units as p = mv Count the blocks Area ≈ 8 blocks x 0.1kg.m/s = 0.8kg.m/s

F (N)

t (s)

4

0 0.8


Impulse vs. Momentum Examine the force over time and the

change is velocity of a cart.

t (s)

v(m/s)

t (s)vf

vi

F (N)


Compare F vs. t w/ v Constant: mass of cart = 0.33 kg vi = - 1.2m/s, vf = + 1.2m/s v = vf-vi =(+1.2m/s-(-1.2m/s)) =

+2.4kg.m/s

t (s)

4

0 0.8

v(m/s)

t (s)

+1.2

F (N)

-1.2


Impulse - Momentum Ft = (0.33kg)vFt = mvMass constantFt = (mv)Ft = porF = p/t

F∆t(N.s)

v(m/s)

0.8

0 2.4

Really just a restatement of Newton’s Second Law. In fact, what Newton really discovered was: F = p/t

Slope = 0.33Ns/m/s = 0.33kg = mcart


Impulse - Momentum F = p/t (mass constant) => F = mv/t => F = ma The form of Newton’s Second Law we

have been using for the past month. Also note that if there is no external

force acting on m, Fext = 0 = p Which is the conservation of linear

momentum!


Clicker ?

1 2 3 4

0% 0%0%0%

1. Smashes2. Cracks3. Doesn’t break4. Has fear

A raw egg is thrown 25 m/s (50 mph) at a blanket. The egg…

60


Applications I Reduce F and increase t. Give six examples of engineering

designed to make “soft landings.”

t (s)

4

0 8

F (N)

Hard Landing

Soft Landing


PRS ?

1. Golfball high, Volleyball low2. Golfball low, Volleyball high3. Golfball and volleyball stay together4. Cataclysm causing the earth to vanish.

What will happen if a 0.03kg golfball atop a 0.3kg volleyball fall 1 meter in concert? Each ball separately rebounds to a medium height. What will happen to the balls if they are stacked on top of each other and allowed to fall?


Make a predictionHurley, at height “zi” above the ground, swings on a rope and collides inelastically with Burley at rest. What height will the pair reach? Start by diagramming the conservation principles needed to solve this problem.

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.

zi


zf=?


Make a prediction1. Conservation of Energy

2. Conservation of linear momentum

3. Conservation of Energy


are needed to see this picture.

zi

Eg Ek


are needed to see this picture.QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.

pfpi


zf=?

EkEintEgEint

Oscillations & Waves

Oscillating Particle Model

Source of Waves:Simple Harmonic Oscillator

Linear Restoring ForceSHO Kinematics

SHO Energy Conservation

Mechanical Waves:Sound Waves

Energy PropagationSuperposition Principle

Doppler Shift

Light Waves & InterferenceDiffraction/Refraction

Polarization/Colors

Classical Mechanics

Particle

Model

KinematicsConstant Velocity

Constant Acceleration

Conservation of Energy

E = W + Q + R

DynamicsFree Particle F=0

Constant Force F=ma

Impulse/MomentumFt = p

ptotal = mv = constantConservation of Linear

Momentum

Central ForceF = (mv2/r) inwardRotational Mechanics

1st Semester

Electricity & Magnetism

Field Model

Particle of Mass mGravitational Field g

Gravitational Force Fg

Potential Energy Eg

Potential Vg &Tools

Particle of Charge qElectric Field E

Electrostatic Force FE

Potential Energy EE

Potential V & Tools

Magnetic dipole µMagnetic Field B

Magnetic Force FB

RHR & Induction

E -> current IOhms Law, Circuits

I -> Magnetic Field B

Multiple Particles

2nd SemesterFluid Statics/Dynamics


Momentum Transfer Use energy conservation to determine initial and

final v before and after the collision with the floor. Ef = mgz = Ei = (1/2)mv2 => v = (2gz)1/2

v=0

m

M

z

z(m) M z

(m) m

vi(m/s)

1.0 -4.5 1.0 -4.5

vfs(m/s)

0.7 +3.7 0.3 +2.4

ps(Ns) +2.4 +.21

vfc(m/s)

? ? 2.0 +6.3

pc(Ns) +2.3 +.32

pc-s(Ns) -0.1 +0.1


Bouncing Ball Data


Unit 8: Consensus

MomentumImpulse


Elastic Collision MM4 Alternative representation: Bar graph to

monitor before, after, and total momentum.

+ = + =

p1i + p2i = pi p1f + p2f = pf

t=0

t=0

1 2

1

3 4

3

4

p2p1

pBefor

ep2p1

pAfter


Inelastic Collision MM4 Alternative representation: Bar graph to monitor

before, after, and total momentum.

+ = =p1i + p2i = pi p1&2f = pf

t=0

t=0

1 2 3 4

p2i=0

p2p1

pBefor

ep1&2

pAfter


Elastic Collision Kinetic energy is conserved (Eki = Ekf) Linear momentum is conserved (pf = pi)

Ekf (J) pf (kg.m/s)

Eki (J)

1

1pi

(kg.m/s)

1

1

Ekf(J) = 0.95 Eki(J) pf(kg.m/s) = 0.95 p

i(kg.m/s)

0 0


Inelastic Collision Kinetic energy is NOT conserved (Eki ≠ Ekf)

Linear momentum IS conserved (pf = pi)

Ekf (J) pf (kg.m/s)

Eki (J)

1

1pi

(kg.m/s)

1

1

Ekf(J) = 0.5 Eki(J) pf(kg.m/s) = 0.95 pf

i(kg.m/s)

0 0


Compare F vs. t w/ v Constant: mass of cart = 0.33 kg vi = - 1.2m/s, vf = + 1.2m/s v = vf-vi =(+1.2m/s-(-1.2m/s)) =

+2.4kg.m/s

t (s)

4

0 0.8

v(m/s)

t (s)

+1.2

F (N)

-1.2


Impulse - Momentum Ft = (0.33kg)vFt = mvMass constantFt = (mv)Ft = porF = p/t

F∆t(N.s)

v(m/s)

0.8

0 2.4

Really just a restatement of Newton’s Second Law. In fact, what Newton really discovered was: F = p/t

Slope = 0.33Ns/m/s = 0.33kg = mcart

Documents

Unit 8: Linear Momentum