Newtonian Mechanics – Momentum, Energy, Collisions

Gernot LaicherFor ACCESS Class 2011

Quick Review: Vectors and Scalars

• Vectors: Magnitude and Direction(e.g., Force, Displacement, Velocity, Acceleration)

• Scalars: Magnitude only(e.g., Temperature, Altitude, Age)

Adding Vectors

1F

2F

Two forces are acting on a mass m:How do we find the total force ?

1F

2F

totalF

1F

2F

21 FFFFi

itotal

Subtracting Vectors

initialv

An object changes it’s velocity.What is it’s change in velocity?

finalv

initialv

finalv

initialfinalinitialfinal vvvvv

initialv

initialv

initialfinalinitialfinal vvvvv

finalv

v

1F

xxxtotal FFF 21

x

y

xF1

yF1

x

y

xF2

yF2

2F

yyytotal FFF 21

xF1

yF1

x

y

xF2

yF2

xxxtotal FFF 21

yyytotal FFF 21

xtotalF

ytotalF

(Note: F2y had a negative value)

totalF

Sir Isaac NewtonGodfrey Kneller’s 1689 portrait of Isaac Newton (age 46)

(Source: http://en.wikipedia.org/wiki/Isaac_Newton)

Newton's First Law : An object at rest tends to stay at rest and an object in uniform motion tends to stay in uniform motion unless acted upon by a net external force.

Newton's Third Law: For every action there is an equal and opposite reaction.

Newton's Second Law: An applied force on an object equals the rate of change of its momentum with time.

rntDisplaceme

:

dt

rdvVelocity

:

dt

rd

dt

d

dt

vdaonAccelerati

:

).(

:

tduringconstvaslongast

r

dt

rdvVelocity

).(

:

tduringconstaaslongast

v

dt

vdaonAccelerati

vmpMomentum

:

sN

ss

mkg

s

mkgs

mkg

mile

m

s

hour

hour

mileskg

hour

mileskgvmp

directionignoringmphgoingcaraofMomentum

5.670

5.6705.670

447.01500

3.1609

360011500

11500

)(1

2

Nss

mkgs

mkg

s

m

g

kgg

s

mgvmp

rifleMsmgoingbulletaofMomentum

7.3

7.3

9480039.0

9481000

9.3

9489.3

)16(/948

Newton’s Second Law

i

inet FF

mmassdt

pd

dt

pdFF

iinet

amvdt

dmdt

vdmv

dt

dm

vmdt

d

dt

pdFF

iinet

For a non-changing mass:

netFm

a

amFnet

0dt

dm

pddtFdt

pdF netnet

Rewriting Newton’s Second Law

)Impulse"("dtFnet

m)Momentum"in Change("pd

Impulse = Change in Momentum

Example: A constant force of 5N applied for 0.5 seconds on a mass of 0.2kg, which is initially at rest.

s

s

p

p

netnet

final

initial

pddtFpddtF5.0

0

sFdtFdtF net

s

s

net

s

s

net 5.05.0

0

5.0

0

initialfinal

p

p

pppdfinal

initial

s

mvvkgNsmvmvsN finalfinalinitialfinal 5.122.05.25.05

In general, the force may depend on time

final

initial

final

initial

t

t

p

p

netnet pddttFpddttF

t

)(tFnet final

initial

t

t

net dttF

Impulse = “Area” under the F(t) curve (Integrate F(t) over time to get the impulse)

A Two-body Collision

1v

1m

2112 FF

Before the collision

2v

2m

During the collision

12 onfromF

2v

1m 2m 1221 onfromonfrom FF

(Newton’s third law)

During the Collision

t

)(tF

)(21 tF

)()( 2112 tFtF Impulse on m2

Impulse on m1

Impulse on m2 = - Impulse on m1

Impulse on m2 = - Impulse on m1

Change of momentum of m2 = - Change of momentum of m1

initialfinalinitialfinal pppp 1122

12 pp

initialinitialfinalfinal pppp 2121

Total momentum after collision = Total momentum before collision

The total momentum of the system is “conserved” during the collision. (This works as long as there are no external forces acting on the system)

1u

1m

After the collision

2u

2m

initialinitialfinalfinal pppp 2121

22112211 vmvmumum

For a 1-dimensional collision we can replace thevector with + or – signs to indicate the direction.

Note: u1, u2, v1, v2 may be positive or negative, depending on direction and depending on your choice of coordinate system.

22112211 vmvmumum

Given: m1 , m2 , v1 , v2

What can we find out about the final velocities u1 and u2 ?

22112211 vmvmumum

Given: m1 , m2 , v1 , v2

What can we find out about the final velocities u1 and u2 ?

Answer: In general there are an infinite number of possible solutions (combinations of the final velocities u1 and u2 thatfulfill the conservation of momentum requirement). Which ofthose solutions really happens depends on the exact natureof the collision.

Excel Program – Using a Solver1) Solving a simple math problem (2x + 6y =14) infinitely many solutions2) Adding a constraint (a second equation) (x+y=3) to get a single solution.

3) Solving a quadratic equation (4x2 + 2x =20) how many solutions?4) Pick different initial conditions for x to find all the solutions.

5) Write an Excel program that has the following input fields: m1, m2 , v1 , v2 , u1 , u2 . 6) Create (and label) fields that calculate p1initial , p2initial , p1final , p2final .7) Create (and label) fields that calculate ptotal initial , ptotal final.8) Create (and label) fields that calculate ptotal final - ptotal initial.9) Fill in these values: m1=1 (kg), m2=1 (kg), v1=1 (m/s), v2=-1 (m/s), u1=some value (m/s), u2=some value (m/s).10) Use a solver that changes u1 and u1 and makes ptotal final - ptotal initial = 0. How many possible solutions can you find?

Mechanical Energy of Two Mass System

222

211 2

1

2

1vmvmEinitial Before the collision

222

211 2

1

2

1umumE final After the collision

Totally Elastic Collision: initialfinal EE

(no mechanical energy is lost)

02

1

2

1

2

1

2

1 222

211

222

211

vmvmumum

0 initialfinal EE

Excel Program – Using a SolverAdd the following to your Excel “Collision-Solver”:

Create fields that calculate Etotal initial , Etotal final , Etotal final - Etotal initial ,% Energy change.

100ChangeEnergy %

initial

initialfinal

E

EE

Now solve your collision problem again. This time, use the constraint Etotal final - Etotal initial = 0Find all possible solutions and interpret their meaning.

Totally Inelastic Collision: finalfinalfinal uuu 21

(The two masses stick together after the collision and move with the same velocity)

221121:momentum ofon Conservati vmvmumm

21

2211

mm

vmvmu

Is there Mechanical Energy Conservation in Totally Inelastic Collisions?

222

211 2

1

2

1vmvmEinitial

2

21

221121

221 )(

2

1)(

2

1

mm

vmvmmmummE final

21

2211

mm

vmvmu

Excel Program – Using a SolverName the Excel worksheet tab for the previously created elasticcollision “Elastic Collision”.

Make a copy of this tab and rename it “Inelastic Collision”

Modify the “Inelastic Collision” tab so that it calculates “u1” from theinitial conditions (masses, initial velocities) according to the formulafor u on the previous page. Then you can make u2=u1. So now both u1 and u2 will always have the same value.

Use a lab notebook to record your findings and investigate several scenarios of collisions

For both elastic and inelastic collisions you should find solutions forthese cases:1) Equal masses, one mass initially at rest.2) Equal masses, equal but opposite velocities (head on collision).3) Very unequal masses, lighter mass at rest.4) Very unequal masses, heavier mass at rest.5) Very unequal masses, equal but opposite velocities (head on

collision)