Lecture 12

Momentum & ImpulseMomentum & Impulse

Lecture 12

Goals:Goals:• Momentum & ImpulseMomentum & Impulse

Solve problems with 1D and 2D Collisions Solve problems having an impulse (Force vs. time)

• EnergyEnergy Understand the relationship between motion and energy Define Potential & Kinetic Energy Develop and exploit conservation of energy principle

Inelastic collision in 1-D: Example

Inelastic means that the objects stick.

A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V :

What is the momentum of the bullet with speed v ?

vV

before after

x

A perfectly inelastic collision in 1-D

Example

What is the momentum of the bullet with speed v ?

Key question: Is x-momentum conserved ?

vV

before after

x

aaaa

vm

V)( 0 M v Mmm BeforeBefore AfterAfter

Exercise Momentum Conservation

A. Box 1

B. Box 2

C. same

Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface.

The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks.

Which box ends up moving fastest ?

1 2

Exercise Momentum Conservation

The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks. Which box ends up moving fastest ? Notice the implications from the graphical solution: Box 1’s

momentum must be bigger because the length of the summed momentum must be the same.

The longer the green vector the greater the speed

1 2

Before After Before After

Ball 1 Ball 1 Ball 2 Ball 2

Box 1 Box 1 Box 2 Box 2

Box 1+Ball 1 Box 1+Ball 1 Box 2+Ball 2 Box 2+Ball 2

A perfectly inelastic collision in 2-D

Consider a collision in 2-D (cars crashing at a slippery intersection...no friction).

vv1

vv2

VV

before after

m1

m2

m1 + m2

If no external force momentum is conserved. Momentum is a vector so px, py and pz

A perfectly inelastic collision in 2-D

vv1

vv2

VV

before after

m1

m2

m1 + m2

x-dir px : m1 v1 = (m1 + m2 ) V cos y-dir py : m2 v2 = (m1 + m2 ) V sin

If no external force momentum is conserved. Momentum is a vector so px, py and pz are conseved

Elastic Collisions

Elastic means that the objects do not stick.

If no external force, then momentum will always be conserved

Start with a 1-D problem.

Before After

Billiards

Consider the case where one ball is initially at rest.

ppa

ppb

FF PPa

beforeafter

The final direction of the red ball will depend on where the balls hit.

vvcm

Billiards: All that really matters is conservation momentum (and energy Ch. 10 & 11)

Conservation of Momentum x-dir Px : m vbefore = m vafter cos + m Vafter cos y-dir Py : 0 = m vafter sin + m Vafter sin

ppafter

ppb

FF PPafter

before after

Force and Impulse (A variable force applied for a given time)

Gravity: At small displacements a “constant” force t

Springs often provide a linear force (-kx) towards its equilibrium position (Chapter 10)

Collisions often involve a varying force

F(t): 0 maximum 0We can plot force vs time for a typical collision.

The impulse, ΔΔρρ, of the force is a vector defined as the integral of the force during the time of the collision.

Force and Impulse (A variable force applied for a given time)

F

ΔΔρ ρ reflects momentum transfer

t

ti tf

t

Impulse ΔΔρρ = area under this curve !

(Transfer of momentum !)

Impulse has units of Newton-seconds

ΔΔρρ

Force and Impulse Two different collisions can have the same impulse since

ΔΔρρ depends only on the momentum transfer, NOT the nature of the collision.

t

F

t

F

tt

same area

t big, FF smallt small, FF big

Average Force and Impulse

t

F

t

F

tt

t big, FFavav smallt small, FFavav big

FFav av

FFav av

Exercise Force & Impulse

A. heavier

B. lighter

C. same

D. can’t tell

Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second.

Which box has the most momentum after the force acts ?

F F light heavy

Boxing: Use Momentum and Impulse to estimate g “force”

Back of the envelope calculation

(1) marm~ 7 kg (2) varm~7 m/s (3) Impact time t ~ 0.01 s

 Question: Are these reasonable? Impulse p ~ marm varm ~ 49 kg m/s  F ~ Δρ/t ~ 4900 N  (1) mhead ~ 6 kg  ahead = F / mhead ~ 800 m/s2 ~ 80 g !  Enough to cause unconsciousness ~ 40% of fatal

blow Only a rough estimate!

الله سبحان

During "collision" with a tree 

ahead ~ 600 - 1500 g

How do they survive? 

• Jaw muscles act as shock absorbers

• Straight head trajectory reduces damaging rotations (rotational motion is very problematic)

Woodpeckers

Home Exercise The only force acting on a 2.0 kg object moving along the x-

axis. Notice that the plot is force vs time. If the velocity vx is +2.0 m/s at 0 sec, what is vx at 4.0 s ?

p = m v = Impulse

mv =+ +

mv = (-8)1 N s + ½ (-8)1 N s + ½ 16(2) N s

m v = 4 N s v = 2 m/s

vx = 2 + 2 m/s = 4 m/s

Energy

We need to define an “isolated system” ?

We need to define “conservative force” ?

Recall, chapter 9, force acting for a period of time gives an

impulse or a change (transfer) of momentum

What if a force acting over a distance:

Can we identify another useful quantity?

Energy

Fy = m ay and let the force be constant y(t) = y0 + vy0 t + ½ ay t2

y = y(t)-y0= vy0 t + ½ ay t2

vy (t) = vy0 + ay t t = (vy - vy0) / ay

Eliminate t and regroup So y = vy0 (vy- vy0) / ay+ ½ ay (vy

2 - 2vy vy0+vy0

2 ) / ay

2

y = (vy0vy- vy02) / ay+ ½ (vy

2 - 2vy vy0+vy0

2 ) / ay

y = ( - vy02) / ay+ ½ (vy

2 - +vy02

) / ay

y = ½ (vy2 - vy0

2 ) / ay

Energy

And now

y = ½ (vy2 - vy0

2 ) / ay

can be rewritten as:

may y = ½ m (vy2 - vy0

2 )

And if the object is falling under the influence of gravity

then

ay= -g

Energy

-mg y= ½ m (vy2 - vy0

2 )

-mg yf – yi) = ½ m ( vyf2 -vyi

2 )

Rearranging to give initial on the left and final on the right

½ m vyi2 + mgyi = ½ m vyf

2 + mgyf

We now define mgy as the “gravitational potential energy”

A relationship between y-displacement and change in the

y-speed

Energy

Notice that if we only consider gravity as the external force then

the x and z velocities remain constant

To ½ m vyi2 + mgyi = ½ m vyf

2 + mgyf

Add ½ m vxi2 + ½ m vzi

2 and ½ m vxf2 + ½ m vzf

2

½ m vi2 + mgyi = ½ m vf

2 + mgyf

where vi2 = vxi

2 +vyi2 + vzi

2

½ m v2 terms are defined to be kinetic energies

(A scalar quantity of motion)

Energy

If only “conservative” forces are present, the total energy If only “conservative” forces are present, the total energy

((sum of potential, U, and kinetic energies, K) of a system) of a system is is conservedconserved

Emech = K + U

K and U may change, but Emech = K + U remains a fixed value.

Emech = K + U = constant

Emech is called “mechanical energy”

Example of a conservative system: The simple pendulum.

Suppose we release a mass m from rest a distance h1 above its lowest possible point.

What is the maximum speed of the mass and where does this happen ?

To what height h2 does it rise on the other side ?

v

h1 h2

m

Example: The simple pendulum.

y

y=0

y=h1

What is the maximum speed of the mass and where does this happen ?

E = K + U = constant and so K is maximum when U is a minimum.

Example: The simple pendulum.

v

h1

y

y=h1

y=0

What is the maximum speed of the mass and where does this happen ?

E = K + U = constant and so K is maximum when U is a minimum

E = mgh1 at top

E = mgh1 = ½ mv2 at bottom of the swing

Example: The simple pendulum.

y

y=h1=h2

y=0

To what height h2 does it rise on the other side?

E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point.

E = mgh1 = mgh2 or h1 = h2

Cart Exercise Revisited: How does this help?

1st Part: Find v at bottom of incline

5.0 m

30°

7.5 m

N

mg

ai = g sin 30°

= 5 m/s2

d = 5 m / sin 30°

= ½ ai t2

10 m = 2.5 m/s2 t2

2s = t

v = ai t = 10 m/s

vx= v cos 30°

= 8.7 m/s

i

j

x

y

Emech is conserved

Ki+ Ui = Kf + Uf

0 + mgyi = ½ mv2 + 0

(2gy)½ = v = (100) ½ m/s

vx= v cos 30°

= 8.7 m/s

One step, no FBG needed

Exercise : U, K, E & Path

A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless

1. Ball is dropped

2. Ball slides down a straight incline

3. Ball slides down a curved incline

After traveling a vertical distance h, how do the speeds compare?

(A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell

h

1 32