Physics 231 Lecture 11

• Main points of today’s lecture:• Examples of energy and power• Examples of energy and power• Impulses: forces that last only a

short time• Moment m• Momentum

• Impulse-Momentum theorem( )F

ΔΔΔ

vp m=

• Momentum conservation

( )ifmm vvvptF −=Δ=Δ=Δ

ffftot ,2,1, ppp +≡

• Momentum and external forcesitotii ,,2,1 ppp ≡+=

t ppF −=Δ itotftotext t ,, ppF −=Δ

Bicycle challenge rules

C t t t t b idi f 20 t P 200 W Th h• Contestants must warm up by riding for 20 s at P > 200 W. Then each contestant is judged by the peak or maximum power generated anytime in the next 10 s. C I h t l t ?• Can I have two volunteers?

• Should the instructor participate?– a) yes– b) no

“Carmichael (Armstrong’s coach) told me that a decent pro cyclist would have put out at (an average power)least four hundred watts (during a 4 hour period), and that the stragglers at the end of the peloton (known as the gruppetto) (which d ’ f l h d f f i i ) ld l k i hdon’t feel as much drag force from air resistance) would clock in at perhaps three hundred and fifty. Armstrong—in top Tour shape—would have come close to five hundred.” (from New Yorker article)

Example

A 50 0 k t d t li b 5 00 l t t t• A 50.0-kg student climbs a 5.00-m-long rope at a constant speed and stops at the top. (a) What must her speed be in order to match the power output of a 200-W light bulb?

P 200 W

yP mgv=P 200 Wm 50 kgvy ?

• (b) Quiz: How much work does she do?– a) 2450 J

b) 245 J

yPv

mg =

2

200W 0.41m / s50kg(9.8m / s )

= =

– b) 245 J– c) 24.5 J– d) 2.45 J

clicker question

A l t if l f t t 29 / i 12 l l l• A car accelerates uniformly from rest to 29 m/s in 12 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if the weight of the car is 1.2x104 N.

) 1 2 103 W– a) 1.2x103 W– b) 6.2x105 W– c) 3.8x103 W

v0` 0v 29m/s

– d) 4.3x104 W WPt

=Δ

W KE KE KE

mg 1.2x104NP ?

f iW KE KE= − fKE=

Therefore:

fKEPt

=Δ

21 mv2 t

=Δ( )24

41 2x10 N 29m / s1

21 mgv2 g t

=Δ

( )( )

42

1.2x10 N 29m / s1 4.3x10 W2 9.8m / s 12s

= =

ExampleA skier starts from rest at the top of a hill that is inclined at 20° with the horizontal The hillside is 200 m long and the coefficient of friction betweenhorizontal. The hillside is 200 m long, and the coefficient of friction between snow and skis is 0.075. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. What is the distance d that the skier glides along the horizontal portion of the snow before coming to rest?g p g

f f 0 0 non cons extKE PE KE PE W W−+ − − = +

0 f extKE 0; KE 0; W 0.= = =0 f ext; ;

of 0PE PE mgh mg(200m)sin(20 ).− = − = −f 0 non consPE PE W .−− =

h non cons f ,hill f ,flatW W W− = +

f ,flat k,flat kW f d mgdμ= − = −o

f ,hill k,hill k kW f (200m) N(200m) mg cos(20 )(200m)μ μ= − = − = −o o

k kmg(200m)sin(20 ) mg cos(20 )(200m) mgdμ μ − = − −

/ /// //. o

o

k

sin(20 ) .342d (200m) cos(20 ) 200m( .940) 724m.0.075μ

= − = − =

Reading Quiz1. Impulse is1. Impulse is

A. a force that is applied at a random time.B. a force that is applied very suddenly.C. the area under the force curve in a

force-versus-time graph.o ce ve sus e g p .D. the interval of time that a force lasts.

Slide 9-5

Impulse: useful concept for forces that last a very short time

Th i hi h f l t h t ti d• There are many processes in which forces last a very short time and are difficult to mathematically describe. Examples are:– Kicking, striking batting, dribbling a ball.– Various types of explosions, firearms, etc.

• The typical time dependence of such forces whose actions can be best described by the associated impulse is described below:

≈0.01s

aveimpulse F t= Δ

Impulse and Momentum• The linear momentum of a particle of mass m is:

• The change in velocity is related to the change in momentum, i.e. impulse and to the average acceleration:

vp m=

p g

• It is related to the average impulsive force:

( )ifif m vvppp −=−=Δ avema t= Δ

g p

ave avet ma tΔ = ΔF

( )v v

( )t m ΔF v v

( )f iave t m t

t−

Δ = ΔΔ

v vF

( )ave f it m Δ = −F v v

ave t Δ = ΔF p

ave tΔ

=ΔpF

Note: the shorter the force acts to achieve the impulse, the larger the impulsive force must be.

Conceptual question

C id t t f d 2 t t i t k• Consider two carts, of masses m and 2m, at rest on an air track. If you push first one cart for 3 s and then the other for the same length of time, exerting equal force on each, the momentum of the light cart isthe light cart is– a) four times– b) twice– c) equal to– d) one-half– e) one-quarter the momentum of the heavy cart.) q y

ave t Δ = ΔF p

f 0f= −p p

0 0=p

ave ft Δ =F p

Partially inelastic collisions: earth and ball• A 0.4 kg ball is dropped from rest at a point 1.5 m above the floor. The h0 1.5 mg pp p

ball rebounds straight upward to a height of 0.8m. How much energy is lost? What is the magnitude and direction of the impulse applied to the ball by the floor? If the ball is in contact with the floor for 0.01

0

hf 0.8 m

m 0.4 kgseconds, what is the impulse force on the ball and on the earth?Choose the gravitation potential energy to vanish at y=0

( ) ( )lost 0 f 0 fEnergy lost: E mgh mgh mg h h 0.4 9.8 1.5 0.8 J 2.7J= − = − = − =

( )after before after beforep p p m v v ; velocities just before and after the collision.Δ = − = −

( ) ( )2before 0 before 0

1 mv mgh v 2gh 2 9.8 1.5 m / s 5.4m / s2

= = − = − = −

( )

( ) ( )before 0 before 0g g2

( ) ( )2after f after f

1 mv mgh v 2gh 2 9.8 0.8 m / s 3.96m / s2

= = = =

( ) ( )after beforep m v v 0.4kg 3.96m / s 5.4m / s 3.75kg m / s upwards Δ = − = − − = ⋅

earth on ballp 3.75kg m / sF 375N upwardsΔ ⋅= = =earth _ on _ ball pt 0.01sΔ

dball _ on _ earthF 375N downwards (from Newton's 3 law)=

Bouncing balls

A i h b ll h h hi h b ll• Assuming each ball has the same mass, which ball experiences the larger impulse?– a) the first balla) the first ball– b) the second ball

Hint: Which ball has the largest change in velocity?

( )ff m vvppp −=−=Δ ( )ifif m vvpppΔ

An elastic collision has twice the impulse as a totally inelastic collisions 0collisions.

f ,elastic i= −v v f ,inelastic 0=v

( )elastic f i f i im 2mΔ = − = − = −p p p v v v ( )elastic f i f i ip p p

( )inelastic f i i im mΔ = − = − = −p p p v v