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PHYSICS 149: Lecture 18 Chapter 7: Linear Momentum 7.2 Momentum 7.3 The Impulse-Momentum Theorem 7.4 Conservation of Momentum Lecture 18 Purdue University, Physics 149 1

# 7.2 Momentum – 7.3 The Impulse-Momentum Theorem – 7.4

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PHYSICS 149: Lecture 18• Chapter 7: Linear Momentum

– 7.2 Momentum

– 7.3 The Impulse-Momentum Theorem

– 7.4 Conservation of Momentum

Lecture 18 Purdue University, Physics 149 1

ILQ 1

Two objects are known to have the same kinetic energy. Do these two objects necessarily have the same momentum?

A) Yes The example of the ice skaters starting from rest and pushing off of each other...the total momentum

B) Noof each other...the total momentum is zero because they travel in opposite directions but the kinetic energy is not zero (they start with energy is not zero (they start with none and gain a bunch)

Lecture 18 Purdue University, Physics 149 2

ILQ 2An astronaut is taking a space walk when his tether breaks He has a big drill How can hetether breaks. He has a big drill. How can he get back to the spaceship?

A) Face away from the ship and throw the drill forward.

B) Face the ship and throw the drill forward.

C) Throw the drill towards the tether.

Lecture 18 3Purdue University, Physics 149

Momentum

• Momentum is a vector quantity• Momentum is a vector quantity.• Units: kg⋅m/s, N⋅s

– Unit conversion: 1 kg⋅m/s = 1 N⋅sMomentum is denoted by p• Momentum is denoted by p.

• Momentum is sometimes referred to as “linear t ” t di ti i h it f l tmomentum” to distinguish it from angular momentum.

Lecture 18 4Purdue University, Physics 149

Momentum is Conserved• Momentum is “Conserved” meaning it can not be

created nor destroyedcreated nor destroyed – Can be transferred

• Total Momentum does not change with time

• Momentum is a VECTOR3 Conservation Laws in one!3 Conservation Laws in one!

Lecture 18 Purdue University, Physics 149 5

Momentum is Conserved• Define Momentum p = m v

– m1 Δv1 = -m2 Δv2 Momentum is conserved – Δp1 = -Δp2

– Δp1 + Δp2 = 0

Momentum is conserved p1f-p1i+p2f-p2i=0

p +p =p +p– Σpinitial = Σpfinal

• Example: Jane pushes Fred so he is going 2.0 m/s. If Fred is twice as heavy as Jane how fast does

p1f+p2f=p1i+p2i

If Fred is twice as heavy as Jane, how fast does Jane end up moving?– Σp = Σp– Σpinitial = Σpfinal

– 0 = mFred vFred + mJane vJane

– vJane = -mFred vFred / mJane = -4 m/s

Lecture 18 Purdue University, Physics 149 6

Jane Fred Fred Jane

Momentum Transfer• During an interaction, momentum is transferred from one

body to another, but the total momentum of the two is y ,unchanged.

ffii pppporpp 212112 +=+Δ−=Δ

• Consider a collision between two spaceships.iv2 fv1 fv2iv1

tFvmptmFtav Δ=Δ=Δ⇒Δ=Δ=Δ 12111

1

1211 tFvmpt

mFtav Δ=Δ=Δ⇒Δ=Δ=Δ 21222

2

2122

)( pvmtFtFvmpFF Δ−=Δ−=Δ−=Δ=Δ=Δ⇒−= 11112212221221 )( pvmtFtFvmpFF Δ=Δ=Δ=Δ=Δ=Δ⇒=

12, ppThus Δ−=ΔLecture 18 7Purdue University, Physics 149

Impulse• Impulse is the product of (average) force and time interval

during which the force acts.

• The change in momentum of an object is equal to the

tΔ= FIg j q

impulse (recall a collision between two spaceships).

)F(Ip tΔ==Δ

• Impulse is a vector quantity

)F(Ip tΔΔ

Impulse is a vector quantity.• Units: N⋅s, kg⋅m/s

– Unit conversion: 1 N⋅s = 1 kg⋅m/s• Impulse is often denoted by I• Impulse is often denoted by I.

Lecture 18 8Purdue University, Physics 149

Impulse Changes Momentum• Define Impulse I = FΔt

Change in momentum requires Force acting– Change in momentum requires Force acting over a time or Impulse

– Recall: mΔv = FΔt– Or Δp = FΔt

Δp = I

Impulse I = FΔtImpulse I = FΔt

Momentum p = mv

Lecture 18 Purdue University, Physics 149 9

ILQBy what factor does an object’s kinetic energy change if its speed is doubled? By what factorchange if its speed is doubled? By what factor does the momentum change?

A) 4, 2 ) ,B) 2, 2 C) 0.5, 4C) 0.5, 4

21p mv K mv2

p mv K mv= =

Lecture 18 Purdue University, Physics 149 10

Impulse and Momentum

I ≡ FaveΔt = pf - pi = Δp

• For single object …

• For collection of objects

– F = 0 ⇒ momentum conserved (Δp = 0)

• For collection of objects …– ptotal = Σp– Internal forces: forces between objects in systemInternal forces: forces between objects in system– External forces: all other forces– ΣFext = 0 ⇒ total momentum conserved (Δptot = 0)

Lecture 18 Purdue University, Physics 149 11

ext ( ptot )– Fext = mtotal a

Conservation of Momentum

• Momentum is a vector quantity, so both the magnitude and the direction of the total momentum at the beginning and end of thedirection of the total momentum at the beginning and end of the interaction must be the same.

• The total momentum of a system is the vector sum of the ymomenta of each object in the system.

• Internal interactions do not change the total momentum of a system.

• External interactions can change the total momentum of a system.

• During a collision, the interaction often takes place during a ffi i l h f i h l f bsufficiently short amount of time that external forces may be

ignored; conservation of momentum may be used.Lecture 18 12Purdue University, Physics 149

ILQTwo identical balls are dropped from the same height onto the floor. In each case they have velocity v downward just before hitting the floor. In case 1 the ball bounces back up, and in case 2 the ball sticks to the fl i h b i I hi h i h i d f h i lfloor without bouncing. In which case is the magnitude of the impulse given to the ball by the floor the biggest?

A) Case 1 Δ mΔ m Δt FΔt IA) Case 1

B) Case 2

C) The same

Δp=mΔv = maΔt = FΔt = I

Bouncing Ball Sticky BallC) The same Bouncing Ball

|I| = |Δp|

| |

Sticky Ball

|I| = |Δp|

| |= |mvfinal – m vinitial|

= |m( vfinal - vinitial)|

= |mvfinal – m vinitial|

= |m(0 - vinitial)|

Lecture 18 Purdue University, Physics 149 13

= 2 m v = m v

ILQTwo identical balls are dropped from the same height onto the floor. In case 1 the ball bounces back up, and in case 2 the ball sticks to the floor without bouncing Incase 2 the ball sticks to the floor without bouncing. In both cases of the above question, the direction of the impulse given to the ball by the floor is the same. What is this direction?

A) Upward

B) Downward

2 1I p mv mv= Δ = −

Lecture 18 Purdue University, Physics 149 14

time

Momentum is a VectorA car with mass 1200 kg is driving north at 40 m/s, and turns east driving 30 m/s. What is the magnitude of the car’s change in momentum?A) 0 B) 12,000 C) 36,000 D) 48,000 E) 60,000

pinitial = m vinitial = (1200 kg) x 40 m/s = 48,000 kg m/s North

pfinal = m vfinal = (1200 kg) x 30 m/s = 36,000 kg m/s East

North-South:Pfinal – Pinitial = (0 – 48000) = -48,000 kg m/s

East-West:

30 m/s

Pfinal – Pinitial = (36000 - 0) = +36,000 kg m/s

Magnitude:

40 m/s

Lecture 18 Purdue University, Physics 149 15

Magnitude:sqrt(P2

North + P2East ) = 60,000 kg m/s

ILQYou drop an egg onto A) the floor B) a thick piece of foam rubber. In both cases, the egg does not bounce.In which case is the impulse greater?

A) Floor

B) Foam

C) the same

I = Δp

Same change in momentumC) the same

In which case is the average force greater

Same change in momentum

A) Floor

B) Foam

Δp = F ΔtF = Δp/Δt

Lecture 18 Purdue University, Physics 149 16

)

C) the same

F Δp/Δtsmaller Δt = large F

Example: Time to BurnFor a safe re-entry into the Earth's atmosphere the pilots of a space capsule must reduce their speedpilots of a space capsule must reduce their speed from 2.6 × 104 m/s to 1.1 × 104 m/s. The rocket engine produces a backward force on the capsule g p pof 1.8 × 105 N. The mass of the capsule is 3800 kg. For how long must they fire their engine? [Hint: Ignore the change in mass of the capsule due to the expulsion of exhaust gases.]

Lecture 18 Purdue University, Physics 149 17

Example: Time to Burnvi=2.6×104 m/s vf=1.1×104 m/sf

F=1.8×105 NM=3800 KgM 3800 KgΔt=?

vvmvmt

vmtF

if )(=

−=

Δ=Δ

Δ=Δ

sNsmkgFF

t

3171/10)1.16.2(3800 4

=××−

=

===Δ

ssKgmN

317/1108.1 25 ×

×

Lecture 18 18Purdue University, Physics 149

Example• A fright train is being assembled in switching yard. Car 1

has a mass of m1=65 ×103 kg and moves at a velocity 0 80 / C 2 h f 92 103 k dv01=0.80 m/s. Car 2 has a mass of m2=92 ×103 kg and

moves at a velocity v02=1.3 m/s and couples to it. Neglecting friction, find the common velocity vf of the cars ft th b l dafter they become coupled.

m2 m1v02 v01

initial2 1

m2 m1vf final

Lecture 18 Purdue University, Physics 149 19

Example• Note the velocity of car 1 increases while the velocity of car 2

decreases. Th l ti d d l ti i b th• The acceleration and deceleration arise because the cars exert internal forces on each other.

• Momentum conservation allows us to determine the change in velocity without knowing what the internal forces are.

• Note that often momentum is conserved but the kinetic energy is not! (Kinetic energy is conserved only in elastic collisions)is not! (Kinetic energy is conserved only in elastic collisions)

m2 m1v02 v01 initial2 1

vf f lLecture 18 Purdue University, Physics 149 20

m2 m1vf final

Example• A fright train is being assembled in switching yard. Car 1

has a mass of m1=65 ×103 kg and moves at a velocity 0 80 / C 2 h f 92 103 k dv01=0.80 m/s. Car 2 has a mass of m2=92 ×103 kg and

moves at a velocity v02=1.3 m/s and couples to it. Neglecting friction, find the common velocity vf of the cars ft th b l dApply conservation of momentum:

Pi=m1vo1+m2v02 Pf=(m1+m2)vf

after they become coupled.

Pi m1vo1 m2v02 Pf (m1 m2)vf

(m1+m2)vf = m1v01+m2v02

1 01 2 02m v m v+1 01 2 02

1 23 3(65 10 )(0 8 / ) (92 10 )(1 3 / )

fvm m

kg m s kg m s

= =+

× + ×

Lecture 18 Purdue University, Physics 149 21

3 3

(65 10 )(0.8 / ) (92 10 )(1.3 / ) 1.1 /(65 10 92 10 )

kg m s kg m s m skg kg

× + ×=

× + ×

ILQAt the instant a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta Which objectthe gun have equal and opposite momenta. Which object has the greater kinetic energy? Or are they the same

A) The same Apply conservation of B) The gun

C) The bullet

Apply conservation of momentum:Pi=0 Pf=MgunVgun+mbulletvbulletPi 0 Pf MgunVgun mbulletvbullet

Vgun=-(mbullet/Mgun)vbullet

2

2 21 12 2

bullet bulletgun gun gun gun bullet bullet

m mK M V M v K

M M⎛ ⎞

= = =⎜ ⎟⎜ ⎟

Lecture 18 Purdue University, Physics 149 22

2 2gun gun gun gun bullet bulletgun gunM M

⎜ ⎟⎜ ⎟⎝ ⎠

Impulse-Momentum Theorem• The “total” impulse on an object is equal to the change in

the object’s momentum during the same time interval.

totalI=

• Example:

total

– Today’s automobiles are designed with a number of safety features, all designed to make the impact last longer (Δt) and so applied force weaker (F). That way we can minimize injury upon collisioncollision.

pi = mvi, pf = mvf = 0 |Δp| = |pf – pi| = |–mvi| = |F|ΔtThe longer Δt the weaker |F|The longer Δt, the weaker |F|.

Lecture 18 23Purdue University, Physics 149

Newton’s Second Law

• The net force is the rate of change of momentum• The net force is the rate of change of momentum.

• This relation is valid even when mass is not constant (just like a rocket engine).g )

Lecture 18 24Purdue University, Physics 149

Newton’s Second Law

F ma=∑

I F t p= Δ = Δ∑

∑New I F t p

pF

= Δ = Δ

Δ∑

∑formulation of N2LpF

t=Δ∑

Lecture 18 Purdue University, Physics 149 25

ILQ• A 3.0 kg object is initially at rest. It then

receives an impulse of magnitude 15 N⋅s Afterreceives an impulse of magnitude 15 N⋅s. After the impulse, the object has

A) a momentum of magnitude 5.0 kg⋅m/s. B) a speed of 45 0 m/sB) a speed of 45.0 m/s. C) a momentum of magnitude 15.0 kg⋅m/s. D) a speed of 7 5 m/sD) a speed of 7.5 m/s.

Lecture 18 26Purdue University, Physics 149

ILQ• An 18-wheeler and a Volkswagen Beetle are

rolling along with the same momentum If yourolling along with the same momentum. If you exert the same force with the brakes to stop each one, which takes a longer time to bring to , g grest?

a) 18-wheelerb) Volkswagen beetle) gc) same for bothd) impossible to sayd) impossible to say

Lecture 18 27Purdue University, Physics 149

Example• A pole-vaulter of mass 60.0 kg vaults to a height of 6.0 m

before dropping to thick padding placed below to cushion her fall.(a) Find the speed with which she lands.– Gravity (conservative force) is the only force acting on her before

landing, so we may use the conservation of mechanical energy g, y gy(recall Chapter 6).

– 0 + mgh = ½ mv2 + 0v = sqrt(2gh) = sqrt(2 ⋅ 9.8m/s2 ⋅ 6.0m) = 10.8 m/s

(b) If the padding brings her to a stop in a time of 0.50 s, what is the average force on her body due to the padding d i h i i l?during that time interval?– Use the impulse-momentum theorem– Δp = pf – pi = mvf – mvi = 60.0 kg [0 – (–10.8 m/s)]

= 648 kg⋅m/s, upward F = Δp/Δt = 648 kg⋅m/s / 0.50 s = 1,296 N, upward

Lecture 18 28Purdue University, Physics 149