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? The example of the ice skaters A) Yes starting from rest and pushing off of each other...the total momentum B) No is zero because they travel in opposite directions but the kinetic energy is not zero (they start with none and gain a bunch)

Lecture 18 Purdue University, Physics 149 2 ILQ 2

An astronaut is taking a space walk when his tether 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 Purdue University, Physics 149 3 Momentum

• Momentum is a vector quantity. • Units: kg⋅m/s, N⋅s – Unit conversion: 1 kg⋅m/s = 1 N⋅s • Momentum is denoted by p.

• Momentum is sometimes referred to as “linear moment”tditiihitftum” to distinguish it from angu lar momen tum.

Lecture 18 Purdue University, Physics 149 4 Momentum is Conserved

• Momentum is “Conserved” meaning it can not be created nor destroyed – Can be transferred

• Total Momentum does not change with time

• Momentum is a VECTOR 3 Conservation Laws in one!

Lecture 18 Purdue University, Physics 149 5 Momentum is Conserved • Define Momentum p = m v –m Δv = -m Δv 1 1 2 2 Momentum is conserved – Δp1 = -Δp2 p1f-p1i+p2f-p2i=0 – Δp1 + Δp2 = 0 p +p =p +p – Σpinitial = Σpfinal 1f 2f 1i 2i • Example: Jane pushes Fred so he is going 2.0 m/s. If Fred is twice as heavy as Jane , how fast does Jane end up moving?

– Σpinitial = Σpfinal

–0 = mFred vFred + mJane vJane

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

Lecture 18 Purdue University, Physics 149 6 Momentum Transfer

• During an interaction, momentum is transferred from one bodyy, to another, but the total momentum of the two is unchanged. G G G G G G Δp2 = −Δp1 or p1i + p2i = p1 f + p2 f • Consider a collision between two spaceships. G G G G v1i v2i v1 f v2 f

G G G G G G F12 G G G G F21 G G Δv1 = a1Δt = Δt ⇒ Δp1 = m1Δv1 = F12Δt Δv2 = a2Δt = Δt ⇒ Δp2 = m2Δv2 = F21Δt m1 m2 G G G G G G G G F21 = −F12 ⇒ Δp2 = m2Δv2 = F21Δt = (−F12 )Δt = −m1Δv1 = −Δp1 G G Thus, Δp2 = −Δp1 Lecture 18 Purdue University, Physics 149 7 Impulse

• Impulse is the product of (average) force and time interval during which the force acts. G G I = FΔt • The changgjqe in momentum of an object is equal to the impulse (recall a collision between two spaceships). G G G Δp = I (= FΔt)

• 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.

Lecture 18 Purdue University, Physics 149 8 Impulse Changes Momentum

• Define Impulse I = FΔt – Change in momentum requires Force acting over a time or Impulse – Recall: mΔv = FΔt –Or Δp = FΔt Δp = I

Impulse I = FΔt Momentum p = mv

Lecture 18 Purdue University, Physics 149 9 ILQ

By what factor does an object’s kinetic energy change if its speed is doubled? By what factor does the momentum change? A)), 4, 2 B) 2, 2 C) 0.5, 4

GG 1 pmv== K mv2 2

Lecture 18 Purdue University, Physics 149 10 Impulse and Momentum

I ≡ FaveΔt= pf -pi = Δp

• For single object … –F = 0 ⇒ momentum conserved (Δp= 0)

• For collection of objects …

–ptotal = Σp – Internal forces: forces between objects in system – External forces: all other forces

– ΣFext = 0 ⇒ total momentum conserved (Δptot = 0)

–Fext = mtotal a

Lecture 18 Purdue University, Physics 149 11 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 the interaction must be the same.

• The total momentum of a system is the vector sum of the momenta 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 suffici ent ly s hort amount o f t ime t hat externa l forces may be ignored; conservation of momentum may be used. Lecture 18 Purdue University, Physics 149 12 ILQ Two 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 floor w it hout bounc ing. In w hic h case is t he magn itu de o f t he impu lse given to the ball by the floor the biggest?

A) Case 1 Δp=mΔv = maΔt = FΔt = I B) Case 2 C) The same Bouncing Ball Sticky Ball |I| = |Δp| |I| = |Δp|

= |mvfinal – m vinitial| = |mvfinal – m vinitial|

= |m( vfinal -vinitial)| = |m(0 - vinitial)| = 2 m v = m v

Lecture 18 Purdue University, Physics 149 13 ILQ Two 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 . 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 G GG G I =Δpmvmv =21 −

time

Lecture 18 Purdue University, Physics 149 14 Momentum is a Vector A 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: P – P = (0 – 48000) = -48,000 kg m/s final initial 30 m/s East-West:

Pfinal –Pinitial = (36000 - 0) = +36,000 kg m/s 40 m/s Magnitude: 2 2 sqrt(P North + P East ) = 60,000 kg m/s Lecture 18 Purdue University, Physics 149 15 ILQ You 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 I = Δp C) the same Same change in momentum

In which case is the average force greater A) Floor Δp= F Δt B) Foam F = Δp/Δt C) the same smaller Δt = large F Lecture 18 Purdue University, Physics 149 16 Example: Time to Burn

For a safe re-entry into the Earth's atmosphere the pilots of a space capsule must reduce their speed from 2.6 × 104 m/s to 1.1 × 104 m/s. The rocket enggpine produces a backward force on the ca psule of 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 Burn 4 vi=2.6×10 m/s 4 vf=1.1×10 m/s F=1.8×105 N M=3800 Kg Δt=?

FΔt = mΔv mΔv m(v − v ) Δt = = f i = F F 3800kg(2.6 − 1.1)×104 m / s 1N = × = 317s 1.8×105 N 1Kgm / s2

Lecture 18 Purdue University, Physics 149 18 Example

• A fright train is being assembled in switching yard. Car 1 3 has a mass of m1=65 ×10 kg and moves at a velocity 3 v01=0800.80 m /s. C2hCar 2 has a mass o f m2=92 ×10 kdkg and moves at a velocity v02=1.3 m/s and couples to it. Neglecting friction, find the common velocity vf of the cars after th ey b ecome coupl e d.

v02 v01 m2 m1 initial

vf m2 m1 final

Lecture 18 Purdue University, Physics 149 19 Example • Note the velocity of car 1 increases while the velocity of car 2 decreases. • The accel erati on and dece lera tion ar ise 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)

v02 v01 m2 m1 initial

v f flfinal m2 m1 Lecture 18 Purdue University, Physics 149 20 Example

• A fright train is being assembled in switching yard. Car 1 3 has a mass of m1=65 ×10 kg and moves at a velocity 3 v01=0800.80 m /s. C2hCar 2 has a mass o f m2=92 ×10 kdkg and moves at a velocity v02=1.3 m/s and couples to it. Neglecting friction, find the common velocity vf of the cars after th ey b ecome coupl e d. Apply conservation of momentum:

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

(m1+m2)vf = m1v01+m2v02 mv101+ mv 202 v f == mm12+ (65×+× 1033kg )(0.8 m / s ) (92 10 kg )(1.3/) 3 m / s ) =1.1ms / (65×+× 1033kg 92 10 kg ) Lecture 18 Purdue University, Physics 149 21 ILQ

At the instant a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. Which object has the greater kinetic energy? Or are they the same A) The same B) The gun Apply conservation of momentum: C) The bullet Pi=0 Pf=MgunVgun+mbulletvbullet

Vgun=-(mbullet/Mgun)vbullet

2 11⎛⎞mm KMVM==22bullet v = bullet K gun gun gun gun⎜⎟ bullet bullet 22⎝⎠MMgun gun

Lecture 18 Purdue University, Physics 149 22 Impulse-Momentum Theorem

• The “total” impulse on an object is equal to the change in the object’s momentum during the same time interval.

G = Itotal

• Example: – 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 collision.

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

Lecture 18 Purdue University, Physics 149 23 Newton’s Second Law

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

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

Lecture 18 Purdue University, Physics 149 24 Newton’s Second Law G G ∑ F = ma G G New IFtp=Δ=Δ∑ formulation of G G Δp N2L F = ∑ Δt

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. After the impulse, the object has

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

Lecture 18 Purdue University, Physics 149 26 ILQ

• An 18-wheeler and a Volkswagen Beetle are rolling along with the same momentum. If you exert the same force with the brakes to stop each one,,g which takes a longer time to bring to rest?

a) 18-wheeler b) Volkswagen beetle c) same for both d) impossible to say

Lecture 18 Purdue University, Physics 149 27 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 ener gy (recall Chapter 6). – 0 + mgh = ½ mv2 + 0 Æ v = 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 diduring t hat t ime interva l? – 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 Purdue University, Physics 149 28