Chapter 7

Linear Momentum

Why study momentum ?- In an isolated system, net force is zero, momentum is a conserved quantity

Applications:- collision problems mass or velocity determination- discovery of missing objects or sub-atomic particles (neutrino中微子 )

http://www.hk-phy.org/articles/neutrino/neutrino.html

Newton’s Second Law

dt

dmvF

dt

dmv

dt

vdm

dt

pd

vmp

Define momentum p

Consider the rate of change of momentum0

Mass of the system does not change in general

Linear Momentum- Like energy, it is conserved

Newton actually stated his Second Law of motion as Net External force equals the change in momentum of a system divided by the time over which it changes.

Collision with a stationary objectAxis-aligning bounding box means that the sides of the box are horizontal and

vertical If the stationary boundary is vertical, vf = [-vix, viy]If the stationary boundary is horizontal, vf = [vix, -viy]For incoming velocity vi = [vix, viy]

The angle that the ball comes in at must equal the angle at which it leaves, that is the angle of incidence (incoming) must equal the angle of reflection (outgoing).

Example• Suppose you are coding a simple Pong game, and you want to model the

collision of the ball with the paddle. If the ball is approaching the paddle with an incoming velocity of [40, 75] when it hits, what should be its resulting velocity be?

AnswerThe final velocity is [-40, 75]

Example 7.1(a) Calculate the momentum of a 110Kg football player running at 8 m/s.(b) Compare the player’s momentum with that of a hard-thrown 0.41Kg

football that has a speed of 25 m/s.m=110Kg, v=8 m/s, mfb=0.41 Kg, vfb=25 m/sp/pfb = ? = 85.9

Example 7.2 What is the average force does exert on a 0.14Kg baseball by a bat, given that

the ball’s initial velocity is 45 m/s and that its final velocity, after a 1.3 ms impact, is 65 m/s in exactly the opposite direction?

m=0.14 Kg, v0=45 m/s, vf=-65 m/s, in 1.3 ms F = ? p/t = (pf – pi)/t = m(vf-v0)/0.0013 = 11800 N ~2600 lb

7.2 ImpluseEffect of a force on an object depends on how long it acts.p = (net F) tThe change in momentum equals the average net external force multiplied by the time this force acts. (net F) t is called the impulse, IImpulse is the same as change in momentum, I=p notice I // p

體操 : for a given p change, net F ↓ if t ↑ less impact

)()( 12

2

1

ttFdttFpIt

t

eff A piano hammer striking a string wouldgenerate a force similar to Factual but its impluse might be the same as that of Feff.

Example 7.3Calculate the final speed of a 110Kg football player running at 8 m/s who collides head on with a padded goalpost and experiences a backward force of 17600 N for 0.055 s.

M=110 Kg, vi=8 m/s, net F=17600 N, t=0.0550 s vf = ?

Use I = (net F) t = M(vf – vi) -17600*0.0550 = 110(vf – 8) vf = -0.800 m/s

The minus sign indicates the player bounces backward.

7.3 Conservation of momentumUnder what condition is momentum conserved ?Net external force = 0 consider a larger system

Consider the impulse p1 = F1 tp2 = F2 tNewton’s Third Law F2 = -F1

p2 = (-F1)t = -p1

p1 + p2 =0

p1 + p2 = constantPtotal = p1 + p2 = p1’ + p2’

Conservation of Momentum (isolated system, net F = 0) Ptotal = constant Ptotal = P’total

- The three dimension in nature is independent- Momentum can be conserved along one direction and not another.

- Momentum is conserved along the X-direction, but not in Y-direction

7.4 Elastic collisions in one dimension

The two collided bodies moving the along the same direction

Elastic collision – both momentum and internal kinetic energy conserved

Very nearly elastic collision because some KE heat, sound

Example 7.4 An elastic collisionCalculate the velocities of two masses following an elastic collision,given that mA = 0.500 Kg, mB = 3.5 Kg, vA = 4.00 m/s, vB = 0.What are the final velocities of mA and mB ? vA’ = ? , vB’ = ?

Since vB=0, pA = pA’ + pB’ mA vA = mA vA’ + mB vB’ ½ mAvA

2 = ½ (mA vA’2 + mB vB’2)Solve the conservation of momentum for vB’ first, and substitute intoconservation of internal KE to eliminate vB’, leaving only vA’ unknown.

There are two solutions for vA’, vA’= 4.00 m/s or -3.00 m/sFirst solution same as initial condition discarded 4.00 m/s vB’ = 1.00 m/sA small mass m, collide with a larger mass M, the larger mass is knocked forward with a lower speed (here mB = 7mA).

7.5 Inelastic collision in one dimensionInelastic collision internal kinetic energy is not conserved,some internal KE may be converted into heat or sound energy

Perfectly inelastic, KE’int = 0

Two equal masses head toward one another at equal speeds andstick together. KEint = mv2, KE’int = 0 internal KE not conservedThe two masses come to rest after collision momentum conserved

Example 7.5 Inelastic collision of puck and goalie(a) Find the recoil velocity of a 70 Kg hockey goalie, originally at rest, who

catches a 0.15 Kg hockey puck slapped at him at a velocity of 35 m/s. (b) How much KE is lost in the collision? Assume friction between the ice and

the puck-goalie system is negligible.m1=0.15 Kg, v1=35 m/s, m2=70.0 Kg, v2 = 0(a) v1’=v2’=v’= ?(b) How much KE is lost in collision ?

(a) m1 v1 + m2 v2 = (m1 + m2) v’ solve for v’ v’ = 0.0748 m/s(b) KE’int – KEint = ½ (m1+m2)v’2 – ½ m1v1

2

= -91.7 J

Example 7.6 A collision that releases stored energy is inelasticTwo carts collide inelastically, and a spring releases its PE and converts it into internal KE. Mass of the cart on the left is m1= 0.35Kg, v1= 2.00m/s. Cart on the right has m2= 0.5Kg, v2= -0.5m/s. After collision, the first cart has a recoil velocity v1’= -4.00m/s.(a) What is the final vel. of the first cart, v2’=? (b) how much energy is released by the spring (assume all internal PE is converted into internal KE)

(a) m1v1 + m2v2 = m1v1’ + m2v2’Solve the conservation of momentum for v2’ first, v2’=3.7m/s(b) KEint = ½ (m1v1

2 + m2 v22)= 0.763 J

KE’int = ½ (m1 v1’2 + m2 v2’2) = 6.22 J KE’int – KEint = 5.46 J

(internal KE increased energy released by the spring

7.6 Collision of point masses in two dimensions

X component : m1v1 = m1v1’cos1 + m2v2’cos2

Y component : 0 = m1v1’sin1 + m2v2’sin2

Complication, if object rotationPoint massesNeglect rotation

Many scatteringexperiments have atarget mass that isstationary in laboratory

Example 7.7 Determine the final velocity of an unseen mass from the scattering of another massA 0.25Kg mass is slid on a frictionless surface into a dark room, where it strikes an object with m2=0.4, v2=0. The 0.25Kg mass emerges from the room at an angle of 45 with its incoming direction, The speed of the 0.25Kg mass was v1=2 m/s, and it is 1.5 m/s after the collision. Calculate the magnitude and direction of the velocity (v’2 and h) of the 0.4Kg mass after the collision.

X : m1v1 = m1v1’cos45 + m2v2’cos2

Y : 0 = m1v1’sin45 + m2v2’sin2

m1= 0.250Kg, m2=0.400Kgv1= 2.00m/s, v2=0v1’=1.5 m/s

solve for sin2 and cos2 tan2

2 = -48.5o

Given v2’ one can determine mass of the unseen mass m2

Elastic Collisions of Two Equal MassesSee p.185- a special case m1 = m2 like the case with billiard balls, or the case with some subatomic collisions- assume v2 =0, and elastic collisions internal KE conserve½ mv1

2 = ½ m (v1’2 + v2’2)X and Y components of momentum conserveX component : m1v1 = m1v1’cos1 + m2v2’cos2

Y component : 0 = m1v1’sin1 + m2v2’sin2

(X component)2 : v12= (v1’cos1 + v2’cos2)2

(Y component)2 : 0 = (v1’sin1 + v2’sin2)2

½ mv12 = ½ m (v1’2 + v2’2) + m v1’v2’ cos(1 + 2)

m v1’v2’ cos(1 + 2)=0(1) v1’= 0 head on collision, incoming mass stop(2) v2’ = 0 no collision, incoming mass unaffected(3) cos(1 + 2) = 0 angle of separation is 90o after collision

Particle Accerlator CERN - European Organization for Nuclear Research

27 Km in diameter

World Class Particle accelerator

• CERN• http://cernenviro.web.cern.ch/CERNenviro/web/main/main

.php• Large Hadron Collider (LHC)• http://livefromcern.web.cern.ch/livefromcern/antimatter/hi

story/historypictures/LHC-drawing-half.jpg• FermiLab• http://www.cs.cmu.edu/~zollmann/pics/2004_10_chicago/

slides/fermilab.html• Stanford Linear Accelerator Center (SLAC) • http://www.pbs.org/wgbh/nova/einstein/toda-ocon-01.html

A submicroscopic particle scatters straight backward from a target particle. In experiments seeking evidence for quarks, electrons were observed t occasionally scatter straight backward from a proton.

7.7 Rockect Propulsion