Chapter 7 Impulse And Momentum Copy - Physics.umanitoba.ca

Chapter 7Impulse and Momentum

1) Linear momentump mv(units: kg m / s)

2) Impulse (produces a finite change in momentum)(a) Constant force:! !J FΔtFrom the 2nd law,!!! Δ(mv)ΔpF ,ΔtΔt! !!J FΔt Δpso(Units: Ns)

(b) Variable force:! !!J FΔt ΔpIf the force is not constant, use the average force

Impulse-momentum theorem! !!J FΔt ΔpImpulseChange in momentum

ExampleC&J 7.9 A space probe is traveling in outer space with amomentum that has a magnitude of 7.5 107 kg·m/s. Aretrorocket is fired to slow down the probe. It applies a forceto the probe that has a magnitude of 2.0 106 N and a directionopposite to the probe’s motion. It fires for a period of 12 s.Determine the momentum of the probe after the retrorocketceases to fire.

3) Conservation of Momentum (andNewton’s laws)Second law:Superposition:! Δp!F Δt!!ΔPsysFsys ΔtIf the net external force on a system is zero:!ΔPsys 0Δt! Psys const

!Psys const Momentum of an isolated system is constant Always conserved; cannot be randomized (internalized) like energy

!!!P0 m1v01 m2 v02Internal forces are equal and opposite, anddo not change momentum of the system.!!!Pf m1v f 1 m2 v f 2!!Pf P0

ExampleImagine two balls colliding on a billiard table that is frictionfree. Use the momentum conservation principle in answeringthe following questions. (a) Is the total momentum of the twoball system the same before and after the collision? (b)Answer part (a) for a system that contains only one of the twocolliding balls.Image reprinted with permission of John Wiley and Sons, Inc.

Example Ice SkatersStarting from rest, two skaterspush off against each other onice where friction is negligible.One is a 54-kg woman andone is a 88-kg man. The womanmoves away with a speed of 2.5 m/s. Find the recoil velocityof the man.

Applying the Principle of Conservation of Linear Momentum1. Decide which objects are included in the system.2. Relative to the system, identify the internal and external forces.3. Verify that the system is isolated.4. Set the final momentum of the system equal to its initial momentum.Remember that momentum is a vector.

ExampleFor tests using a ballistocardiograph, a patient lies on ahorizontal platform that is supported on jets of air. Each timethe heart beats, blood is pushed out from the heart in adirection that is nearly parallel to the platform. The body andplatform recoil, and this recoil can be detected. Suppose that0.050 kg of blood is pushed out of the heart with a velocity of0.25 m/s and that the mass of the patient platform is 85 kg.Assuming that the patient does not slip wrt the platform, andthat the patient and platform start from rest, determine therecoil velocity of the platform.

Example (Homework)Two friends, Al and Jo, have a combined mass of 168 kg.At an ice skating rink they stand close together on skates, atrest and facing each other, with a compressed spring betweenthem. The spring is kept from pushing them apart becausethey are holding each other.When they release their arms, Al moves off in one direction ata speed of 0.90 m/s, while Jo moves off in the oppositedirection at a speed of 1.2 m/s. Assuming that friction isnegligible, find Al's mass.

4) 1-d collisions!v1!v2!′v1!′v2!!!′!′m1v1 m2 v2 m1v1 m2 v2One equation, two unknowns;initial conditions are not enough (even in 1d)

a) Completely inelastic collisions!!!′!′m1v1 m2 v2 m1v1 m2 v2!′ !′ !v1 v2 v′

Examplem1 1.0 kg, v1x 10 m/sm2 2.0 kg, v2x -8 m/sIf they stick together, find final velocity:

Equal masses, opposite momentavvmmv’ 0All kinetic energy lostOpposite of explosion

Equal masses, one at restvmmv’ v / 21/2 kinetic energy is lost

m1 m2vm1m2 1000 m1m1v′ vm1 m21 thousandth of KE survives

b) Completely elastic collisionsKE conserved

Elastic collision soluble:!v1!v2!′v2!′v1Conservation of momentum:!!!′!′m1v1 m2 v2 m1v1 m2 v2Conservation of energy:12m v m v m1v1′ m2 v2′21 11222 212212Two equations, two unknowns (v1’, v2’)2

!v1!v2!′v1!′v2Result for elastic collision in one dimension with v2 0.m1 m2v1′ v1m1 m22m1v2′ v1m1 m2(x-components,not magnitudes)

elastic collision with m1 m2, v2 0!v1m1 m2v1′ v1 0m1 m22m1v2′ v1 v1m1 m2

Newton’s cradleConservation of momentum:v1 v1′ v2′v1Conservation of mech energy:v v1′ v2′212v2′v1′2v1v2′v1′For both to be true, one of the final velocities must be zero.

elastic collision with m1 m2, v2 0m1 m2v1′ v1 v1m1 m22m1v2′ v1 0m1 m2Larger mass acquires negligible KE

Bouncing ball (floor has infinite mass)hh’Energy conserved mgh mgh’If h’ h, some mechanical energy lost in the collisionVelocities:v 2ghv' 2gh'h ′ v′ h v 2

elasticinelasticcompletelyinelastic

elastic collision with m1 m2, v2 0v1m1 m2v1′ v1 v1m1 m22m1v2′ v1 2v1m1 m2

Example Sling-shot effectIf a small mass with speed v collideselastically with a large mass at speedV, find the final speed of the smallmass.Answer: v 2V

The trajectories that enabled NASA's twin Voyager spacecraft totour the four gas giant planets and achieve velocity to escape oursolar system (http://en.wikipedia.org/wiki/Slingshot effect)

ExampleIf a small ball is dropped with and above a larger ballfrom a height h, and all collisions are elastic, howhigh does the smaller ball rebound?A)B)C)D)E)h2h3h6h9h

ExampleA 1055 kg van, stopped at a traffic light, is hit directly in therear by a 715 kg car traveling with a velocity of 2.25 m/s.Assume that the transmission of the van is in neutral, thebrakes are not being applied and that the collision is elastic.What is the final velocity of the car and the van?m1 m2v1′ v1m1 m2 -0.43 m/s2m1v2′ v1m1 m2 1.82 m/s

ExampleBallistic PendulumThe mass of the block of woodis 2.50-kg and the mass of thebullet is 0.0100-kg. The blockswings to a maximum height of0.650 m above the initial position.Find the initial speed of thebullet.

6) Conservation of momentum in 2d! !!′ ! ′p1 p2 p1 p2v1!!!′!′m1v1 m2 v2 m1v1 m2 v2v2m1v1x m2 v2 x m1v1x ′ m2 v2 x ′m1v1y m2 v2 y m1v1y′ m2 v2 y′If initial conditions are known, thisgives 2 equations, with 4 unknowns,so more information is needed.v1’v2’

a) Inelastic collision: p1’ p2’ p’ reduces unknowns to px’and py’Example: Find v’ if m1 1450 kg, m2 1750 kg, v1 11.5 m/s, v2 15.5 m/ss conservation of x momentumm2m1v1x m2 v2 x m1v1x ′ m2 v2 x ′m1v1x 0 (m1 m2 )vx ′v2m1xvx ′ v1θyv ' vx ′ 2 vy′ 2 9.95 m/stan θ vy ′vx ′ θ 58.4ºv’m1v1x 5.21 m/sm1 m2 conservation of y momentumm1v1y m2 v2 y m1v1y′ m2 v2 y′0 m2 v2 y (m1 m2 )vy′vy ′ m2v2 y 8.48 m/sm1 m2

b) Elastic collision: Energy conservation adds 3rdequation:12mv12 12 mv2 2 12 mv1′ 2 12 mv2′ 2The last condition is determined by the shape & location of impact:e.g.FFor a billiard ball collision, the angle of the object ball isdetermined by the line through the centres at the point of contact.

Example: Cue ball angle:Fv1 identical masses elastic collision m2 initially at restθ1Find Φ θ1 θ 2θ2?Conservation of momentum! !′ ! ′v1 v1 v2Φv1’v2’v1v2’v1’Conservation of energyv12 v1′ 2 v2′ 2Therefore, by PythagorasΦ 90º

Example: Cue ball angle:v2’ identical masses elastic collision m2 initially at restFv1Conservation of momentum! !′ ! ′v1 v1 v2Φv1’v2’v1v1’Conservation of energyv12 v1′ 2 v2′ 2Therefore, by PythagorasΦ 90º

Example2 pucks collide on an air hockey table.mA 0.025 kg and A is initially moving with a velocity of 5.5 m/s.It collides with B (mass 0.050 kg) which is initially at rest.After the collision the 2 pucks fly apart with the angles shown. Find the finalspeeds of A and B.

Example2 pucks collide on an air hockey table. mA 0.025 kg and A isinitially moving with a velocity of 5.5 m/s. It collides with B (mass0.050 kg) which is initially at rest. After the collision the 2 pucks flyapart with the angles shown. Find the final speeds of A and B.Cons y momentummA vAy′ mB vBy′ 0(1)Cons x momentummA vAx ′ mB vBx ′ mA vAx(1) — mA vA′ sin 65º mB vB′ sin 37º 0 — (2) — m v ′ cos 65º m v ′ cos 37º m vA AB BA A— m v ′ sin 65º mA vA′ cos 65º mB A A cos 37º mA vA mB sin 37º (2)′ sin 65ºmvAAvB ′ mB sin 37º

m v ′ sin 65º mA vA′ cos 65º mB A A cos 37º mA vA mB sin 37º Solve forvA′ 3.4 m/sThen′ sin 65ºmvAAvB ′ mB sin 37º 2.6 m/s

8) Centre of Massa) Acceleration and force:The centre-of-mass of a system of particles (or 3d object)reacts to the total force like a point particle with a mass equalto the total mass.!!Ftotal mtotal aCMIf the total force is zero, the centre-of-mass does notaccelerate.

b) Position of the centre-of-mass!xCM ! mi xi miFor 2 masses:!xCM!!m1 x1 m2 x2 m1 m2In one dimension,For m1 5.0 kg, m2 12 kg, x1 2.0 m, and x2 6.0 m,xCM 4.8 m

!xCM!!m1 x1 m2 x2 m1 m2b) Velocity of the centre-of-massFor 2 masses:!vCM!!Δx1Δx2!m m12ΔxCMΔtΔt Δtm1 m2!vCM!!m1v1 m2 v2 m1 m2numerator is total momentum(vCM constant if ext force is zero)CM momentum is simply equal to the total momentum:!!!(m1 m2 )vCM m1v1 m2 v2

!vCM!!m1v1 m2 v2 m1 m2b) Acceleration of the centre-of-massFor 2 masses:!aCM!!Δv1Δv2!m m12ΔvCMΔtΔt Δtm1 m2!aCM!!m1a1 m2 a2 m1 m2satisfies the definition:!!Ftotal mtotal aCMnumerator is total force

!xCM ! mi xi miFor 2 masses:!xCM!vCM!aCM!!m1 x1 m2 x2 m1 m2!!m1v1 m2 v2 m1 m2!!m1a1 m2 a2 m1 m2

ExampleTwo people are standing on a 2.0-m-long platform, one at each end.The platform floats parallel to the ground on a cushion of air, like ahovercraft.One person throws a 6.0-kg ball to the other, who catches it.The ball travels nearly horizontally.Excluding the ball, the total mass of the platform and people is 118 kg.Because of the throw, this 118-kg mass recoils.How far does it move before coming to rest again?

ExampleC&J 7.35 A projectile (mass 0.20 kg) is fired andembeds itself in a target (mass 2.50 kg). The target,with the projectile in it, flies off after being struck.What percentage of the projectile’s incident KE doesthe target (with the projectile in it) carry off afterbeing struck?

ExampleC&J 7.61 Three guns are aimed at the centre of a circle, and eachfires a bullet simultaneously. The directions in which they fireare 120 apart. Two of the bullets have the same mass of4.50 10-3 kg and the same speed of 324 m/s. The other bullethas an unknown mass and a speed of 575 m/s. The bulletscollide at the centre and mash into a stationary lump. What is theunknown mass?

ExampleHans Brinker is on skates and there is no friction.He has two identical snowballs, and wants to get fromA to B just by throwing the balls.Is it better to throw them together, or one after theother?Assume that the relative velocity after release is thesame whether he throws one or two snowballs.

For simplicity, consider the mass of the twosnowballs and HB to be the same: mThe relative velocity (separation speed) is vCase 1: throwing them together.v1 2v/3HB(All v’s representmagnitudes.)Case 1:mv1 (2m)v2v v1 v2v 2v2 v2 3v2vHB 2v/3 4v/6v2 v/3! ""! ""! ""!v v21 v2 g vg1! "! "!v v2 v1

For simplicity, consider the mass of the twosnowballs and HB to be the same: mThe relative velocity (separation speed) is vCase 1:vHB 2v/3 4v/6Case 2: throwing them sequentially.v1 v/3HBFirst throwv2 2v/3HBv/3Case 2:v/2HB Second throwvHB v/3 v/2 5v/6v/2

For simplicity, consider the mass of the twosnowballs and HB to be the same: mThe relative velocity (separation speed) is vCase 1:vHB 2v/3 4v/6Case 2:vHB v/3 v/2 5v/6sequentially is better

Applying the Principle of Conservation of Linear Momentum 1. Decide which objects are included in the system. 2. Relative to the system, identify the internal and external forces. 3. Verify that the system is isolated. 4. Set the final momentum of the system equal to its initial momentum. Remember that momentum is a vector.