Physics IMr. Chou

Name---------------Pcriod _

Momentum

I. Dcfinition of MomentumMomentum is the quantity of motion an object has. The momentum of an object

is defined by thc equation of _momentum_ = _mass_ times _vclocity_ or _mv_.Momentum is a _vector _ (vcctor/sealar) quantity. Since the mass of an object isalways positive, the dircction of the momcntum vcctor is always the _same_ as thedircetion of the _velocity_ vector. The direction of the velocity vector is always the sameas the direction of _motion_ (which way the objcct is currently heading).

. E'hICI . ompanson WIt KmetIc nergvMomentum Kinetic Energy

Depends on Mass Massvelocity speed

Vector or Scalar vector scalarUscd in analyzing collisions Single or multiple objects

Given that p = mv, momentum would have units of_kg mls__

Law of Conservation of I\lomentum

Assume no _extcmal_ forces (friction, applied force, etc) act on a system of two ormore objects. This situation is also known as an _isolated_ system. Then the totalmomentum in a collision is conservcd. That is, the total momentum stays _constant_.The momentum one object loses must be -llaincd _ by some other object in the collision.

Problem solvin~ for collisions:I) A 1S-kg medicine ball is thrown at a velocity of 20 km/hr to a 50-kg person who isat rest on ice. The person catches the ball and subsequently slides with the ballacross the ice. Determine the velocity of the person and the ball after the collision.

Momentum = mass * velocity (kg m/s)

Before collision After collision

.20% 8@o----?>'"

ball 15*20 = 300 15 VI

person 0 60 VI

Total 300 kg km/hr 300 kg kmlhrImomentum

75 vf= 300VI= 300/75 =4 km/hr

Physics IMr. Chou

Name---------------Period----

2) A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. Theimpact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that frictioneffects from the road are negligible, determine the velocity of the truck immediatelyafter the collision.

Momentum = mass * velocity (kg m/s)

Before collision After collision

truek 3000*10 = 30000 3000 Vf

car 0 1000 * 15 = 15000

Total 30000 kg m1smomentum

30000 = 3000 Vr + 150003000 Vf = 15000

Vf= 5 m/s

3) A 3000-kg truck moving rightward with a speed of 5 km/hr collides head-on witha 1000-kg car moving leftward with a speed of 10 km/hr. The two vehicles sticktogether and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck. (CAREFUL:Be cautious of the +/- sign on thevelocity of the two vehicles.)

Momentum = mass * velocity (kg m/s)

Before collision After collision

truck 3000*5 = 15000 3000 * Vf

car 1000* (-10) = -10000 1000 * vI'

Total 5000 kg kmlhr 4000 vI'momentum

5000 = 4000 Vf

Vf = 1.25 kmlhr

Momentum and Collisions Name:

A-Ftef

'11= ?>

IOkV = lovSoV

Isoo

Momenlum Before Momentum AfterCollision Collision

Momentum Before Momenlum AfterCollision Collision

) *' I./-!; -=- 2f-~ I *" V = IV

025*0= 0 oZS"~V=.25V

Ball

Skater

Large Fish

Small Fish

Collision AnalysisRead from Lesson 2 of the Momentum and Collisions chapter at The Physics Classroom:

hIIp:/lwww.physicsdassroom.comIClasslmomenlum/u412d.hlmlhIIp:/Iwww.physicsdassroom.comIClasslmomenlu m/u412e.hlml

MOPkConnection: Momentum and Collisions: sublevels 7, 8 and 915~~ V~O=-@ 1. A lamedicine ball is thrown at a velocity of~ 15 r to a 50-Jigskater who is al resl on ice.=- 1'he skaler catc1leSffieball and subsequently/0 k~ slides with the ball across the ice.

Consider the &atlfa'n'll'1fi:6all as two separateparts of an isolated system. (no external forces).Fill in the before. and afler-collision table below.

Total I~o-+o = ISO loV+sW=6D VDetermine the velocity of medicine ball and t~kater ~the ~ion. PSYW

~o ~ t V ~ [-V-=--l-.-~-Yl'1-~-J~ :Al f'h'th flk'" AffF-r.~arge 's WI a mass 0 • g IS In motion al45crn/s when it encounters a smaller fish D.l) I:::d-"(m=O.25kg) which is al rest. The large fishswallows the smaller fish and continues inmotion at a reduced speed. Fill in the before-and after-collision table below.

Total Lf-5 1'1+.~SV -= l.l>V

Determine the velocity of the large and Ihe st fish after the collisi n, PSYW

32-- I. 2~ V ?(v:= S6~J/. 2~ /, "2S" )~ ._---

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Momenlum and Collisions

3. A 0.150-kg baseball moving al a speed of 45.0 m/s crosses Iheplate and strikes the 0.250-kg catcher's mitt (originally at rest).The catcher's mitt immediately recoils backwards (at the samespeed as the ball) before the catcher applies an external force tostop its momentum. If the catcher's hand is in a relaxed state at /the time of the collision, it can be assumed that no net external tJ. IS- IC(j-force exists and the law of momentum conservation applies to thebaseball-catcher's mitt collision. Fill in the before- and after-collision table below.

Momentum Before Momenlum AfterCollision Collision

Baseball ~l~~'f~""6:~ "

V.lS'

Calcher's Mill .ZS*,O-=- ~ .'DV "/'

Tolal 6.1.r 6>% V .~

c:;;;"~

Momentum AfterCollision

1800 1/

/2-00 V/2.Car

Total

Truck

600 VDetermine Ihe velocily of the lruck and car immedialely afler Ihe collision. PSYW

-- I~:::' ~":,O/)IV ~ O.1f '"~J

Momenlum BeforeCollision

A 4BOO-kgtruck traveling with avelocity of +4.0 m/ s collides head-onwith a 1200-kg car have traveling witha velocity of -12 m/ s. The truck andcar entangle and move together aflerthe collision. Fill in the before- andafter-collision table below.

Determine the velocity of the baseball/ catcher's mill immediately after the collision. PSYW, -

4.

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Physics IMr. Chou

NamePeriod

Date-------Impulse and Momentum Change

An object with momentum must be m 0 v ( ;[t, The more momentum an object has,the more difficult it will be hard to stop the obJ ct. Thus, It would reqUIre a greateramoUl;lt..of force or a longer amount of time or both to bring such an object to a halt. Asthe To rGL- acts upon the object for a given amount of T:""~ , the object'svelocity is changed; and hence, the object's f1.1 "~lU".jS changed.

If the force acts opposite the object's motion, the lorce Sro'-V~ d"w he object.If a lorce acts in the same direction as the object's motion, then the force rp-a-eck '-< Pthe object.

f:"vtF =

Recall, Newton's Second Law states that an unbalanced causes an object to

L\ ~ d..o-Awz.

Ju..<-Ilji Sld ••..S

do,"" "

,- !'vi * AV-f~ -t-F If t =~

~ ::vImpulse = Force. time = change in momentum

'C"'ll. ppl,/ "- .f.,rCL OIK..r f2L'tQd .,-f-f~The above equation is known as the Impulse-Momentum Theorem, where impulse isdefined as the force multiplied by time. .r- 10 <>- \ \" q-r,

, I,)~~ qMomentum also can be expressed as UnIts of F "I-t. II" 71 J

N s, ,J ~-i. \Implications of the Impulse-Momentum Theorem 0 . \First, an impulse acting on an object causes a change in the I~~/) Imomentum. ~ I ~

sa nJ/. :::? /,// /'/ /1. , I ::;From the Impulse-Momentum Theorem, we can solve for "" . pl. e c.a.'J /~ti' v 5."'\the average force exerted on the object da,,'r! ~ I d~iJ\ V

.. :::, V F/ ~Force. time = change in momentum ('\ tn '" Drealer

Tl""'- -r.~ fl1.01".4t""Q •Force = change in momentum / time of impact . '.~. ~ [I\< Mt-d,,;:G,l 'J

So lor a given amount of change in momentum, the greaterthe time of impact, the smaller the average force acting onthe object. J' cJ.u,...,.. ,'.... J!.'1,D~ haA

TO r c.e =- ? ~..~ S "-' -F" CR..

T."--<- h,. (..u a (I II~ 3 {'Or-O. i.f f•..•....•.Si'W'-S "-'4.(( ~ cf f~ I~

Momenlum and Collisions Name:

Simple Computations with Impulse = Momentum Change

Read from Lesson 1 of the Momenlum and Collisions chapter al The Physics Classroom:

hllp:/Iwww.physicsdassroom.comIClasslmomenlumlu411 b.hlmlhIIp:/Iwww.physicsdassroom.comIClasslmomen lumlu411c.hIml

A car with a mass of 1000 kg is al resl al a sloplight. When Ihe lighllums green, it is pushed by anel force of 2000 N for 10 s.

v<1c.h.~ ,'>--10 ;;

1 [ - Fllt= ]ooorJ*IO-t""'f'''-U- . \= (~QI 000 N .~ .\

Whal is the value of the change in momentum tliatthe car experiences?:- --'T

I....p •.• 1u ~ ct.a .,~~ {" J'J1. 0".~t ~ : r20, 000

Whal is Ihe final velocily of Ihe car allhe end of 10 seconds?

Whal is the value of the acceleralion Ihal the car experiences?

a.c~1 = het -{or'ce 2000 N _ ~I:l-r-~-I""1<', rs - i 00 0 ,":) - S",

Whal is the value of the change in velocity thai the car experiences?

a.Cc.d '" c..h.~ '" v~ I (-b~ J

Whal is the value of Ihe impulse on Ihe car?

5.

4.

3.

2.

1.

vel

6.

-ft"c.1 vel -: Onj vel 1-The car continues allhis speed for a while.

Whal is the value of the change in momentum Ihe car experiences as il continues at this velocity?

7.D

Whal is the value of Ihe impulse on the car as il conlinues allhis velocily?

DThe brakes are applied 10 Ihe car, causing it 10 come 10 resl in 4 s.

8. Whal is the value of Ihe change in momenlum Ihallhe car experiences?

El:x 1000 leo ( () - 20') - -2-0,000 iPu "'"1

9. Whal is Ihe value of the impulse on the car?

10. Whal is the value of the force (average) that causes the car to stop?

11. What is the acceleration of the car as it stops?

@ The Physics Classroom, 2009 Page 1

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Momentum and Collisions

BEWARE

T ere is a disease known as formula fixatmn w i is common among p ysics students. Itparticularly infects those who perceive physics as an applied math course where numbersand equations are simply combined to solve algebra problems. However, this is Ill!1the truena ture of physics. Physics concerns itself with ideas and concepts which provide areasonable explanation of the physical world. When students divorce the mathematics fromthe ideas, formula fixation takes root and even mathematical problem-solving can becomedifficult. Do you have formula fixation? Test your health by trying these computational

roblems.

12. A force of SOON causes na SO-kgfullback to change his velocity by 10 ml s. Determine the impulseexperienced by the fuIIback. PSYW ~r '

~ CAe«'- Y' L .,

T,...,f ~ Iu. -: Yh t'/f "(' ~ == Rb k~ 't ( 10111'1) -=v-e

13. A O.So-kgsoccer ball experiences an impulse of 25 N .S. Determine the momentum change of thesoccer ball. PSYW

T. ...lu -= /11 ewe.t ":;,J.j N • .s""1' Cl..~

14. A 12Do-kgcar is brought from 25 ml s to 10 mls over a time period of S.Oseconds. Determine theforce experienced by the car. PSYW

-Foru.. = 1'l1.x- (Vf - V,,) == l2-0o/cd ,*. (JO ~ - 2)"'-1' )~-S606 1'1\-t r ac

15. A 90-kg tight end moving at 9.0 m/ s encounters a 400 N.s impulse. Determine the velocity changeof the tight end. PSYW 7""r-!u ~ tntlss -j( vel c.h"'--1f'-

Jfoo Ns '= ,!() ~ b\l

(4v .;:--;;If m/s \16. A O.lo-kg hockey puck decreases its speed from 40 m'ts-to 0 ml sin 0.025 s. Determine the force

whichitexperien~:.~s:v O;.II~ (-.lfO''''S) -:. (--/60 NI' 0.02, S - )

17. A Real Brain Twister: A O.lO-kghockey puck is at rest. It encounters a force of 20 N for 0.2seconds which sets it into motion. Over the next 2.0 seconds, it encounters 0.4 Newtons of resistanceforce. Finally, it encounters a final force of 24 N for 0.05 seconds in the direction of motion. What isthe final velocity of the hockey puck. PSYW

You may ave been tricked, but ose were not intended as trick questions. T e questions wereintended to test your understanding of the concepts of momentum change, impulse, mass, force, timeand veloci chan e. How is our understandin level ro ressin ? Do au have formula fixation?

<!:l The Physics Classroom, 2009 Page 2

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