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PH1011 Tut 4: Forces, Momentum and Impulse

1) [‘A’ J86/II/8]Whenabodymovesthroughafluid,aretardingforceduetoturbulencemaybeexperienced.Inthecaseofasphereofradius𝑟movingwithspeedvthroughastationaryfluidofdensity𝜌whichisatrest,thisforceis

𝐹 = 𝑘𝜌 𝑟!𝑣!

where𝑘isaconstant.(NotethatthisformuladiffersfromStoke'slawinthelecturenotes).When spherical raindrops fall through still air, all but the smallest experience a retardingforce given by the equation above. It is found that drops of a given radius approach thegroundwithanapproximatelyconstantspeed,which is independentofthecloud inwhichtheyareform.

(a)Findanexpressionforthisterminalspeed𝑣!intermsoftheconstant𝑘,theradius𝑟ofthedrop, itsdensity𝜌!,thedensityofair𝜌!oftheairandtheaccelerationoffreefallg.(Youmayneglecttheupthrustduetoair).Sketchthevelocity-timegraphoftheobjectfromthemomentitisreleasedintothefluid.

(b) The terminal speed of the raindrop of radius 1mm is approximately7 𝑚𝑠!!. In freakstorms,hailstoneswithradiiaslargeas20mmmayfall.Estimatethespeedwithwhichsuchstonesstriketheground.[Takethedensityofwateras1×10! 𝑘𝑔 𝑚!!andthedensityoficeas9×10! 𝑘𝑔 𝑚!!.

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2) [Activity]a) Thedensityoficeis0.917g/cm3,whereasthatofseawateris1.025g/cm3.Whatpercentof

an iceberg is above the surface of the water? (Discussion) If some ice and seawater are

placedinameasuringcylinder,willtheseawaterlevelrise,droporstaythesamewhenall

theicemelts?

b) Whenacrownofmass14.7kgissubmergedinwater,anaccuratescalereadsonly13.4kg.Isthecrownmadeofgold?

c) Asphericalironballbearingofdiameter4mmisplacedonapieceofStyrofoamfloatinginacylinder of oil as shown in the figure below. You may assume that the weight of the

Styrofoamisnegligible.Densityofironis7.86gcm-3;densityofoilis0.83gcm-3.

i) CalculatetheupthrustactingonthepieceofStyrofoam.ii) WhatisthevolumeofoildisplacedbythepieceofStyrofoamandtheironballbearing?iii) TheballbearingrollsoffthepieceofStyrofoamandfallsintotheoil.Willtheoillevelbe

higher,lowerorthesameasbefore?Explainyouranswer.

Ironballbearing

Styrofoam

Oil

Originaloillevel

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3) [S9.15] The magnitude of the net force exerted in the x direction on a 2.50-kg particle varies in time as shown in the figure below. Find (a) the impulse of the force over the 5.00-s time interval, (b) the final velocity the particle attains if it is originally at rest, (c) its final velocity if its original velocity is −2.00 ! m/s, and (d) the average force exerted on the particle for the time interval between 0 and 5.00 s.

4) [G9.32] (a)Calculatethe impulseexperiencedwhena65-kgperson landsonfirmgroundafterjumpingfromaheightof3.0m.(b)Estimatetheaverageforceexertedontheperson’sfeetbythegroundifthelandingisstiff-legged,andagain(c)withbentlegs.Withstifflegs,assumethebodymoves1.0cmduringimpact,andwhenthelegsarebent,about50cm.[Hint:Theaveragenetforceonherwhichisrelatedtoimpulse,isthevectorsumofgravityandtheforceexertedbytheground.]

5) Aneutron collideselasticallywithaheliumnucleus (at rest initially)whosemass is four times

that of the neutron. The helium nucleus is observed to move off at an angle𝜃!"! = 40∘ .Determinetheangleoftheneutron,𝜃!! andthespeedsofthetwoparticles,𝑣!! and𝑣!"! ,afterthecollision.Theneutron’sinitialspeedis6.2×10! 𝑚/𝑠.

6) [G9.59]Aneonatom(m=20.0u)makesaperfectlyelasticcollisionwithanotheratomat rest.

Aftertheimpact,theneonatomtravelsawayata55.6°anglefromitsoriginaldirectionandtheunknownatomtravelsawayata–50.0°angle.What is themass (inu)of theunknownatom?[Hint:Youcouldusethelawofsines.]

7) [G9.103]A0.25-kgskeet(claytarget)isfiredatanangleof28°tothegroundwithaspeedof25

m/s.When it reaches themaximum height, h, it is hit from below by a 15-g pellet travellingverticallyupwardsata speedof230m/s.Thepellet isembedded in the skeet. (a)Howmuchhigherh’doestheskeetgoup?(b)HowmuchextradistanceΔxdoestheskeettravelbecauseofthecollision?

Answers to selected questions

1. (a) 𝑣! =!!!!"#!!!!

(b) 30 m/s

2. (c)(i) 2.58×10!!𝑁 (ii) 0.317 𝑐𝑚! (iii) Lower 3. (a) 𝐼 = 12.0 𝑁𝑠 𝚤; (b) 𝑣! = 4.80

!!𝚤; (c) 𝑣! = 2.80

!!𝚤; (d) 𝐹!"# = 2.40 𝑁 𝚤

4. (a) (magnitude) 499 Ns, (b) 301 mg = 191737 N, (c) 7 mg = 4459 N 5. 𝑣!"! = 1.899×10!𝑚/𝑠; 𝑣!! = 4.899×10!𝑚/𝑠;𝜃!! = 85.6∘ 6. 39.9 u 7. (a) 8.6 m; (b) 38.4 m