phy 231 1 physics 231 lecture 13: keeping momentum remco zegers walk-in hour: tue 4-5 pm helproom...
DESCRIPTION
PHY Chapter 6 Momentum & Collisions When a bullet hits the wall, its velocity is very much reduced. The wall does not move, although the force on the ball is the same as the force on the wall (Newton’s 3rd law: F wall-bullet =-F bullet-wall ). F wall-bullet =m bullet a bullet F bullet-wall =m wall a wall M bullet > a wallTRANSCRIPT
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PHY 2311
PHYSICS 231Lecture 13: Keeping momentum
Remco ZegersWalk-in hour: Tue 4-5 pm helproom
BPS1248
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PHY 2312
quiz (extra credit)
At which point are the kinetic and potential energy highest,respectively?a) A and Bb) D and Ec) D and Ad) B and Ae) D and D
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PHY 2313
Chapter 6 Momentum & Collisions
When a bullet hits the wall, its velocityis very much reduced. The wall does notmove, although the force on the ballis the same as the force on the wall(Newton’s 3rd law: Fwall-bullet=-Fbullet-wall).
Fwall-bullet=mbulletabulletFbullet-wall=mwallawall
Mbullet << Mwall
|abullet|>> awall
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PHY 2314
Is it only the mass???
Vbullet=100 m/s
Vbullet=200 m/sWhen the bullet gets stopped in the wall, it deaccelerates from its initial velocityto 0. So, its acceleration is vbullet/t,with t some small time (independent of v).
Second law: Fwall-bullet=mbulletabullet=mbulletvbullet/tThe force also depends on the velocity of the bullet!
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PHY 2315
More general…and formal.F=ma Newton’s 2nd lawF=mv/t a=v/tF=m(vfinal-vinital)/tDefine p=mv p: momentum (kgm/s)F=(pfinal-pinitial)/tF=p/t
The net force acting on an object equalsthe change in momentum (p) in a certaintime period (t). Since velocity is a vector, momentum is alsoa vector, pointing in the same direction as v.
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PHY 2316
ImpulseF=p/t Force=change in (mv) per time
period (t).p=Ft The change in momentum equals
the force acting on the object timeshow long you apply the force.
Definition: p=ImpulseWhat if the force is not constant within the timeperiod t?
ts s s
F1
F2
F3
p=Ft=(F1s+F2s+F3s)== t(F1s+F2s+F3s)/t= tFaverage
p=Faverage t
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PHY 2317
car hitting haystack
car hitting wallThe change in momentum (impulse) is the same, but the forcereaches a much higher value when the car hits a wall!
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PHY 2318
Some examplesA tennis player receives a shot approaching him (horizontally) with 50m/s and returns the ball in the opposite direction with 40m/s. The mass of the ball is 0.060 kg.A) What is the impulse deliveredby the ball to the racket?B) What is the work done by theracket on the ball?
A) Impulse=change in momentum (p). p=m(vfinal-vinitial)=0.060(-40-50)=-5.4 kgm/sB) W= KEfinal-KEinitial=½mvfinal
2-½mvinital2 (no PE!)
=½0.060([-40]2-[50]2)=-27 J
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PHY 2319
Child safetyA friend claims that it is safe to go on a car trip with yourchild without a child seat since he can hold onto your 12kg child even if the car makes a frontal collision (lasting 0.05s and causing the vehicle to stop completely) at v=50 km/h(about 30 miles/h). Is he to be trusted?
F=p/t force=impulse per time period =m(vf-vi)/t vf=0 and vi=50km/h=13.9 m/s m=12kg t=0.05sF=12(-13.9)/0.05=3336 NThis force corresponds to lifting a mass of 340 kg or about 680 pounds! DON’T TRUST THIS GUY!
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PHY 23110
question
The velocity change is largest in case: A B
The acceleration is largest in case: A B
The momentum change is largest in case: A B
The impulse is largest in case: A B
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PHY 23111
accident
Consider the cases A and B in which a car crashes intoa wall. In which case is the likelihood that the passengersare severely hurt greater?
A: the change in impulse is largest (A:9mcar B:5mcar)and thus the force acting on the passengers largest
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PHY 23112
Momentum p=mv
F=p/t
Impulse (the change in momentum) p= Ft
Demo: the ‘safe’ bungee jumper
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PHY 23113
Conservation of MomentumF21t = m1v1f-m1v1iF12t = m2v2f-m2v2i
Newton’s 3rd law:F12=-F21
(m1v1f-m1v1i)=-(m2v2f-m2v2i)
Rewrite:
m1v1i+m2v2i=m1v1f+m2v2fp1i+p2i=p1f+p2f
CONSERVATION OF MOMENTUM
CLOSED SYSTEM!
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PHY 23114
momentum conservation
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PHY 23115
Moving in spaceAn astronaut (100 kg) is drifting away from the spaceship with v=0.2 m/s. To getback he throws a wrench (2 kg) in the direction away from the ship. With whatvelocity does he need to throw the wrenchto move with v=0.1 m/s towards the ship?
Initial momentum: maivai+mwivwi =100*0.2+2*0.2=20.4 kgm/sAfter throw: mafvaf+mwfvwf=100*(-0.1)+2*vwf kgm/s
Conservation of momentum: maivai+mwivwi= mafvaf+mwfvwf
20.4=-10+2*vwf vwf=15.7 m/s
a) 0.1 m/s b) 0.2 m/s c) 5 m/s d) 16 m/se) this will never work?
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PHY 23116
Types of collisionsInelastic collisions Elastic collisions•Momentum is conserved•Some energy is lost in the collision: KE not conserved•Perfectly inelastic: the
objects sticktogether.
•Momentum is conserved•No energy is lost in the collision: KE conserved
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