PHYSICS 231 Lecture 13: Collisions

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PHYSICS 231Lecture 13 Collisions

• Remco Zegers
• Walk-in hour Thursday 1130-1330 am
• Helproom

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Previously momentum and impulse
Momentum pmv F?p/?t Impulse (the change in
momentum) ?p F?t
Demo the safe bungee jumper
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Conservation of Momentum
F21?t m1v1f-m1v1i F12?t m2v2f-m2v2i Newtons
3rd law F12-F21 (m1v1f-m1v1i)-(m2v2f-m2v2i)
Rewrite m1v1im2v2im1v1fm2v2f p1ip2ip1fp2f
CONSERVATION OF MOMENTUM
CLOSED SYSTEM!
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Moving in space
An astronaut (100 kg) is drifting away from the
spaceship with v0.2 m/s. To get back he throws a
wrench (2 kg) in the direction away from the
ship. With what velocity does he need to throw
the wrench to move with v0.1 m/s towards the
ship?
Initial momentum maivaimwivwi 1000.220.220
.4 kgm/s After throw mafvafmwfvwf100(-0.1)2
vwf kgm/s Conservation of momentum
maivaimwivwi mafvafmwfvwf 20.4-102vwf
vwf15.7 m/s
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Types of collisions
Inelastic collisions
Elastic collisions

• Momentum is conserved
• Some energy is lost in the collision KE not
  conserved
• Perfectly inelastic the objects stick
• together.

• Momentum is conserved
• No energy is lost in the collision KE conserved

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Perfectly inelastic collisions
Conservation of P m1v1im2v2im1v1fm2v2f After
the collision m1 and m2 form one new object with
mass Mm1m2 m1v1im2v2ivf(m1m2) vf(m1v1im2v2
i)/ (m1m2)
after
before
Demo perfect inelastic collision on airtrack
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Perfect inelastic collision an example
50 m/s
20 m/s
25 m/s
A car collides into the back of a truck and their
bumpers get stuck. What is the ratio of the mass
of the truck and the car? (mtruckcmcar) What is
the fraction of KE lost?
m1v1im2v2ivf(m1m2) 50mc20cmc25(mccmc) so
c25mc/5mc5 Before collision
KEi½mc502½5mc202 After collision
KEf½6mc252 Ratio KEf/KEi (6252)/(5025202)0
.83 17 of the KE is lost (damage to cars!)
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Ballistic Pendulum
How high will the block go?
Mblock1 kg

• There are 2 stages
• The collision
• The Swing of the block

Mbullet0.1 kg Vbullet20 m/s
h
The collision The bullet gets stuck in the block
(perfect inelastic collision). Use conservation
of momentum. m1v1im2v2ivf(m1m2) so
0.12010vf(0.11) vf1.8 m/s
The swing of the block Use
conservation of Mechanical energy. (mgh½mv2)start
of swing (mgh½mv2)at highest
point 0½1.1(1.8)21.19.81h so h0.17 m
Why cant we use Conservation of ME right from
the start??
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Elastic collisions
Conservation of momentum m1v1im2v2im1v1fm2v2f
Conservation of KE ½m1v1i2½m2v2i2½m1v1f2½m2v2f
2 Rewrite conservation of KE
a) m1(v1i-v1f)(v1iv1f) m2(v2f-v2i)(v2fv2i) R
ewrite conservation of P b) m1(v1i-v1f) m2(
v2f-v2i) Divide a) by b) (v1iv1f)(v2fv2i) re
write (v1i-v2i)(v2f-v1f)
Use in problems
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elastic collision of equal masses
Given m2m1. What is the velocity of m1 and m2
after the collision in terms of the initial
velocity of m2 if m1 is originally at rest?
m1v1im2v2im1v1fm2v2f
m2v2im1v1fm2v2f v2iv1fv2f
(v1i-v2i)(v2f-v1f) -v2i
v2f-v1f
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elastic collision of unequal masses
Given m23m1. What is the velocity of m1 and m2
after the collision in terms of the initial
velocity of the moving bullet if a) m1 is
originally at rest b) m2 is originally at rest
A) m1v1im2v2im1v1fm2v2f
3m1v2im1v1fm2v2f 3v2iv1f3v2f
(v1i-v2i)(v2f-v1f) -v2i
v2f-v1f
B) m1v1im2v2im1v1fm2v2f
m1v1im1v1fm2v2f v1iv1f3v2f
(v1i-v2i)(v2f-v1f) v1i
v2f-v1f