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```							Chapter 10
The Take-Home Message on
Chapter 10
   If two bodies are exerting STRONG forces on each
other over a short period of time, then you have a
collision problem.
   Linear Momentum is ALWAYS conserved.
   Kinetic Energy is NOT always conserved.
– Kinetic Energy is conserved during an elastic collision.
– Kinetic Energy is not necessarily conserved during an
inelastic collision.
   Impulse-Linear Momentum Theorem
– Relates strength and duration of collision force to
change in momentum
– Similar to Work-Kinetic Energy Theorem
Equations
   Δp = pf - pi = J       (10-4)
 J = F Δt                 (10-8)
 F 
n
 mv           (10-10)
t
(m1  m2 )
v1 f              v1i   (10-18)
(m1  m2 )
2m1
2f              v1i
v                         (10-19)
(m1  m2 )
More Equations
   .                                        (10-28)

.       2m1              (m2  m1 )
   v2 f               v1i              v2i (10-29)
(m1  m2 )         (m1  m2 )
   .m v  (m  m )V                         (10-34)
1      1   2

   . 1v1  m2v2  (m1  m2 )V
m                                        (10-36)
Last One
P      m1v1i  m2v2i
   .
vcm            
m1  m2     m1  m2
(10-30)
Problem Types
   Impulse and Linear Momentum
– Series of Collisions
   Elastic Collision
– 1-D
– 2-D
– Moving Target vs. Stationary Target
   Inelastic Collision
– 1-D
– 2-D
How to Solve Them
   Impulse and Linear Momentum
– Use Conservation of Momentum
– Impulse-Linear Momentum Theorem
   Series of Collisions
– Use Equation (10-10)
   Elastic Collisions
– Use Conservation of Momentum
– Use Conservation of Kinetic Energy
– In Two Dimensions Treat Momentums as
Vector Quantities (x,y), and Solve as Usual
Problem #1
   A 283g air-track glider
moving at 69 m/s on a 24 m
long air track collides
elastically with a 467g glider
at rest in the middle of the
horizontal track.

The end of the track over which the struck glider moves
is not frictionless, and the glider moves with a
coefficient of friction =0.02 with respect to the track.
Will the glider reach the end of the track? Neglect the
length of the glider.
Problem #1 Solution
   m1 = 0.283 kg
   m2 = 0.467 kg
   Kinetic Energy Conserved (Elastic Collision)
   Momentum Conserved
   Ki=Kf
1                    1               1
Ki  (m1 )( v1i )  K f  (m1 )( v1 f )  (m2 )( v2 f )2
2                     2

2                    2               2
(0.283 kg)(0.69 m/s)2 = (0.283 kg)(v1f)2 + (0.467 kg)(v2f)2

   0.1445(v1f)2 + 0.2335(v2f)2 = 0.0674
Problem #1 Solution (cont.)
   Pi = Pf
m1v1i = m1v1f + m2v2f
(0.283 kg)(0.69 m/s) = (0.283 kg)(-v1f) + (0.467 kg)(v2f)
 v1f = -0.69 m/s + (1.65)v2f

   Two Equations and Two Unknowns
0.1415 (-0.69 m/s + 1.65v2f)2 + 0.2335(v2f)2 = 0.067
   Solve to find that v2f = 0.525 m/s
   Substitute this to Solve for Kinetic Energy of Struck Glider
K = 1/2 m2(v2f)2 = 1/2 (0.467)(0.525)2
KE of struck glider = 0.064 J
Problem #1 Solution (cont.)
   Ffriction = μN = μm2g
= (0.02)(0.467 kg)(9.8 m/s2)
= 0.0915 N
   Work Done by Friction = (0.0915 N)(1.2 m) =
0.11 J
   K < Wfriction
   Glider Does Not Reach the End of the Track
Problem #2
   Wile E. Coyote, having just
passed Physics 207 at VCU,
splurges on a 500 kg ACME
rocket. He reads the rocket
specs to discover a relative
exhaust speed of 600 m/s and a
fuel consumption rate of 2
kg/s.
   He launches horizontally after the roadrunner and flies for twenty
seconds before the road turns and the rocket does not work. He
collides with the 4000 kg ACME delivery truck (initially
stationary). Neglecting gravity and air resistance, what is the
relative velocity of the truck just after the totally inelastic
collision? Assume the super genius to have a mass of 25 kg.
Problem #2 (cont.)

   Rocket Science is needed for this problem:
– Use the second rocket equation to find the
rocket speed at impact
mi
 vf - vi = u ln
mf
500kg  25kg
   vf = (600 m/s) ln
485kg

   vf = 47.55 m/s
Problem #2 (cont.)
   Now solve for vtruck using conservation of momentum:
– P i = Pf
– mrocket ivrocket i + mtruck i vtruck i = mrocket f vrocket f + mtruck f vtruck f

mrocketi vrocketi       (485kg)( 47.55m / s )
– vf =                                
mrocket f  mtruck f        485kg  4000kg

= 5.14 m/s
Recommendations for Further
Study
Mon          Tue       Wed         Thu      Fri

7:00a

   If you have any          8:00

9:00

more questions          10:00

11:00                Lunch                 Lunch

PHYS
Lunch
PHYS
Lunch

1:00                                               Carlisle
expensive textbook            PHYS207
Recitation
491

Dept.
491
Group
Meeting
2:00                          Curriculu

– Go see our Web
PHYS207 Faculty        m
Recitation Meeting     Committe
e
3:00                Joe
Study Guide                   Scott
Walton
Tom
Carlson

– Go harass Carlisle
Turner                                      PHYS
4:00           Office                     Office   690
PHYS207 Hour                       Hour     Colloqui
Recitation                                  um
(see attached          5:00
PHYS      Scott
PHYS
6:00                                      207E

schedule)              7:00
207E      Wen

```
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