1. Collision and Explosion
PHYSICS 220
Lecture 12
Collision and Explosion
Lecture 12

2. iClicker S Pinitial = S Pfinal M v = M vf + M vf v = 2vf vf = v/2
A railroad car is coasting along a horizontal track with speed v when it runs into and connects with a second identical railroad car, initially at rest. Assuming there is no friction between the cars and the rails, what is the speed of the two coupled cars after the collision?
A) v
B) v/2
C) v/4
D) 2v
S Pinitial = S Pfinal
M v = M vf + M vf
v = 2vf
vf = v/2
Lecture 12

3. Type of Collision Elastic Collisions: Inelastic Collisions:
collisions that conserve mechanical energy
Inelastic Collisions:
collisions that do not conserve mechanical energy
Completely Inelastic Collisions:
objects stick together
Completely Inelastic
Elastic
Inelastic
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4. Explosion “before” M Example: m1 = M/3 m2 = 2M/3 v1 v2 “after” m1 m2
Which block has larger |momentum|?
Each has same |momentum|
Which block has larger speed?
mv same for each  smaller mass has larger velocity
Which block has larger kinetic energy?
K = mv2/2 = m2v2/2m = p2/2m
 smaller mass has larger kinetic energy
Is mechanical (kinetic) energy conserved? NO
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5. Explosion and Collision
Procedure
“before”
“after” M m1 m2
Draw “before” and “after”
Define system so that Fext = 0
Set up a coordinate system
Compute ptotal “before”
Compute ptotal “after”
Set them equal to each other
Collision
“before”
“after” m1 m2
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6. 2-D Problems Ptotal,x and Ptotal,y independently conserved
after
before
Ptotal,x and Ptotal,y independently conserved
Ptotal,x,before = Ptotal,x,after
Ptotal,y,before = Ptotal,y,after
Lecture 12

7. Completely Inelastic Collisionx
before m2 v2 m1 v1
after
m1+m2 vf 
Work on overhead -- symbols only.
Find vf and 
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8. Completely Inelastic Collision in Two Dimensions
Find vf and  x
after
m1+m2 vf  y
Work on overhead -- symbols only.
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9. Center of Mass Center of Mass L m Example 1: xCM = (0 + mL)/2m = L/2 L
X=L
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10. Lecture 12

11. Center of Mass
For symmetric objects that have uniform density the CM will simply be at the geometrical center! + CM
Lecture 12

12. Exercise The disk shown below (1) clearly has its CM at the center.
Suppose the disk is cut in half and the pieces arranged as shown in (2): Where is the CM of (2) as compared to (1)?
A) higher B) lower C) same
X CM
(1)
(2)
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13. Exercise
The CM of each half-disk will be closer to the fat end than to the thin end (think of where it would balance).
The CM of the compound object will be halfway between the CMs of the two halves. X
X CM
This is higher than the CM of the disk X X
(1)
(2)
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14. Center of Mass
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15. Dynamics of Many Particles
ptot = mtotVcm
FextDt = Dptot = mtotDVcm
or Fext = mtotacm
So if Fext = 0 then Vcm is constant
Center of Mass of a system behaves in a SIMPLE way
- moves like a point particle! - velocity of CM is unaffected by collision if Fext = 0
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16. Astronauts & Rope
Two astronauts at rest in outer space are connected by a light rope. They are at a distance L and they begin to pull towards each other. Where do they meet?
A) L/ B) 2L/ C) 1/5 L
M = 1.5m m L
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17. Astronauts & Rope... They start at rest, so VCM = 0.
VCM remains zero because
there are no external forces.
So, the CM does not move!
They will meet at the CM.
M = 1.5m m CM L
x=0
x=L
Finding the CM:
If we take the astronaut on the left to be at x = 0:
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18. Elastic Collision Assuming Collision is elastic (KE is conserved)
Balls have the same mass
One ball starts out at rest pf pi
vcm F Pf
before
after
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19. iClicker Which of these is possible? (Ignore friction and gravity) A BM
“before”
“after” A B
Which of these is possible? (Ignore friction and gravity) A B
C = both
D = neither
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20. Example m2 m1 v02 v01 initial m2 m1 vf final
A freight train is being assembled in switching yard. Car 1 has a mass of m1=65 103kg and moves at a velocity v01=0.80 m/s. Car 2 has a mass of m2=92 103kg and moves at a velocity v02=1.3 m/s and couples to Car 1. Neglecting friction, find the common velocity vf of the cars after they become coupled. m2 m1
v02
v01
initial m2 m1 vf
final
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21. Example
A fright train is being assembled in switching yard. Car 1 has a mass of m1=65 103kg and moves at a velocity v01=0.80 m/s. Car 2 has a mass of m2=92 103kg and moves at a velocity v02=1.3 m/s and couples to it. Neglecting friction, find the common velocity vf of the cars after they become coupled.
Apply conservation of momentum:
Pi=m1vo1+m2v Pf=(m1+m2)vf
(m1+m2)vf = m1v01+m2v02
Lecture 12