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2D Collisions In all collisions, momentum is conserved In elastic collisions, kinetic energy is also conserved As momentum is a vector, we can break momentum into components and employ the conservation of momentum
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2D CollisionsPhysics 12 Adv
2D Collisions In all collisions, momentum is conserved In elastic collisions, kinetic energy is also
conserved As momentum is a vector, we can break
momentum into components and employ the conservation of momentum
2D Collision Two cars approach an intersection; the first
car is travelling east at a velocity of 15m/s and the car has a mass of 1000.kg. The second car is travelling north at a velocity of 10.m/s and has a mass of 1200.kg. If the cars collide and stick together, determine the following: The velocity immediately after the collision The direction of motion immediately after the
collision
2D Collision
A
B
v=15m/s
v=10.m/s
AB
v=?
2D Collisions
skgmxp
vmpskgmxp
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smv
smv
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toty
BBtoty
totx
AAtotx
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oB
oA
B
A
/102.1
/105.1
'90,/.10
0,/15
.1200
.1000
4
4
o
o
y
y
yBAtoty
x
x
xBAtotx
totytoty
totxtotx
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smvsmsmv
smv
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39,/7.8'
39/8.6/5.5tan
/7.8')/5.5()/8.6('
/5.5'
').2200(/102.1
')(/8.6'
').2200(/105.1
')(
'/102.1
'/105.1
1
22
4
4
4
4
Elastic Collisions A proton travelling with speed 8.2x105m/s
collides elastically with stationary proton in a hydrogen target. One of the protons is observed to be scattered up at a 60.° angle. At what angle will the second proton be
scattered? What will the speed of each of the protons be
after the collision?
Elastic Collisions
1
2
1
2
v1= 8.2x105m/s
v’1= ?
v’2=?
Before
After
Kinetic Energy
22
21
21
22
21
21
2121
''
'21'
210
21
'''
vvv
mvmvmv
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tottot
Momentum
sin'sin'0
''0
''0cos'cos'
''''0
'
21
21
21
211
211
211
vv
vv
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pp
yy
yy
xx
xx
Three Equations, Three Unknowns
??'?'
.60
0,/102.8
)3(sin'sin'0)2(cos'cos'
)1(''
2
1
51
21
211
22
21
21
vv
smxv
vvvvv
vvv
o
o
In Equations 2 and 3, take v’1 to left and square both sides
222
221
21
21
222
22111
21
211
211
sin'sin'
sin'sin'sin'sin'0
cos'cos'cos'2
cos'cos'cos'cos'
vv
vvvv
vvvvv
vvvvvv
Add the final steps from the last step then substitute equation 1
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vvvvv
vvv
vvvvv
vvvvvvv
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vvvvv
/101.4'
cos2'2cos'2'2
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sin'sin'
cos'cos'cos'2
51
11
1121
22
21
21
22
2111
21
22
222
222
221
22111
21
222
221
222
22111
21
Use the answer for v’1 to solve for v’2 and angle
o
vv
smxv
vvv
vvv
.30
)50.0(sin
sin''sin
/101.7'
''
''
1
2
1
52
21
212
22
21
21
Ballistic Pendulum In a ballistic pendulum, there are
components where energy is conserved and components where it is lost
It is therefore important that we analyze each component of the system correctly
Inelastic Collision The first part of the analysis involves the
inelastic collision between the projectile and the bob
We know that as it is an inelastic collision, momentum is conserved but energy is not
Pendulum After the collision, the projectile and bob act
as a pendulum and will swing to a maximum height
If this height can be measured, then through the conservation of energy, we can determine the speed of the projectile and bob immediately after the collision
Question A forensic expert needed to find the velocity of a
bullet fired from a gun in order to predict the trajectory of a bullet. She fired a 5.50g bullet into a ballistic pendulum with a bob that had a mass of 1.75kg. The pendulum swings to a height of 12.5cm above its rest position before dropping back down. What was the velocity of the bullet before it hit and became embedded in the bob?
Analysis
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ghMmvMm
EE
bottom
bottom
bottom
toppbottomk
/567.12
)()(21 2
)()(
Analysis
smxv
vmMmv
vMmvm
b
bottomb
bottomb
/1000.5
)(
2
Practice Problems Page 509
Questions 35-37 Page 515
Questions 39-40 Page 524
Questions 41-45
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