AP Physics B
Impulse and Momentum
Using Physics terms, what put the egg in motion?
Once the egg was moving, why did it keep moving?
Momentum equals mass times velocity.
Unit:
Using Physics terms, how did you stop the egg?
Then if you multiply both sides by “t”, then
Notice the right side of the equation,What physics term is defined by that part of the equation?
The quantity Ft is called an Impulse.
Impulse = Change in Momentum
Units of Impulse: Units of Momentum:
ExampleA 100 g ball is dropped from a height of h = 2.00 m above the floor. It
rebounds vertically to a height of h'= 1.50 m after colliding with the floor. (a) Find the momentum of the ball immediately before it collides with the floor and immediately after it rebounds, (b) Determine the average force exerted by the floor on the ball. Assume that the time interval of the collision is 0.01 seconds.
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EE
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o
o
AB
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21 2
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o
AB
/4.55.1*8.9*22
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after
before
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Impulse is the AreaSince J=Ft, Impulse is the AREA of a Force vs.
Time graph.
How about a collision?Consider 2 objects speeding
toward each other. When they collide......
Due to Newton’s 3rd Law the FORCE they exert on each other are EQUAL and OPPOSITE.
The TIMES of impact are also equal.
Therefore, the IMPULSES of the 2 objects colliding are also EQUAL
21
21
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JJ
FtFt
ttFF
How about a collision?If the Impulses are equal then
the change in MOMENTUMS are also equal!
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2211
21
21
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Momentum is conserved!The Law of Conservation of Momentum: “In the
absence of an external force (gravity, friction), the total momentum before the collision is equal to the total momentum after the collision.”
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total
car
truck
totalo
caro
otrucko
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/*15003*500
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Several Types of collisionsSometimes objects stick together or blow apart.
In this case, momentum is ALWAYS conserved.
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When 2 objects collide and DON’T stick
When 2 objects collide and stick together
When 1 object breaks into 2 objects
Elastic Collision = Kinetic Energy is ConservedInelastic Collision = Kinetic Energy is NOT Conserved
ExampleA bird perched on an 8.00 cm tall swing has a mass of 52.0 g, and the base of the swing has a mass of 153 g. Assume that the swing and bird are originally at rest and that the bird takes off horizontally at 2.00 m/s. If the base can swing freely (without friction) around the pivot, how high will the base of the swing rise above its original level?
How many objects due to have BEFORE the action?
How many objects do you have AFTER the action?
1
2
swing
swing
ToT
AB
v
v
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)2)(052.0()153.0()0)(205.0( )(1
2211
-0.680 m/s
6.19
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22
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0.024 m
Example Granny (m=80 kg) whizzes around the rink with a velocity of 6 m/s. She suddenly collides with Ambrose (m=40 kg) who is at rest directly in her path. Rather than knock him over, she picks him up and continues in motion without "braking." Determine the velocity of Granny and Ambrose.How many objects do I have before the collision?
How many objects do I have after the collision?
2
1
T
T
TToo
ab
v
v
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120)0)(40()6)(80(2211
4 m/s
What happens if we have two unknowns?At an amusement park, a 96.0 kg bumper
car moving with a speed of 1.24 m/s bounces elastically off a 135 kg bumper car at rest. Find the final velocities of the cars.
1 1 2 2 1 1f 2 2fm v m v m v m v
2 2 2 21 1 2 2 1 1f 2 2f
1 1 1 1.
2 2 2 2m v m v m v m v
1 1 1f
2 2f 2
( )1.
( )
m v v
m v v
Let subscript 1 refer to the 96 kg car and subscript 2 refer to the 135kg car.Use momentum conservation.
Use conservation of kinetic energy.
Rearranging the first equation gives
2 21 1 1f 1 1 1f 1 1f
2 22 2f 2 2f 22 2f 2
( ) ( )( )1 .
( )( )( )
m v v m v v v v
m v v v vm v v
1 1f
2f 2
( )1,
( )
v v
v v
2f 1 1f 2or .v v v v
Rearranging the second equation gives
Comparing these two equations implies that
2fv 1f .v
1 1 2 2 1 1f 2 1 1f 2
1 2 1 2 2 1 2 1f
( )( ) 2 ( )m v m v m v m v v vm m v m v m m v
1 2 21f 1 2
1 2 1 2
2m m mv v v
m m m m
2f 1 1f 2 1f 2f 2 1Since , .v v v v v v v v
Substitute for in the first equation and solve for
1fv 2f .v
1 1 2 2 1 2f 2 1 2 2f
1 1 2 1 2 1 2 2f
( )2 ( ) ( )m v m v m v v v m vm v m m v m m v
Substitute for in the first equation and solve for
1 2 12f 1 2
1 2 1 2
2
2(5400kg) m 0.150kg 5400kg m4.30 8.11
5400kg 0.150kg s 5400kg 0.150kg s
17 m/s
m m mv v v
m m m m
2D Inelastic Collisions must rely on the Conservation of Momentum:Example: A car with a
mass of 950 kg and a speed of 16 m/s approaches an intersection, as shown. A 1300 kg minivan traveling at 21 m/s is heading for the same intersection. The car and minivan collide and stick together. Find the speed and direction of the wrecked vehicles just after the collision, assuming external forces can be ignored.
Collisions in 2 DimensionsThe figure to the left shows
a collision between two pucks on an air hockey table. Puck A has a mass of 0.025-kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.050-kg and is initially at rest. The collision is NOT head on. After the collision, the two pucks fly apart with angles shown in the drawing. Calculate the speeds of the pucks after the collision.
vA
vB
vAcos
vAsin
vBcosvBsin
Collisions in 2 dimensions
vA
vB
vAcos
vAsin
vBcosvBsin
)37cos)(050(.)65cos)(025(.0)5.5)(025.0( BA
xBBxAAoxBBoxAA
xox
vv
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BA vv 040.00106.01375.0
AB
AB
BA
yBByAA
yoy
vv
vv
vv
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757.0
0227.00300.0
)37sin)(050.0()65sin)(025.0(0
0
Collisions in 2 dimensions
AB vv 757.0
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v
vv
vv
A
A
AA
AA
/84.2
04845.01375.0
03785.00106.01375.0
)757.0)(050.0(0106.01375.0
BA vv 040.00106.01375.0
smvB /15.2)84.2(757.0
For 2D Elastic Collisions, KE is also conserved:Example: The collision of two 7 kg curling stones.
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