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```Momentum and
Collisions
Momentum
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The linear momentum of an object of
mass m moving with a velocity v is the
product of the mass and the velocity.
Momentum = mass X velocity
p=mv
Units kg●m/s
Momentum Practice
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A 2250 kg pickup truck has a velocity of
25 m/s to the east. What is the
momentum of the truck?
Given: m=2250 kg
v= 25 m/s
p=mv 2250kg x 25 m/s=
5.6x104 kg●m/s
Changing Momentum
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A change in momentum takes force and
time
F=m(Δv/Δt)
Newtons original formula
F=Δp/Δt
Δp=mvf-mvi
Force= change in momentum during time
interval
Impulse
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Impulse is the force for the time interval.
Impulse-momentum theorem is the
expression FΔt= Δp
If you extend the time of impact you
reduce the amount of force.
Force and Impulse
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A 1400 kg car moving westward with a
velocity of 15 m/s collides with a utility
pole and is brought to rest in .30 s. Find
the force exerted on the car during the
collision.
Given m=1400 kg, Δt= .30 s
vf=0 m/s vi=15 m/s
What formula can I use?
Impulse-Momentum
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FΔt= Δp
Δp=mvf-mvi
FΔt=mvf-mv I am looking for force
F=mvf-mv/ Δt Now I just plug in my
numbers
(1400*0-1400*-15)/.30= 7.0*104 N to the
east
Stopping Distance
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A 2240 kg car traveling west slows down
uniformly from 20.0 m/s to 5.0 m/s How
long does it take the car to decelerate if
the force on the car is 8410 to the east?
How far does the car travel during the
deceleration?
What am I given?
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M= 2240 kg
vi= 20.0 m/s to the west = -20.0 m/s
vf= 5.0 m/s to the west = -5.0 m/s
F= 8410 N to the east so it stays positive
Unknown Δt and Δx
Impulse Momentum Theorem
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FΔt= Δp
Δp=mvf-mvi
Δt=mvf-mvi /F
((2240*-5)-(2240*-20))/8410
Δt=4.0s
Displacement of the vehicle
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To find the change in distance
Average velocity= change in d/change in t
Average velocity =(initial +final)/2
Since these are equal I can combine them
Δx/Δt=(vf+ vi)/2
Solving for x=((vf+ vi)/2) *Δt
x=((-20+-5)/2)*4
x=-50 so it would be 50m to the west
Conservation Of Momentum
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When I have two objects A and B
p(Ai)+p(Bi)=p(Af)+p(Bf)
Total initial momentum=total final
momentum
Conservation of Momentum
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A 76 kg boater, initially at rest in a
stationary 45 kg boat, steps out of the
boat and onto the dock. If the boater
moves out of the boat with a velocity of
2.5 m/s to the right what is the final
velocity of the boat?
Given: m of person= 76 kg vi=0
m/s,vf=2.5 to the right
m of boat= 45 kg vi=0 vf=?
m1v1i+m2v2i=m1v1f+m2v2f
Because the boater and the boat are initially
at rest m1v1i + m2v2i=0
Rearrange to solve for final velocity of boat
v2f = (-m1/m2)/v1f
-76/45 * 2.5=4.2 to the right -4.2
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Two train cars of mass 8000 kg are
traveling toward each other. One at a
speed of 30 m/s to the west and the other
at a speed of 25 m/s to the east. They
crash and become entangled, what is their
final velocity and direction?
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m1 =8000 kg
m2= 8000 kg
v1=-30m/s
v2=25 m/s
m1v1+m2v2=mtotvtot
```
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