Transcript MOMENTUM AND COLLISIONS

LINEAR MOMENTUM
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Momentum = Mass x Velocity
 p=mv
The SI unit for momentum is kg·m/s
 Momentum and velocity are in the same
direction
 Is a vector
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Using the equation
p=mv
 At the same velocity, as mass increases
– momentum increases
 At the same mass, as velocity increases
– momentum increases
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Example
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You are driving north, a deer with mass
of 146 kg is running head-on toward you
with a speed of 17 m/s. Find the
momentum of the deer.
p=mv
 p=146 kg * 17 m/s
 p= 2500 kg·m/s to the south
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CHANGING MOMENTUM
A change in momentum takes time and
force
 For example soccer – when receiving a
pass it takes force to stop the ball
 It will take more force to stop a fast moving
ball than to stop a slow moving ball
 A toy truck and a real truck moving at the
same velocity, it will take more force to stop
the real truck than to stop the toy truck
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IMPULSE
Impulse is the applied force times the
time interval FΔt
 Impulse = FΔt
 Force is reduced when the time interval
increases
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 Example – giving a little when you catch a
ball
 Trampoline
IMPULSE-MOMENTUM THEOREM
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From Newton’s second law
And the equation for acceleration: a=v/t
We can find the equation for force in terms
of momentum
F=Δp/Δt
Force = change in momentum / time
interval
We can also rearrange this equation to find
the change in momentum in terms of
external net force and time
Δp=FΔt, Δp=FΔt=mvf-mvi
(it will never go away)
Example
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A 0.50 kg football is thrown with a
velocity of 15 m/s to the right. A
stationary receiver catches the ball and
brings it to rest in 0.020 s. What is the
force exerted on the ball by the
receiver?
p=mv
 Δp=FΔt=mvf-mvi
 vi= 15 m/s
 vf=0.0 m/s
 t=0.020 s
 F*0.020 s = 0 – 0.50 kg*15 m/s
 F= 380 N
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IMPULSE-MOMENTUM THEOREM
AND STOPPING DISTANCE
The impulse-momentum theorem can be
used to determine the stopping distance
of a car or any moving object
 Use Δp=FΔt
 So Δt=Δp/F=mvf-mvi/F
 Then when you find time you can use
 distance = average velocity * time
 Δx=1/2(vf+vi) Δt
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Example
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A 2240 kg car traveling to the west
slows down uniformly to rest from 20.0
m/s. The decelerating force on the car is
8410 N to the east. How far would the
car move before stopping?
Δt=Δp/F=mvf-mvi/F
Δt=0-2240kg*20.0m/s / 8410 N
 Δt = 5.33 s
 Δx=1/2(vf+vi) Δt
 Δx=1/2(0+20.0m/s)*5.33
 x=53.3 m to the west
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What happens to momentum when two or
more objects interact?
First you have to consider total momentum
of all objects involved. This is the sum of
all momentums
Like energy momentum is conserved
Conservation of Momentum:
Total initial momentum = total final
momentum
m1v1i + m2v2i = m1v1f + m2v2f
Example
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A boy on a 2.0 kg skateboard initially at
rest tosses an 8.0 kg jug of water in the
forward direction. If the jug has a speed
of 3.0 m/s relative to the ground and the
boy and the skateboard move in the
opposite direction at 0.60 m/s, find the
mass of the boy.
m1v1i + m2v2i = m1v1f + m2v2f
 m1=mass of jug = 8.0 kg
 m2= mass of boy and skateboard= 2.0 kg + x
 V1i = 0, v2i = 0
 V1f = 3.0 m/s, v2f=0.60 m/s
 0=8.0kg*3.0m/s forward + (2.0kg +x)*.60m/s back
 24 kg·m/s = (2.0 kg + x)*.60 m/s
 40 kg = 2.0 kg + x
 x= 38 kg
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Newton’s Third law and Collisions
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If the forces exerted in a collision are equal and
opposite
And the time each force is exerted would be the
same
Than the impulse on each object in a collision
would be equal and opposite
Since impulse is equal to the change in
momentum than the change in momentum would
be equal and opposite
So if one object gained momentum after a
collision than the other object must lose the same
amount of momentum
Collisions
When two objects collide and then move
together as one mass, the collision is called a
 Perfectly inelastic collision
 These are easy situations because the two
objects become pretty much one object after
the collision
 So the conservation of momentum equation
becomes
 m1v1i + m2v2i = (m1 + m2)vf
 The two objects will have the same final
velocity
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Example
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A grocery shopper tosses a 9.0 kg bag
of rice into a stationary 18.0 kg grocery
cart. The bag hits the cart with a
horizontal speed of 5.5 m/s toward the
front of the cart. What is the final speed
of the cart and the bag?
m1v1i + m2v2i = (m1 + m2)vf
 0 + 9.0kg * 5.5 m/s = (18kg + 9kg)vf
 50. kg m/s = 27*vf
 1.9 m/s
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Kinetic Energy and Inelastic
Collisions
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Total kinetic energy is not conserved in inelastic
collisions, it does not remain constant
Some of the energy is converted into sound
energy and internal energy as the objects
deforms during the collision
This is why it is called inelastic, elastic usually
means something that can keep its shape or
return to its original shape
In physics elastic means that the work done to
deform an object is equal to the work done to
return to its original shape
In inelastic collisions some of the work done on
the inelastic material is converted to other forms
of energy such as heat or sound
Example
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A 0.25 kg arrow with a velocity of 12 m/s
to the west strikes and pierces the
center of a 6.8 kg target.
 What is the final velocity of the combined
mass?
 What is the decrease in kinetic energy
during the collision?
m1v1i + m2v2i = (m1 + m2)vf
 0 + 0.25 kg *12 m/s = (0.25 kg+ 6.8 kg)v
 vf=0.43 m/s to the west
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KEi = 0+ ½ 0.25*(12m/s)2
= 18 J
KEf = ½ (7.1 kg)*(0.43 m/s)2
= .66 J
Change in kinetic = 17 J
Elastic Collisions
In an elastic collision, two objects collide
and return to their original shapes with
no loss of total kinetic energy.
 The two objects move separately
 Both total momentum and total
kinetic energy are conserved
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Since both are conserved than these
equations apply
 m1v1i + m2v2i =m1v1f +m2v2f
 ½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2
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Example
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A 16.0 kg canoe moving to the left at
12.5 m/s makes an elastic head-on
collision with a 14.0 kg raft moving to the
right at 16.0 m/s. After the collision, the
raft moves to the left at 14.4 m/s.
Disregard any effects of the water.
 Find the velocity of the canoe after the
collision.
 Verify your answer by calculating the total
kinetic energy before and after the collision.
Collisions in Two or more
Dimensions
Conservation of momentum still applies
 Use your vectors.
 Momentum is conserved in all directions
 pxi=pxf
 pyi=pyf
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mAvAxi +mBvBxi = mAvAxf +mBvBxf
 mAvAyi +mBvByi = mAvAyf +mBvByf
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Center of Mass (CM)
The point at which all mass of an
object can be considered to be
located
 An object can be considered a point or
small particle no matter what the size or
shape of the object
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Center of mass moves just as a particle
moves no matter what the object does
Finding the center of mass of an
object
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Define a coordinate system
(preferably one that would make the math easy)
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Center of mass can be found by finding
the sum of all the masses times their
respective distance from the defined
origin and dividing by the sum of all the
masses
m A x A  m B xB
x cm 
m A  mB
Center of gravity (CG)
The point at which the force of gravity
can be considered to act
 Gravity acts on all parts of the object
 But for determining translational motion
we can assume that gravity acts in one
particular spot
 Same spot as the center of mass
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