Conservation of Momentum
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Transcript Conservation of Momentum
Physics 11 - Chapter 7
Review: Momentum
• Momentum
• Impulse
p mv
J Ft mv p
Conservation of momentum:
• Momentum is conserved:
pI pF
p p'
• This is an expression of Newton’s first law:
– “An object at rest or in uniform motion will remain at rest or in
uniform motion unless acted on by an external force.”
• External forces can change the momentum of a system
(Impulse)
J Ft mv p
Conservation of momentum:
• In interactions between two bodies/objects,
momentum of one object can change, but the
total momentum of the system remains constant.
p1 p2 p1 ' p2 '
m1v1 m2v2 m1v1 'm2v2 '
Law of Conservation of Momentum
(continued…)
• The total momentum of objects before a collision
is the same as the total momentum of the same
objects after they collide.
• The change in momentum in an isolated system
is zero. The objects within the system may
interact and exchange momentum, but the total
momentum does not change!
Collisions:
• There are three
kinds of collision
– Elastic collision
– Inelastic collision
– Partially elastic
collision
Elastic Collision
The two
identical balls
hit each other
and bounce
back to the
same position
Partially Elastic Collision
The balls
bounce back
but not to
the same
original
position
Inelastic Collision
The balls do
not bounce
i.e. they stick
together
Types of Momentum Problems:
1. Elastic collisions
Initial
Final
2. Inelastic collisions
Initial
Final
3. Explosions
Initial
Final
Collisions in One Dimension:
• These are the only type we’ll be looking at!
• Momentum is a vector quantity, so both
magnitude and direction of the momentum must
be conserved.
Example #1:
A 1.75x104 kg boxcar is rolling down a track toward a
stationary boxcar that has a mass of 2.50x104 kg. Just
before the collision the first car is moving east at
5.45m/s. When the boxcars collide, they lock together
and continue down the track. What is the velocity of the
2 box cars after the collision?
Step 1= draw a diagram!
m1v1 m2v2 m1v1 'm2v2 '
•
•
What we know before the collision: m1=1750 kg, m2=2500 kg,
v1 = 5.45 m/s, v2= 0
What we know after the collision: m1=1750 kg, m2=2500 kg,
the 2 cars are now stuck together so they have the same
velocity! Velocity is what we are solving for.
m1v1 + m2v2 = m1v1' + m2v2‘
(1750)(5.45) + (2500)(0) = 1750v + 2500v
9537.5 = 4250v
2.24 m/s = v
Another type of conservation of
momentum problem: Recoil
Recoil = the interaction that occurs when
two stationary objects push against each
other and then move apart
Ex: Recoil
• A Barrett M82 is a high calibre sniper rifle.
Below are it’s specifications:
–
–
–
–
Barrel length: 73.7 cm
Rifle weight: 14.0 kg
Bullet velocity: 853 m/s
Typical bullet weight: 50.0g
• Calculate the magnitude force exerted on
the person shooting the gun.
Explosion: Recoil
How to go about the problem:
1. Calculate the momentum of the rifle knowing
the momentum of the bullet
2. Calculate the impulse imparted to the
riflemen to stop the gun.
•
•
Impulse is change in momentum
Impulse is force multiplied by time
– Need to know the time (how long) the explosion takes.
Calculate the velocity of the rifle using the
conservation of momentum:
What we know: initially neither the bullet or the rifle are moving,
rifle mass: 14.0 kg, bullet velocity: 853 m/s, bullet mass:
50.0g
m1v1 m2 v2 m1v1 ' m2 v2 '
0 m1v1 ' m2 v2 '
m1v1 ' m2 v2 '
m
(14.0kg)(v1 ' ) (0.0500kg)(853 s )
v1 ' 3.05 ms
To calculate the impulse:
• We know the person stops the gun, so to
find the force, we need to know the
interaction time.
J Ft mv p
p Ft
Calculating the time (way back to
kinematics!):
What we know: Barrel length (distance the bullet
travels): 73.7 cm, initial velocity (bullet) is 0m/s
and final velocity (bullet): 853 m/s
M
L
m
vI
vF
1
d vF vI t
2
2d
2(0.737 m)
t
0.00173s
m
m
vF vI 853 s 0.00 s
Almost done!
What we know: mass rifle (14.0kg), rifle velocity (3.05m/s), t = 0.00173s
pRifle p f pI
p m1v1 0
p (14.0kg)( 3.05 ms 1 ' )
p 42.65
kg m
s
p Ft
p
42.65
F
2.46 x10 4 N
t 0.00173
Try it :
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