Transcript Linear Momentum

Linear Momentum
• Momentum is a measure of how hard it is to stop or
turn a moving object.
• What characteristics of an object would make it
hard to stop or turn?
• For one particle
– p = mv
– Note that momentum is a vector with the same direction
as the velocity!
– NO VELOCITY = NO MOMENTUM
• For a system of multiple particles
– p = Σpi --- add up the vectors
• The unit of momentum is…
– kg m/s or Ns
• Calculate the momentum of a 65-kg sprinter running east
at 10 m/s.
p = mv
p = 65 Kg ·10 m/s = 65 kg·m/s east
• Calculate the momentum of a system composed of a 65-kg
sprinter running east at 10 m/s and a 75-kg sprinter
running north at 9.5 m/s.
Px
2
2
2
A +b =c
System: 2 runners P
y
P12 + P22 = P2
P=√(P12 + P22)
P=√(ΣPx)2 + (ΣPy)2)
P1 + P2 ≠ P
p = mv
P=√((mxvx)2 + Σ(myvy)2)
P=√((10m/s·65 kg)2 + (9.5 m/s·75 kg)2) =
964.45 N·s
• Calculate the momentum of a 65-kg sprinter running east
at 10 m/s.
p = mv
p = 65 Kg ·10 m/s = 65 kg·m/s east
• Calculate the momentum of a system composed of a 65-kg
sprinter running east at 10 m/s and a 75-kg sprinter
running north at 9.5 m/s.
Px
System: 2 runners P
y
θ
Now need direction
because momentum is a
vector
θ = Tan-1 (O/A)
p = mv
P1 + P2 ≠ P
θ = Tan-1 (ΣPx / ΣPx )
θ = Tan-1 (mxvx / myvy )
θ = Tan-1 (65kg · 10m/s / 75kg ·
9.5m/s ) = 42.37º E of N
• Like any change, change in momentum is calculated by
looking at final and initial momentums.
• Δp = pf – pi
– Δ p: change in momentum
– pf: final momentum
– pi: initial momentum
• Using only a meter stick, find the momentum change of
each ball when it strikes the desk from a height of exactly
one meter.
• Which ball, tennis or golf, has the greatest change in
momentum?
•In which case is
the magnitude of
the momentum
change greatest?
• Impulse is the product of an external force and
time, which results in a change in momentum of a
particle or system.
• J = F t and J = ΔP
• Therefore Ft = ΔP
• Units: N s or kg m/s (same as momentum)
• To increase momentum: exert more
force or exert force for a longer period of
time
• Ex: follow through in sports – tennis,
hockey, etc.
• To decrease momentum: force must be
in opposite direction of motion (negative)
• Ex: friction, air resistance, tree slowing
down a car
• Changes in momentum caused by cushy
objects have less force and more time
than those caused by hard items.
• Double time, cut force in half
• Ex: Jump off a 3 meter platform onto concrete
– time to slow down your motion is very small
once you contact the cement so the force is
great.
• Jump off a 3 meter platform onto a thick mat –
time to slow down your motion is extended
because the matt gives so the force is lessened.
• Ex: Car crumple zones increase the time of
impact so that less force is exerted causing
fewer injuries to passengers.
• Ex: bending legs when landing, moving away
from a punch, stretchy cord when bungee
jumping
Usually high magnitude,
short duration.
Suppose the ball hits the
bat at 90 mph and leaves
the bat at 90 mph, what is
the magnitude of the
momentum change?
What is the change in the
magnitude of the
momentum?
F(N)
Impulse on a graph
4000
3000
Area under curve is the impulse
2000
1000
0
0
1
2
3
4
5
t(ms)
• Suppose a 1.5-kg brick is dropped on a glass table top from
a height of 20 cm.
– What is the magnitude and direction of the impulse
necessary to stop the brick?J = ΔP Vf=Vi + at
d = 0.5at2 t= √(2·0.2m/9.8m/s2) = 0.202s Vf=0m/s + (-9.8 m/s2· 0.202s) = -1.98
m/s
t= √(2d/a) J = P - P
J = mvf – mvf J = 1.5kg· 0m/s – (-1.5kg· 1.98m/s) =
f
i
+2.97 kg· m/s
– If the table top doesn’t shatter, and stops the brick in
0.01 s, what is the average force it exerts on the brick?
Ft = ΔP
F = ΔP/t
F = -2.97 kg· m/s / 0.01s = +296.94 N
– What is the average force that the brick exerts on the
table top during this period?
F = 2.97 kg· m/s / 0.01s = -296.94 N
F(N)
3000
2000
1000
0
0
0.2
0.4
0.6
0.8
t(s)
This force acts on a 1.2 kg object moving at 120.0 m/s. The
direction of the force is aligned with the velocity. What is the
new velocity of the object?
F = 2500 N m = 1.2 kg
Ft = mv
t = 0.1s vi = 120m/s
v = Ft/m
v = 208.3 m/s
vf = ?
v = 2500N· 0.1s/1.2Kg
v = 208.3 m/s + 120m/s
=328.3 m/s
• A 75-kg man sits in the back of a 120-kg canoe that
is at rest in a still pond. If the man begins to move
forward in the canoe at 0.50 m/s relative to the
shore, what happens to the canoe?
System
Before
After
Vboat
Vperson
mman = 75kg mcanoe = 120kg
ΣPi = ΣPf
vman = 0.5m/s vcanoe = ?
0=vcanoemcanoe + vmanmman
vcanoe = - vmanmman / mcanoe
vcanoe = - 0.5 m/s· 75 kg / 120kg = - 0.3215 m/s
• External forces: forces coming from outside the
system of particles whose momentum is being
considered. External forces change the momentum
of the system.
• Internal forces: forces arising from interaction of
articles within a system. Internal forces cannot
change momentum of the system.
EXTERNAL FORCE EXAMPLE
The club head exerts
an external impulsive
force on the ball and
changes its
momentum.
• The acceleration of the
ball is greater because its
mass is smaller.
Pi ≠Pf
System
INTERNAL FORCE EXAMPLE
•The forces the
balls exert on each
other are internal
and do not change
the momentum of
the system.
• Since the balls have
equal masses, the
magnitude of their
accelerations is equal.
Pi = Pf
System
• Explosions
– When an object separates suddenly, as in an explosion,
all forces are internal.
– Momentum is therefore conserved in an explosion.
– There is also an increase in kinetic energy in an
explosion. This comes from a potential energy decrease
due to chemical combustion.
• Recoil
– Guns and cannons “recoil” when fired.
– This means the gun or cannon must move backward as
it propels the projectile forward.
– The recoil is the result of action-reaction force pairs,
and is entirely due to internal forces. As the gases from
the gunpowder explosion expand, they push the
projectile forwards and the gun or cannon backwards.
• Collision
– When two moving objects make contact with each other,
they undergo a collision.
– Conservation of momentum is used to analyze all
collisions.
– Newton’s Third Law is also useful. It tells us that the
force exerted by body A on body B in a collision is equal
and opposite to the force exerted on body B by body A.
Collisions Continued
•During a collision,
external forces are
ignored.
• The time frame of the
collision is very short.
• The forces are
impulsive forces (high
force, short duration).
System
• Collision Types
– Elastic collisions
• Also called “hard” collisions
• No deformation occurs, no kinetic energy lost
– Inelastic collisions
• Deformation occurs, kinetic energy is lost
– Perfectly Inelastic (stick together)
• Objects stick together and become one object
• Deformation occurs, kinetic energy is lost
• (Perfectly) Inelastic Collision
– Simplest type of collisions.
– After the collision, there is only one velocity, since there
is only one object.
– Kinetic energy is lost.
– Explosions are the reverse of perfectly inelastic
collisions in which kinetic energy is gained!
• Example Problem Inelastic collision (stick together)
•An 80-kg roller skating
grandma collides
inelastically with a 40-kg
kid.
•What is their velocity
after the collision?
• How much kinetic
energy is lost?
a. No external force = 0
ΣPi = Σ Pf
m1iv1i + m2iv2i = m1fv1f + m2fv2f
m1iv1i + m2iv2i = (m1f + m2f)vf
vf = m1iv1i + m2iv2i / (m1f + m2f)
System (Grand mother and child
vf = (80 kg · 6 m/s+ 40 kg
·0m/s)/ (80 kg + 40 kg) = 4
m/s
• Example Problem Inelastic collision (stick together)
•An 80-kg roller skating
grandma collides
inelastically with a 40-kg
kid.
•What is their velocity
after the collision?
• How much kinetic
energy is lost?
b. ΔK = Kf - Ki
System (Grand mother and child
ΔK = 1/2mfv2f – 1/2miv2i
ΔK = 1/2(m1+m2)fv2f – 1/2mi1v2i
ΔK = 1/2(80 kg+40 kg)· (4 m/s)2 – 1/2· 80 kg· (6 m/s)2 = -480J
• A car with a mass of 950 kg and a speed of 16 m/s
to the east approaches an intersection. A 1300-kg
minivan traveling north at 21 m/s approaches the
same intersection. The vehicles collide and stick
together. What is the resulting velocity of the
vehicles after the collision?
C
Collide
V
• A car with a mass of 950 kg and a speed of 16 m/s
to the east approaches an intersection. A 1300-kg
minivan traveling north at 21 m/s approaches the
same intersection. The vehicles collide and stick
together. What is the resulting velocity of the
vehicles after the collision?
P = √((mxvx)2 + (myvy)2)
ΣPi = Σ Pf
Sum your momentum in the x
and y direction
P
Px
P = √((1300kg · 21 m/s)2 + (950kg · 16
m/s )2) = 31246.28 Ns
mcvc + mvvv = v(mc+mv)
v = (mcvc + mvvv) / (mc+mv)
(mcvc + mvvv) = P = 31248.28 Ns
Py
P2 = Px2 + Py2
P = √(Px2 + Py2)
v = (mcvc + mvvv) / (mc+mv)
v = P / (mc+mv)
v = 31248.28 Ns / (950 kg+1300 kg) = 13.88 m/s
• A car with a mass of 950 kg and a speed of 16 m/s
to the east approaches an intersection. A 1300-kg
minivan traveling north at 21 m/s approaches the
same intersection. The vehicles collide and stick
together. What is the resulting velocity of the
vehicles after the collision? Solution
v 2
vy = 12.13333
m/s
ΣPi = Σ Pf
ΣPix = Σ Pfx
ΣPiy = Σ Pfy
mcvc = (mc+mv)v
mvvv = (mc+mv)v
vx = mcvc / (mc+mv)
vy = mvvv / (mc+mv)
vx = 950 Kg · 16 m/s / (950
kg + 1300 kg) = 6.7556m/s
vx = 6.7556 m/s
v2 = vx2 + vy2
v = √(vx2 + vy2)
v = √((6.7556 m/s)2 +
(12.13333 m/s)2) = 13.89 m/s
vy = 1300 Kg · 21 m/s / (950 kg +
1300 kg) = 12.1333 m/s
• A car with a mass of 950 kg and a speed of 16 m/s
to the east approaches an intersection. A 1300-kg
minivan traveling north at 21 m/s approaches the
same intersection. The vehicles collide and stick
together. What is the resulting velocity of the
vehicles after the collision? Solution
v 2
vy = 12.13333
P
m/s
Px = myvy
vx
Py = mxvx
Tan θ = O/A
θ = Tan (O/A)
θ = Tan -1 (myvy / mxvx)
θ = Tan -1 (1300 kg· 21 m/s /
950 kg · 16 m/s) = 60. 89º
-1
= 6.7556 m/s
Tan θ = O/A
θ = Tan -1 (O/A)
θ = Tan -1 (12.13333 m/s /
6.7556 m/s ) = 60.89º
•
•
•
•
•
Conservation of Momentum
For a closed system (no external forces),
total momentum remains the same.
Momentum cannot be created or
destroyed, but can be transferred from
one object to another.
Ex:
Cannon with cannon ball ready to fire
has 0 kgm/s of momentum.
Cannon firing –-> cannon exerts force on
cannon ball, cannon ball exerts same
force back on the cannon.
• These forces occur over the same time
period.
• Cannon ball and cannon have equal and
opposite impulses () – they add up to
zero.
• Because the cannon has more mass, it
accelerates less than the cannon ball.
• The acceleration of the cannon is called
recoil – it moves backwards because the
cannon ball is shot out forward.
• Suppose a 5.0-kg projectile launcher shoots a 209 gram
projectile at 350 m/s. What is the recoil velocity of the
projectile launcher? ΣPi = Σ Pf ΣPi = 0
Σ Pf = mlavla + mpvp
vla = - mpvp / mla
0 = mlavla + mpvp
mlavla = - mpvp
vla = - 0.209 kg · 350 m/s / 5.0kg = 14.63 m/s
• An exploding object breaks into three fragments. A 2.0 kg
fragment travels north at 200 m/s. A 4.0 kg fragment
travels east at 100 m/s. The third fragment has mass 3.0
kg. What is the magnitude and direction of its velocity?
ΣPx = mxvx
Explosion
ΣPy = myvy
• An exploding object breaks into three fragments. A 2.0 kg
fragment travels north at 200 m/s. A 4.0 kg fragment
travels east at 100 m/s. The third fragment has mass 3.0
kg. What is the magnitude and direction of its velocity?
P
Tan θ = O/A
Px = myvy
Py = mxvx
P2 = Px2 + Py2
θ = Tan -1 (O/A)
θ = Tan -1 (myvy / mxvx)
θ = Tan -1 (2.0 kg· 200 m/s /
4.0 kg · 100 m/s) =
P = √(Px2 + Py2)
P = √((mxvx)2 + (myvy)2)
P = √((4 kg · 100 m/s)2 + (2.0 kg ·
200 m/s)2) = 565.69 Ns
P3 frag =
565.69 Ns
45º
P3 frag = m3v3
v3 = P3 frag / m3
v3 = 565.69 Ns / 3.0 kg = 188.56 m/s
• In elastic collisions, there is no deformation of colliding
objects, and no change in kinetic energy of the system.
Therefore, two basic equations must hold for all elastic
collisions
– ΣPb = Σ Pa (momentum conservation)
– Σ Kb = Σ Ka (kinetic energy conservation
• A 500-g cart moving at 2.0 m/s on an air track elastically
strikes a 1,000-g cart at rest. If the 500 g cart has a velocity
of 1 m/s after the collision, what are the resulting velocities
of the two carts?
m v =m v -m v
ΣPb = Σ Pa
m1bv1b + m2bv2b = m1av1a +
m2av2a
V2b = 0 m/s
m1bv1b = m1av1a + m2av2a
2a 2a
1b 1b
1a 1a
v2a = (m1bv1b - m1av1a )/ m2a
v2a = (0.5 kg· 2 m/s – 0.5 kg· 1
m/s )/ 1 kg = 0.5 m/s
• 2 – D collisions
– Momentum in the x-direction is conserved.
• ΣPx (before) = ΣPx (after)
– Momentum in the y-direction is conserved.
• ΣPy (before) = ΣPy (after)
– Treat x and y coordinates independently.
• Ignore x when calculating y
• Ignore y when calculating x
– Let’s look at a simulation:
• http://surendranath.tripod.com/Applets.html
• Calculate velocity of 8-kg ball after the collision.
3 m/s
2 m/s
y
y
8 kg
x
2 kg
θ1f = 50º
θ2f = ?
2 kg
x
0 m/s
Break down velocity of ball 1 in
x and y components
v = 2 m/s
θ = 50º
vx1 = vcos θ
vy1 = vsin θ
v2f = ?
8 kg
vy1 =2m/s sin 50º = 1.53 m/s
vx1 =2m/s cos 50º = 1.29 m/s
• Calculate velocity of 8-kg ball after the collision.
3 m/s
2 m/s
y
y
8 kg
x
2 kg
θ1f = 50º
θ2f = ?
2 kg
x
0 m/s
vy1 = 1.53 m/s
vx1 = 1.29 m/s
Determine v2xf
v2f = ?
8 kg
Determine v2yf
m1bv1b + m2bv2b = m1av1a + m2av2a
m1bv1b + m2bv2b = m1av1a + m2av2a
V2b = 0m/s
v2a = (m1bv1b - m1av1a ) / m2a
V2b = 0m/s
V1b = 0m/s
v2a = (- m1av1a ) / m2a
v2a = (2 kg· 3 m/s – 2 kg · 1.29 m/s
) / 8 kg = 0.4275 m/s
v2a = (– 2 kg · 1.53 m/s ) / 8 kg =
- 0.3825 m/s
• Calculate velocity of 8-kg ball after the collision.
3 m/s
2 m/s
y
y
8 kg
2 kg
θ1f = 50º
x
θ2f = ?
2 kg
0 m/s
v2f = ?
8 kg
v2x = 0.4275 m/s
θ= ?
v= ?
v2y = 0.3825 m/s
x
Tan θ = O/A
θ = Tan -1 (v2y/v2x)
v2 = v2x2 + v2y2
v =√( v2x2 + v2y2 )
θ = Tan -1 (0.3825 m/s /
0.4275 m/s) = 41.82º
v =√( (0.4275 m/s)2 + (0.3825 m/s)2 ) = 0.574 m/s