Linear Momentum
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p = mass X velocity
p = mv
Unit = kg-m/s
Vector Quantity
What is the momentum of a 100.0 kg football
player running at 6.0 m/s?
p = mv = (100.0 kg)(6.0 m/s) = 600 kg-m/s
If a freshman running at 2 m/s has a
momentum of 80 kg m/s, what is his mass?
Momentum and Force
Second Law of Motion
“rate of change of momentum of a body is equal
to the net force applied on it.”
SF = Dp
Dt
(this is the more general form)
Derivation
SF = Dp
Dt
=
SF = m(v-vo)
Dt
SF = m Dv
Dt
SF = ma
mv-mvo
Dt
Impulse
• Impulse = Dp
Dp = FDt
• Usually occurs
over a very short
timeframe
Force
area = impulse
time
Impulse: Example 1
A 50-g golf ball is struck with a force of 2400 N.
The ball flew off with a velocity of 44 m/s. How
long was the club in contact with the ball?
Dp = pf – pi
Dp = mvf – 0 (ball was still)
Dp = (0.050 kg)(44 m/s) = 2.2 kg-m/s
SF = Dp/Dt
Dt = Dp/SF
Dt = 2.2 kg-m/s/(2400 kg-m/s2) = 9.1 X 10-4 s
Impulse: Example 2
In a crash test, a 1500 kg car moving at –15.0 m/s
collides with a wall for 0.150 s. The bumpers on
the car cause it to rebound at +2.60 m/s.
Calculate the impulse, and the average force
exerted on the car.
Impulse = Dp
pi = mvi = (1500 kg)(-15.0 m/s) = -22,500 kg-m/s
pf = mvf = (1500 kg)(+2.60 m/s) = +3900 kg-m/s
Dp = pf – pi
Dp = 3900 kg-m/s – (-22,500 kg-m/s)
Dp = 26,400 kg-m/s
To calculate the force:
SF = Dp
Dt
SF = 26,400 kg-m/s =
0.150 s
176,000 N
Impulse and Area
• Impulse = Area under Force vs. Time graph
• Can use the average Force to approximate
an answer
Average Force
Area = FaveDt
Connection to Calculus
• You integrate to get
area
Dp = FDt
dp = Fdt
tf
p = F dt = J
to
Calculate the impulse (Dp) given the
following graph.
Force vs. Tim e
12
10
Force(N)
8
6
4
2
0
0
2
4
6
Tim e (s)
8
10
12
Calculate the impulse (Dp) given the
following graph.
Calculate the impulse (Dp) given the
following graph.
A 150.0 g baseball is thrown with a speed of
20.0 m/s. The interaction force vs. time is
shown in the following graph.
a. Calculate the impulse. (4.50 kg m/s)
b. Calculate the return speed of the baseball. (10.0
m/s)
A 100.0 g golf ball is dropped from a height of 2.00
meters from the floor.
a. Calculate the speed of the ball just as it hits the
ground. (6.26 m/s)
b. Calculate the impulse. (1.2 kg m/s)
c. Calculate the return speed of the ball just as it
bounces back.(5.74 m/s)
d. Calculate the return height of the ball. (1.68 m)
The Law of Conservation of Momentum
“The total momentum of an isolated system of
bodies remains constant
m1v1 + m2v2 = m1v1’ + m2v2’
momentumbefore = momentumafter
A 0.015 kg bullet is fired with a velocity of
200 m/s from a 6 kg rifle. What is the
recoil velocity of the rifle (consider it’s
direction)?
A 10g and a 30 g ball are placed inside a tube
with a compressed spring between them.
When the spring is released, the 10 g ball
exits the tube at 6.0 m/s.
a. Calculate the speed of the 30 g ball.
b. Does it have to be in the opposite
direction?
A 10,000 kg railroad car moving at 24.0 m/s
strikes an identical car that is stationary. They
lock together. What will be there common speed
afterwards?
Mr. West (75 kg) sees a stationary cart (25.0
kg) 8.00 m ahead of him in the hallway. He
accelerates at 1.00 m/s2 and jumps on the
cart.
a. Calculate his speed just as he jumps on the
cart
b. Calculate his speed as he and the cart zoom
down the hall.
Rocket Ship
• Can use the Third Law (“equal and opposite
forces”)
• Can also use Law of Conservation of
momentum
 Dp = 0 (since neither the gas or rocket were
moving initially)
• Which moves more, the rocket or the gas?
A 0.145 kg ball is thrown at +40.2 m/s. The bat
has a mass of 0.840 kg.
a. How fast must you swing the bat to return the
ball at 20.1 m/s? (-10.4 m/s)
b. How fast must you swing the bat to return the
ball at 40.2 m/s? (-13.9 m/s)
Types of Collisions
• Elastic
– KE and momentum conserved
• Inelastic
– Momentum conserved
– KE not conserved
– Also, if energy is added (like in a chemical
reaction)
• Perfectly inelastic
– Two objects stick together after the collision
Perfectly Inelastic Collision: Ex 1
An 1800-kg Cadillac is stopped at a traffic light.
It is struck in the rear by a 900-kg Ion moving
at 20.0 m/s. The cars become entangled.
Calculate their velocity after the collision.
KE is not conserved
m1v1 + m2v2 = m1v1’ + m2v2’
m1v1 + m2v2 = (m1 + m2)vf
0 + (900 kg)(20.0 m/s) = (1800kg + 900kg)vf
vf = +6.67 m/s
How much KE was lost as a result of the
collision?
KEi = ½ (900 kg)(20 m/s)2 = 180,000 J
KEf = ½ (2700 kg)(6.67 m/s)2 = 60,000 J
DKE = 60,000 J – 180,000 J = -120,000J
Perfectly Inelastic Collision: Ex 2
Two balls of mud collide head on and stick
together. The first ball of mud had a mass of
0.500 kg and was moving at +4.00 m/s. The
second had a mass of 0.250 kg and was moving
at –3.00 m/s. Find the velocity of the ball after
the collision.
m1v1 + m2v2 = m1v1’ + m2v2’
m1v1 + m2v2 = (m1 + m2)vf
(0.500 kg)(4.00 m/s)+(0.250 kg)(-3.00 m/s) = (0.750 kg)(vf)
vf = +1.67 m/s
A lab experiment, a 200 g air track glider and
a 400 g glider collide and stick with velcro.
The 200 g glider was moving to the right at
3.00 m/s. Afterwards, the system is moving
to the left at 0.40 m/s. Calculate the initial
speed of the 400 g glider. (-2.1 m/s)
A uranium atom (238 amu) decays into a small
fragment and a “daughter” nucleus. The small
fragment is ejected at 1.50 X 107 m/s and the
daughter nucleus at 2.56 X 105 m/s. Calculate the
mass of both products. (234 amu, 4 amu)
Collisions in 2 or 3 Dimensions
(Glancing collisions)
• A moving marble collides with a stationary
marble.
• Which of the following situations is/are not
possible after:
2-D Collisions: Example 1
At an intersection, a 1500-kg
car travels east at 25 m/s.
It collides with a 2500-kg
van traveling north at 20
m/s. If the vehicles stick
together afterwards,
calculate the magnitude
and velocity of the cars
after the collision.
Let’s first work with the initial components
Spix = (1500 kg)(25 m/s)
Spix = 37,500 kg-m/s
Spiy = (2500 kg)(20 m/s)
Spiy = 50,000 kg-m/s
After the collision:
v
vy = v sinq
q
vx = v cosq
After the collision:
Spix = Spfx
37,500 kg-m/s = mvx
37,500 kg-m/s = (1500 kg + 2500 kg)vx
37,500 kg-m/s = (4000 kg)vcosq
vcosq = 9.375 m/s
Spiy = Spfy
50,000 kg-m/s = mvy
50,000 kg-m/s = (1500 kg + 2500 kg)vy
50,000 kg-m/s = (4000 kg)vsinq
vsinq = 12.5 m/s
Divide the two equations
vsinq = 12.5 m/s
vcosq = 9.375 m/s
sinq = 1.33
cosq
tanq = 1.33
Now solve for v
vsinq = 12.5 m/s
v = (12.5 m/s)/sin (53o)
v = 16 m/s
q = 53o
A 5000 kg van travelling south at 10.0 m/s collides
with a 2500 kg SUV travelling west. After the
collision, the vehicles stuck together and made a
45o angle with the horizontal in a south-west
direction.
a. Calculate the initial speed of the 2500 kg car.
b. Calculate the speed just after the collision.
(Remember, two equations, two unknowns).
2-D Collisions: Example 3
A pool ball moving at +3.0 m/s strikes another pool
ball (same mass) that is initially at rest. After the
collision, the two balls move off at + 45o.
Calculate the speeds of the two balls. (2.1 m/s)
vred
q = +450
q = -450
vgreen
Let’s first work with the initial components
Spix = (m)(3 m/s)
Spiy = 0
After the collision (x-direction):
vredx = vredcos(45o)
vgreenx = vgreencos(-45o)
Spix = Spfx
Spix = predfx + pgreenfx
(m)(3 m/s) = mvredcos(45o) + mvgreencos(-45o)
3 m/s = vredcos(45o) + vgreencos(-45o)
After the collision (y-direction):
vredy = vredsin(45o)
vgreeny = vgreensin(-45o)
Spiy = Spfy
Spiy = predfy + pgreenfy
0 = mvredsin(45o) + mvgreensin(-45o)
mvredsin(45o) = -mvgreensin(-45o)
vredsin(45o) = -vgreensin(-45o)
Here are the two equations we will solve:
3 m/s = vredcos(45o) + vgreencos(-45o)
vredsin(45o) = -vgreensin(-45o)
vred = -vgreensin(-45o)
sin(45o)
vred = vgreen
3 m/s = vredcos(45o) + vredcos(-45o)
3 m/s = vred(cos(45o) + cos(-45o))
3 m/s = 1.414vred
vred = 2.1 m/s = vgreen
Two zombie heads of unequal mass sit on an
icerink. The first head (m1 = 4.00 kg) is
propelled toward the stationary second head
(m2 = 2.00 kg) at a velocity of v1= 4.00 m/s.
After the collision, both heads move off at 35o
to the x-axis. Calculate their final speeds.
1.63 m/s, 3.25 m/s
Two different pucks are placed on an air hockey table.
A 15.0 g red puck is pushed to the right at 1.00 m/s.
A 20.0 g green puck is pushed to the left at 1.20 m/s.
After the collision, the green puck travels at 1.10 m/s
at an angle of 40.0o south of the horizontal.
a) Calculate the x and y components of the green puck’s
velocity after the collision. (-0.843 m/s, -0.707 m/s)
b) Calculate the x and y components of the green puck’s
momentum after the collision. (-0.0169 kg m/s,
-0.0141 kg m/s)
c) Find the speed and direction of the red puck. (1.08
m/s, 60.7o north of east)
d) Can you express your answer to (c) in “i and j”
notation? ((0.527 i + 0.940j) m/s)