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Momentum (vector quantity)
• p = mass X velocity
• p = mv
• Unit = kg m/s
(p comes from “impetus”)
What is the momentum of a 100.0 kg football
player running at 6.0 m/s?
The Law of Conservation of Momentum
If Mr. West is running at 2 m/s and has a
momentum of 80 kg m/s, what is his mass?
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)?
m1v1 + m2v2 = m1v1’ + m2v2’
The Rhino has a mass of 900 kg and hits the Hulk
running at 2 m/s. After the collision, the Hulk is
moving at 1.3 m/s and Rhino stops. What is the
Hulk’s mass?
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Two boats meet in the middle of a lake. Boat A has a
mass of 150 kg, and boat B a mass of 250 kg. The
person in Boat A pushes against Boat B and has a
speed of -1.4 m/s.
a) What is the initial momentum of Boat A?
b) Calculate the velocity of Boat B.
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 their common speed
afterwards? (Ans: 12 m/s)
Rocket Ship
• Before moving, p = 0
• Which moves faster, the rocket or the gas?
A 940-kg Mazda Miata collides into the rear of a
2460-kg pick-up truck which was at rest at a light.
The Mazda’s pre-collision speed was 12.5 m/s.
Determine the post-collision speed of the two
entangled cars.
1. A 0.145 kg ball is thrown at +40.2 m/s. The bat
has a mass of 0.840 kg. How fast must you
swing the bat to return the ball at 20.1 m/s?
Assume the bat stops when you hit the ball.
(ANS: -10.4 m/s)
2. How fast must you swing the bat to return the
ball at 40.2 m/s? (ANS: -13.9 m/s)
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If a 5-kg bowling ball is projected upward with a
velocity of 2.0 m/s, then what is the recoil velocity of
the Earth (mass = 6.0 x 1024 kg).
A 25.0-gram bullet enters a 2.35-kg watermelon (at rest)
with a speed of 217 m/s and exits the opposite side with a
speed of 109 m/s.
a. Draw a picture and explain what will happen?
b. Was momentum conserved?
c. Calculate the final speed of the watermelon.
v = 1.67 X 10-24 m/s (downward)
A 3000-kg truck moving rightward with a speed of 5
km/hr collides head-on with a 1000-kg car moving
leftward with a speed of 10 km/hr. The two vehicles stick
together and move with the same velocity after the
collision. Determine the post-collision velocity of the car
and truck.
Types of Collisions
• Elastic
– KE and momentum are conserved
• Inelastic
– Momentum is conserved
– KE is not conserved
• Lost (heat, sound)
• Added (chemical reaction)
• Perfectly inelastic
– Momentum is conserved
– Two objects stick together after the collision
Elastic Collisions
• Both KE and momentum are conserved
m1v1i + m2v2i = m1v1f + m2v2f
m1v1i - m1v1f = m2v2f - m2v2i
m1(v1i - v1f) = m2(v2f - v2i)
½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2
m1v1i2 - m1v1f2 = m2v2f2 - m2v2i2
m1(v1i2 - v1f2) = m2(v2f2 - v2i2)
Remember that (a2 – b2) = (a + b)(a – b)
m1(v1i + v1f) (v1i - v1f) = m2 (v2i + v2f)(v2f - v2i)
Divide the yellow box equations:
m1(v1i + v1f) (v1i - v1f) = m2 (v2i+ v2f)(v2f - v2i)
m1(v1i - v1f)
m2(v2f - v2i)
v1i + v1f = v2i + v2f
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Elastic Collisions: Example 1
A cue ball moving at 2.0 m/s strikes the red ball.
What is the speed of both balls after the elastic
collision if they have equal mass?
Elastic Collisions: Example 2
Two pool balls of equal mass collide. One is
moving to the right at 20 cm/s, and the other to
the left at 30 cm/s. Calculate their velocities
after they collide elastically.
+20 cm/s
-30 cm/s
vwf = -50 cm/s + vbf
vwf = -50 cm/s + (+20 cm/s) = -30 cm/s
+20 cm/s
-30 cm/s
-30 cm/s
+20 cm/s
mvci + mvri = mvcf + mvrf all masses are equal
mvci + 0 = mvcf + mvrf
The red ball is still
vci = vcf + vrf
Factored out the m
2.0 m/s = vcf + vrf
Need another equation
vci + vcf = vri + vrf
2.0 m/s + vcf = 0 + vrf`
Substitute
vrf` = 2.0 m/s + vcf
2.0 m/s = vcf + 2.0 m/s + vcf
0 = 2vcf or vcf = 0
2.0 m/s = vcf + vrf = 0 + vrf
vrf = 2.0 m/s
mvwi + mvbi = mvwf + mvbf
vwi + vbi = vwf + vbf
20 cm/s + (-30 cm/s) = vwf + vbf
-10 cm/s = vwf + vbf
Need another equation
vwi + vwf = vbi + vbf
KE is conserved
vwf = vbi + vbf - vwi
vwf = -30 cm/s + vbf – 20 cm/s
vwf = -50 cm/s + vbf
Now substitute
-10 cm/s = -50 cm/s + vbf + vbf
40 cm/s = 2vbf
vbf = +20 cm/s
Elastic Collisions: Example 3
A proton of mass 1.00 amu moving at 3.60 X 104
m/s elastically collides head on with a still
Helium nucleus (4.00 amu). What are the
velocities of the particles after the collision?
vpf = -2.16 X 104 m/s
Note how they exchanged velocities (because they have equal mass)
vhf = 1.44 X 104 m/s
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vpf = vpi(mp – mh) = 3.60 X 104 m/s(1.01 amu-4,00 amu)
(mp + mh)
(1.01 amu + 4.00 amu)
vpf = -2.15 X 104 m/s
A proton of mass 1.00 amu moving at 3.60 X 104
m/s elastically collides head on with a Helium
nucleus (4.00 amu) moving at -1.80 X 104 m/s.
Calculate the velocities of the particles after the
collision?
vpi + vpf = vhf
vhf = 3.60 X 104 m/s + (-2.15 X 104m/s)
vhf = 1.45 X 104 m/s
(vp = -5.04 X 104 m/s, vHe 3.6 X 103 m/s)
(1.00)(3.60X104) +(4.00)(-1.8X104) = 1.00v1f + 4.00v2f
-3.6X104 = 1.00v1f + 4.00v2f
A 2.00 kg mass moves east at 5.00 m/s and collides
with a 3.00 kg travelling west at 5.00 m/s. Caculate
the final speed of both masses.
3.60X104 +v1f = -1.8X104+ v2f
5.40X104 +v1f = v2f
5.40X104 = -v1f + v2f
-3.6X104 = 1.00v1f + 4.00v2f
5.40X104 = -v1f + v2f
1.8 X 104 = 5v2f
v2f = 3.6 X 103 m/s
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 Cube
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?
Ki = ½ (900 kg)(20 m/s)2 = 180,000 J
Kf = ½ (2700 kg)(6.67 m/s)2 = 60,000 J
DK = 60,000 J – 180,000 J = -120,000J
What percent of the initial kinetic energy was
lost?
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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.
a. Find the velocity of the ball after the collision.
b. Calculate the percentage of kinetic energy lost.
95.0 kg Big Frank running at 3.50 m/s collides with a
120.0 kg tackle running in the opposite direction at
4.00 m/s. They grapple and stick together.
a) Calculate the velocity after the collision. (-0.69 m/s)
b) Calculate the kinetic energy of the system before the
collision. (1542 J)
c) Calculate the kinetic energy after the collision. (51 J)
d) What percent of the kinetic energy was lost in the
collision? (97%)
vf = +1.67 m/s , 79.5%
Ballistic Pendulum
Perfectly Inelastic Collision: Ex 3
A 5.00 gram bullet is fired into a 1.00 kg block of
wood. The wood-bullet system rises 5 cm.
Calculate the initial velocity of the bullet.
Used to determine
velocities
Perfectly Inelastic
collisions (KE
converts to PE)
m1v 1 + m 2v 2 = m 1v 1’ + m 2v 2’
m1v1 + m2v2 = (m1 + m2)vf
0 +(0.005 kg)(vi) = (1.00 kg + 0.005 kg)(v f)
(0.005 kg)(vi) = (1.005 kg)(vf)
A 7.00 g bullet is fired into a 0.950 kg ballistic
pendulum. The bob rises to a height of 22.0 cm.
Calculate the initial speed of the bullet.
Moment of impact
We need a second equation (two unknowns)
½ mv21 + mgy1 = ½ mv22 + mgy2
½(1.005 kg)(vf2) +0 =0 + (1.005 kg)(9.8 m/s2)(0.05 m)
vf = 0.990 m/s
(0.005 kg)(vi) = (1.005 kg)(0.990 m/s)
vi = 199 m/s
(284 m/s)
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An 8.00 g bullet is fired at 350 m/s into a ballistic
pendulum bob of 2.00 kg.
a. Calculate how high the coupled bullet and bob
will rise. (9.9 cm)
b. Calculate the angle and the horizontal distance
the bob travels. The length of the pendulum is 80
cm. (28.8o, 38.5 cm)
A bullet of mass 10.0 grams and a muzzle velocity
350.0 m/s is fired into a block of mass
500.0 grams. The string is 7.00 m long.
a. Calculate the velocity of the block and bullet
after the impact. (6.86 m/s)
b. Calculate the height to which the pendulum will
rise. (2.40 m)
A 10.0 g bullet is fired into a 1.20 kg ballistic
pendulum at a speed of 320 m/s. The string on
the pendulum is 150 cm.
a. Calculate the height that the pendulum rose.
(35.6 cm)
b. Calculate the angle the pendulum rises. (40.3o)
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
Derivation
SF = Dp
Dt
=
mvf-mvi
Dt
• Impulse = Dp
(this is the more general form)
Impulse
Dp = FDt
• Usually occurs over a very short timeframe
SF = m(vf-vi)
Dt
SF = m Dv
Dt
SF = ma
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A 50-g golf ball is struck with a force of 2400 N.
The ball flew off with a velocity of 44 m/s.
a) Calculate the impulse (2.2 kg-m/s)
b) How long was the club in contact with the ball?
(9.2 X 10-4 s)
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
A 0.144 kg baseball is moving toward homeplate at
-43.0 m/s. It is bunted with a force of 6500 N for 1.30 ms.
a) Calculate the initial momentum of the ball. (-6.19 kg m/s)
b) Calculate the impulse (the change in momentum) (8.45 kg
m/s)
c) Calculate the final momentum (2.26 kg m/s)
d) Calculate the final velocity of the ball. (+15.7 m/s)
Impulse and Area
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.
a) Calculate the impulse (26,400 kg-m/s)
b) Calculate the average force exerted on the car.
(176,000 N)
A 50.0 kg physics student jumps for joy over
her grade. She jumps up with a speed of 2.1
m/s for 0.36 s.
a) Calculate her impulse (105 kg m/s)
b) Calculate the force she exerts on the floor (292
N)
c) Calculate the maximum height of her jump
(use conservation of energy or kinematics).
(22.5 cm)
• Can use the average Force to approximate
an answer
• Impulse = Area under Force vs. Time graph
Average Force
Area = FaveDt
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Connection to Calculus
• You integrate to get
area
Dp = FDt
dp = Fdt
Calculate the impulse (Dp) given the
following graph.
Force vs. Tim e
12
10
p = F dt
Force(N)
8
tf
6
4
to
2
0
0
2
4
6
8
10
12
Tim e (s)
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)
Calculate the impulse (Dp) given the
following graph.
A 100.0 g golf ball is dropped from a height of 2.00
meters from the floor.
a. Calculate the velocity 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)
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a. Calculate the
impulse delivered
by the force
graphed in the
figure above
between t = 0 and
t = 5.
b. If the object was a
5 kg box,
calculate the final
speed of the box.
Collisions in 2 or 3 Dimensions
(Glancing collisions)
A 10.0 g bouncey ball is
dropped from a height of
1.50 meters from the floor.
a. Calculate the velocity of the
ball just as it hits the ground.
(-5.42 m/s)
b. Calculate the impulse.
(0.075 kg m/s)
c. Calculate the return speed of
the ball just as it bounces
back.(2.08 m/s)
d. Calculate the return height of
the ball. (22.1 cm)
If both masses are equal, then the angles of
deflection add to 90o.
• A moving marble collides with a stationary
marble.
• Which of the following situations is/are not
possible after:
Collisions in 2 or 3 Dimensions
Problem Solving:
1. Resolve all vectors into their x and y
components
2. Spix = Spfx
Spiy = Spfy
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 speed and
direction of the cars after
the collision.
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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:
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) vx
vx = 9.375 m/s
After the collision:
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) vy
vy= 12.5 m/s
v
vy = v sinq
q
vx = v cosq
An 8.00 kg mass collides with a 5.00 kg mass that
is at rest. Initially, the 8.00 kg mass was traveling
to the right at 4.50 m/s. After the collision, it is
moving with a speed of 3.65 m/s and at an angle
of 27.0° to its original direction. The 5.00 kg
mass was travelling at 3.32 m/s, 53.0o below the
horizontal. Prove that momentum was conserved.
Now solve for v
v2 = vx2+vy2
v = 16 m/s
Divide the two equations
tanq = vy = 12.5
vx
9.38
q = 53o
2-D Collisions: Example 1
A pool ball moving at +3.00 m/s strikes another
pool ball (same mass) that is initially at rest.
After the collision, the two balls move off at
45.0o angles. Calculate the speeds of the two
vred
balls.
q = +450
x coordinate
mgvg + mrvr = mgvg + mrvr
3m + 0 = mgvg + mrvr
3 = vg + vr
3 = vgcos(45o) + vrcos(-45o)
3 = 0.707vg + 0.707vr
4.24 = vg + vr
q = -450
vgreen
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y coordinate
0 = mgvg + mrvr
0 = mgvg + mrvr
0 = vg + vr
0 = vgsin(45o) + vrsin(-45o)
0 = 0.707vg - 0.707vr
vg = vr
Two equations, two unknowns
4.24 = vg + vr
vg = vr
2.12 m/s
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.
2.44 m/s, 4.88 m/s
Two 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) Find the speed and direction of the red puck. (1.08
m/s, 60.7o north of east)
b) Can you express your answer to (b) in “i and j”
notation? ((0.527 i + 0.940j) m/s)
A 100.0 kg man running east at 1.00 m/s collides
with a 56.0 kg man running west at 2.00 m/s.
After the collision, the 100.0 kg man is now
running at 0.800 m/s, 30.0o north of east.
Calculate the speed and the angle of the 56.0 kg
man.
A 1000 kg car travelling south at 15.0 m/s is hit
by a 1500 kg van travelling west at 20.0 m/s.
They stick together.
a) Calculate the final speed of the cars. (13.4 m/s)
b) Calculate the direction that the cars will travel
in. (26.6o South of West)
c) Express your answer in “i and j” notation.
(1.62 m/s, 26.2o S of W)
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A 2000 kg truck heading south at 25 m/s collides
with a 1000 kg car moving east at 25 m/s. The
two become entangled. Calculate the final
velocity (speed and direction). Also, express your
answer in “i” and “j” notation.
(18.6 m/s, 63.5o S of E, (8.3i – 16.6j) m/s
Center of Mass
Center of mass – one point on a particle that
follows the same path.
An 8.00 kg mass collides with a 5.00 kg mass that is
at rest. Initially, the 8.00 kg mass was traveling to
the right at 4.50 m/s. After the collision, it is
moving with a speed of 3.65 m/s and at an angle of
27.0° to its original direction. Calculate the speed
and direction of the 5.00 kg ball after the collision.
(3.32 m/s, 53.0o below the horizontal)
General Motion
1. Translational Motion
1. all points of an object follow the same path
2. Sliding a book across a table
2. Rotational Motion
3. General Motion – combination of
translational and rotation motion
Center of Mass
1. Stand perpendicular to the wall. Both feet
must be against the baseboard. Now lift
your outer leg.
2. Stand with your back against the wall. Be
sure your heels touch the baseboard. Now
try to pick up an eraser without bending
your knees.
Translational
Translational and Rotational
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Center of Mass
3. Get down on your elbows and knees on the
floor. Place your elbows right in front of
your knees and then move your hands
behind you back. Try to move an eraser
with your nose.
Can even be outside of the object’s body.
(Fosbury flop, boomerang, hula hoop)
Calculating Center of Mass: Ex 1.
xCM = m1x1 + m2x2
m1 + m2
Three people each massing about 60 kg sit on a
banana boat as shown below. Calculate the
center of mass.
xCM = m1x1 + m2x2 + m3x3
m1 + m2 + m3
xCM = (60 kg)(1m) + (60 kg)(5 m) + (60 kg)(6m)
180 kg
xCM = 4.0 m
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Where is the center of mass for the Earth-Moon system.
Assume the center of the Earth is the origin. Some
values you need are:
mEarth = 5.97 X 1024 kg
mMoon = 7.35 X 1022 kg
Earth-Moon distance = 3.84 X 108 m
Calculate the center of mass for the Earth-Sun system.
Assume the center of the Sun is the origin. Some
values you need are:
mEarth = 5.97 X 1024 kg
mSun = 2.00× 1030 kg
Earth-Sun distance = 1.50 X 1011 m
(try it with the moon as the origin.)
(4.48 X 105 m)
Find the center of mass of the CN- ion if carbon
has a mass of 12.0 u and nitrogen has a mass of
14.0 u. The bond length is 1.136 Angstroms.
A 150 kg man and a 450 kg owlbear sit on
opposite sides of a 4 m plank. Calculate
where the plank should be pivoted to
produce a seesaw.
(0.612 A from the carbon)
(3 m from the man)
Repeat the calculation, but now assume the
plank has a mass of 10.0 kg and all the mass
of the plank acts at the center, 2.00 m.
(2.98 m from the man)
Calculate the center of mass of the water
molecule. Hydrogen’s mass is 1.00
g/mole and oxygen’s is 16.0 g/mole. The
distance between the two nuclei is 0.942
Angstroms. You may do all your
calculations in these units.
(0, -0.0641)
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Now calculate the center of mass of ozone:
(0.447, 0 (left O as origin))
Calculate the center of mass of SO2, whose
bond angle is about 120o and whose bond
length is 1.43 Angstroms.
(0.357, 0)
2. -0.720 m/s opposite the direction of package
4. 4.8 m/s
6. 1.6 X 104 kg
8. 510 m/s
10. 4200 m/s
12. a) 4700 m/s and 6900 m/s b) 5.90 X 108 J
14. 130 N, not enough
16. a) 2.0 kg m/s
b) 400 N
18.a) 460 kg m/s(E) b) -460 kg m/s(W)
c) 460 kg m/s (E) d) 610 N (E)
22. v1’ = -1.00 m/s (rebound)
v2’ = 2.00 m/s
24. v1’ = 0.70 m/s
v2’ = 2.20 m/s
26.a) v1’ = 3.62 m/s v2’ = 4.42 m/s
b) -396 kg m/s, +396 kg m/s
30. 0.61 m
32. 3000 J and 4500 J
38. 60o and 6.7 m/s
40. a) 30o
b) v1’ = v/(3 ½ )
c) 2/3
44. A = 0o, v1’ = 3.7 m/s, v2’ = 2.0 m/s
46.6.5 X 10-11 m
48. 3.83lo from edge of
56. 4.66 X 106 m
smallest cube
11.m1v1 = m1v1 + m2v2
(0.013)(230) = (0.013)(170) + 2v2
v2 = 0.39 m/s
25.(0.220)(5.5) = (0.22)(-3.7) + mTvT
2.024 = mTvT
5.5 – 3.7 = 0 +vT
vT = 1.8 m/s
mT = 1.12 kg
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26.(450)(4.5) + (550)(3.7) = 450v1 + 550v2
4060 = 450v1 + 550v2
4.5 + v1 = 3.7 + v2
0.8 + v1 = v2
v1 = 3.62 m/s v2 = 4.42 m/s
b) Dp1 = (450)(3.62) – (450)(4.5) = -396 kg m/s
Dp2 = (550)(4.42) – (550)(3.7) = +396 kg m/s
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K=½ m(p2/m2)
K = p2/(2m)
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4. A softball of mass 0.200 kg is moving at a speed
of 5.50 m/s. It collides elastically with another
ball that is initially at rest. Afterward it is found
that the incoming softball has bounced backward
with a speed of 3.70 m/s. Calculate the velocity of
the target ball and the mass of the target ball.
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