Linear Momentum
Momentum (vector quantity)
• p = mass X velocity
• p = mv
• Unit = kg-m/s
Momentum: Example 1
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
Momentum: Example 2
If Mr. West is running at 2 m/s and has a
momentum of 80 kg m/s, what is his mass?
The Law of Conservation of Momentum
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)?
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?
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?
m1v1 + m2v2 = m1v1’ + m2v2’
m1v1 + m2v2 = (m1 + m2)vf
(10,000 kg)(24.0 m/s) + 0 = (20,000 kg)(vf)
vf = +12.0 m/s
(Notice that the speed was cut exactly in half. Why?)
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.
Rocket Ship
• Before moving, p = 0
• Which moves more, the rocket or the gas?
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)
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
– 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
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?
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
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
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
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
Note how they exchanged velocities (because they have equal mass)
Elastic Collisions: Example 3
A proton of mass 1.01 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 = -2.15 X 104 m/s
vhf = 1.45 X 104 m/s
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
vpi + vpf = vhf
vhf = 3.60 X 104 m/s + (-2.15 X 104m/s)
vhf = 1.45 X 104 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
v = +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
Perfectly Inelastic Collision: Ex 3
A 95.0 kg running back running at 3.50 m/s collides with
a 120.0 kg linebacker 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%)
Ballistic Pendulum
Used to determine
velocities
Perfectly Inelastic
collisions (KE
converts to PE)
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.
m1v1 + m2v2 = m1v1’ + m2v2’
m1v1 + m2v2 = (m1 + m2)vf
0 +(0.005 kg)(vi) = (1.00 kg + 0.005 kg)(vf)
(0.005 kg)(vi) = (1.005 kg)(vf)
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
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.
(284 m/s)
An 8.00 g bullet is fired at 350 m/s into a
ballistic pendulum bob of 2.00 kg.
Calculate how high the coupled bullet and
bob will rise.
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.
(35.1 cm)
b. Calculate the angle the pendulum rises. (40.5o)
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:
If both masses are equal, then the angles of
deflection add to 90o.
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 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) vx
vx = 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) vy
vy= 12.5 m/s
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
q = -450
vgreen
x coordinate
mgvg + mrvr = mgvg + mrvr
3m + 0 = mgvg + mrvr
3 = v g + vr
3 = vgcos(45o) + vrcos(-45o)
3 = 0.707vg + 0.707vr
4.24 = vg + vr
y coordinate
0 = mgvg + mrvr
0 = mgvg + mrvr
0 = v g + 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
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) Can you express your answer in “i and j”
notation?
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)
Center of Mass
Center of mass – one point on a particle that
follows the same path.
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.
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.
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
Translational
Translational and Rotational
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
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
(try it with the moon as the origin.)
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
Find the center of mass of the CO molecule if
carbon has a mass of 12.0 u and oxygen has a
mass of 16.0 u. The bond length is 1.13
Angstroms.
(0.645 A from the carbon)
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.
(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.
(0.745, 0.512)
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.
Can even be outside of the object’s body.
(Fosbury flop)
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
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?
(9.1 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
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 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. (-0.144
m/s)
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)
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
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)
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
56. 4.66 X 106 m