Transcript Momentum - curtehrenstrom.com

Linear Momentum of a Particle
Momentum is defined as the product of mass and
velocity:
p = m• v
Units: kg • m/s
Scalar times a vector that produces a vector
answer!
A net force acting upon a particle will change its
net momentum in the direction of that net force:
∑F = dp
dt
A 4.88 kg object with a speed of 31.4 m/s strikes a
steel plate at an angle of 42.0˚ with the horizontal
and rebounds at the same speed and angle:
What is the change (magnitude and direction) of
the linear momentum of the object?
v0 = 31.4 m/s @ -42.0˚
v = 31.4 m/s @ 42.0˚
m = 4.88 kg
p0 = mv0cosø1 + mv0sinø1 = (114i – 103j) kg-m/s
p = mvcosø2 + mvsinø2 = (114i + 103j) kg-m/s
∆p = p – p0 = (0i + 206j) kg-m/s
∆p =
px2 + py2
= 206 kg-m/s
Ø = tan-1 (py/px) = 90.0˚
p
∆p
po
42.0˚
Impulse and Momentum
A collision occurs when a relatively large force
acts over a relatively short period of time.
A net force will produce acceleration, and
therefore a change in momentum:
F = dp
dt
Fdt = dp
The net force acting over time will produce a
change in momentum-- this is called impulse (J)!
The impulse-momentum theorem states that the
net force acting on a particle during any given
time interval is equal to the change in momentum
of the particle during that time period:
For a constant force: J = ∆p = pf - pi
For a varying force: J = ∫ F(t)dt = ∆p
A baseball (m = .14 kg) is thrown horizontally
with a speed of 42 m/s is struck by a bat. The ball
leaves the bat with a velocity of 50.0 m/s at an
angle of 35˚ with the horizontal. A) What is the
impulse of the force exerted on the ball? B)
Assuming the collision lasts for 1.5 ms, what
average force acts upon the ball? C) Find the
change in momentum of the bat.
A) J = ?
J = p – p0
J
p
α
ø
po
Solve using components (or trig):
pxo = mvxo = (.14 kg)(- 42 m/s) = - 5.9 kg• m/s
pyo = 0
px = mvxcosø = (.14 kg)(50 m/s)(cos 35˚)
= 5.7 kg • m/s
py = mvysinø = (.14 kg)(50 m/s)(sin 35˚)
= 4.0 kg • m/s
Jx = px - pxo = (5.7) - (- 5.9) = 11.6 kg • m/s
Jy = py - pyo = (4.0) - 0 = 4.0 kg • m/s
J=
Jx2 + Jy2
= 12.3 kg • m/s
α = tan-1 (Jy / Jx ) = 19˚
J = 12.3 kg • m/s at 19˚
B) J = Fav • ∆t
Fav = J / ∆t = 12.3 kg • m/s / .0015 s = 8200 N
about 1840 lbs of force!
acting in the same direction as the impulse!
C) the ∆p of bat must follow ∆p = ∆p1 + ∆p2 = 0
Therefore, ∆p for bat must equal - ∆p for ball
∆pbat = (- 11.6i + - 4.0j ) kg• m/s
or:
= 12.3 kg• m/s at 199˚
Conservation of momentum in collisions
If there are no external forces acting upon a
system of particles, then there can be no net
change in total momentum of the system:
∆P = ∆p1 + ∆p2 + … = 0
Conservation of momentum in collisions
In any type of collision, linear momentum is
always conserved!
Total energy is also always conserved, but
often times mechanical energy will be
converted to internal, rotational, radiant, etc.,
forms of energy.
• If the mechanical energy is conserved, then
the collision is known as an elastic collision.
• If Ki  Kf , then it is an inelastic collision.
Elastic Collisions:
p is always conserved:
m1v1i + m2 v2i = m1v1f + m2 v2f
K is also conserved:
.5m1v1i2 + .5m2v2i2 = .5m1v1f2 + m2v2f2
P eq. can be expressed: m1(v1i - v1f ) = m2(v2f - v2i )
K eq. can be:
m1(v1i2 - v1f2 ) = m2(v2f2 - v2i2 )
Dividing these equations:
v1i + v1f = v2f + v2i
The following can then be derived:
v1f = (m1 – m2)(v1i)
m1 + m2
v2f =
2m1 (v1i)
m1 + m 2
+
+
2m2 (v2i)
m1 + m2
(m2 – m1) (v2i)
m1 + m2
If the target particle is at rest:
v1f = (m1 – m2)(v1i)
m1 + m2
v2f =
2m1 (v1i)
m1 + m 2
Inelastic Collisions:
Momentum is conserved (as always), but K is
not, so we examine one special case:
• In a completely
inelastic collision,
the two particles
stick together and
have a common
velocity afterward:
v1f = v2f = vf
Two - Dimensional Collisions
Momentum is always conserved in collisions:
v=0
ø2
pi = pf
ø1
m1v1ix = m1v1f cosø1 + m2v2f cosø2
m1v1iy = m1v1f sinø1 + m2v2f sinø2
Remember: if the collision is elastic Ki = Kf
A 55 kg skater head north at 8.8 km/h and
collides with an 83 kg skater heading east at 6.4
km/h. They embrace in a completely inelastic
collision. A) What is the magnitude and
direction of the velocity of the couple after the
collision? B) What is the fractional change in
kinetic energy of the skaters due to the
collision?
V
ø
A
B
Momentum is conserved in the x and y direction:
mAvA = (mA + mB)(Vcosø)
mBvB = (mA + mB)(Vsinø)
Eq. (2)  (1) :
tanø = mBvB
ø = tan-1 . 911 = 42.3˚
V=
mAvA
(mA + mB)(cosø)
mAvA
(1)
(2)
= . 911
= 5.21 km/h
Loss of K = Kf - Ki
Ki
Ki = .5mAvA2 + .5mBvB2 = 3830
Kf = .5(mA + mB)V2 = 1870
Loss of K = 1870 - 3830
= - 0 . 51
3830
51% of K is dissipated in some form or
other
Ballistic Pendulum
used to measure
projectile speeds
before electronic
timing!
∆h
A bullet (m = 9.5 g) is fired into the block (M =
5.4 kg) and they rise to a height of 6.3 cm.
What was the initial speed of the bullet?
Momentum is conserved:
mv = (M + m)V
Total energy of the swinging bullet + block
pendulum is conserved:
.5(m +M)V2 = (m + M)g∆h
V = 2 g∆h
v = (M + m) 2g∆h
m
= 630 m/s
Show how this is obviously an inelastic
collision:
Ki = Kb = .5mv2 = 1900 J
Kf = K b + B = .5(m + M)V2 = U b + B
U b + B = (m + M)g∆h = 3.3 J
Ki ≠ K f
Only 3.3 J out of 1900 J, or 0.2%, is
transferred to the mechanical energy of the
pendulum!
The rest would be transferred as internal
energy-- heat, sound, etc.
A steel ball of mass .514 kg is fastened to a cord
68.7 cm long and released when the cord is
horizontal. At the bottom of its path, the ball strikes
a 2.63 kg steel block initially at rest on a frictionless
surface. The collision is elastic. Find A) the speed
of the ball and B) the speed of the block, both just
after the collision. C) Suppose that on collision,
one-half the mechanical kinetic energy is converted
to internal energy and sound energy. Find the final
speeds.
L
m1
L
d
m2
Two pendulums of equal
mass and length are
situated as shown. The
first pendulum is released
from height d and strikes
the second.
Assume the collision between the pendulums
is completely inelastic and neglect the mass of
the strings and any frictional effects. How
high will the center of mass rise after the
collision?
Particle Systems and Center of Mass
In this case, the motion of the two particles is
complicated, but the overall motion of the system
is relatively simple.
The center of mass (cm)– the “average” location
of the masses follows a relatively simple motion
of a single particle.
The location of the center of mass of the system is
proportional to the each mass’ location:
M = m1 + m2
xcm =
1
M
(m1x1 + m2x2)
For a many particle system:
xcm = 1/M(m1x1 + m2x2 + …. mnxn)
M = m1 + m2 + …. mn = ∑mn
xcm = (1/M)∑mnxn
ycm = (1/M)∑mnyn
zcm = (1/M)∑mnzn
rcm = (1/M)∑mnrn
vcm = (1/M)∑mnvn
acm = (1/M)∑mnan
Three particles are arranged as follows:
1
2
3
m1 = 4.1 kg @ (-2 cm, 3 cm)
m2 = 8.2 kg @ ( 4 cm, 2 cm)
m3 = 4.1 kg @ ( 1 cm, -2 cm)
What is the
magnitude and
direction from
the origin of the
cm of the
system?
xcm = (1/M)(m1x1 + m2x2 + m3x3)
xcm = (1/16.4 kg)[(4.1 kg)(-2 cm) + (8.2 kg)(4 cm)
+ (4.1 kg)(1 cm)] = 1.8 cm
ycm = (1/M)(m1y1 + m2y2 + m3y3)
ycm = (1/16.4 kg)[(4.1 kg)(3 cm) + (8.2 kg)(2 cm)
+ (4.1 kg)(-2 cm)] = 1.3 cm
rcm = (1.8 cm)2 + (1.3 cm)2 = 2.22 cm
ø = tan-1 (y/x) = tan-1 (1.3/1.8) = 36˚
If each side measures L, where is the cm located?
3M
M
M
Newton’s Laws then hold for each individual
parts as well as the center of mass for the system.
Therefore, if the net external forces acting on the
parts of the system are not balanced, there will
be a net force acting on the center of mass of the
system:
∑Fext = macm
If the net external force acting upon the parts of
the system is balanced, then the center of mass of
the system will move at constant speed:
∑Fext = 0
Three particles are arranged as follows:
6N
1
2
3
14 N
m1 = 4.1 kg @ (-2 cm, 3 cm)
m2 = 8.2 kg @ ( 4 cm, 2 cm)
m3 = 4.1 kg @ ( 1 cm, -2 cm)
12 N @ 45˚
What is the
magnitude and
direction from
the origin of the
acceleration of
the system?
Fx = F1x + F2x + F3x =
- 6 N + cos45˚(12 N) + 14 N = 16.5 N
Fy = F1y + F2y + F3y =
0 + sin45˚(12 N) + 0 = 8.5 N
Fext = Fx2 + Fy2 = 18.6 N
ø = tan-1(Fy /Fx)= 27˚
a=F
18.6 N
16.4 kg
M
= 1.1 m/s2
Two masses, one of 1.10 kg and the other of 1.40
kg, are connected by a light string and passed over
a frictionless pulley of radius 25.0 cm (Atwood
machine). The lighter mass is held in place on the
left level with the heavier mass. A) Where is the
center of mass of the two masses at this time? The
lighter mass is then released. B) What is the
acceleration of the center of mass of the two
masses?
A shell is fired with a velocity of 466 m/s at an
angle of 57.4˚ with the horizontal. At the top of
its trajectory the shell explodes into two
fragments of equal mass. One fragment, whose
speed immediately after the explosion is zero,
falls straight down. How far from the cannon will
the other shell land, assuming it lands at the same
elevation as which it was fired from?
vcmx = vcosø = 251 m/s
vcmy = vsinø = 393 m/s
vcmx = (1/M)(m1v1 + m2v2)
∆t =
vy - vyo
g
= 393 - (-393)
9.80
m1 = m2 = m
vcmx = mv2
= v2
2m
2
v2 = 502 m/s
251 m/s = .5v2
x2 = vcmx(.5∆t) + v2 (.5∆t) = 30, 200 m
= 80.2s
mr = 93 kg
mb = 52 kg
xro = 1.5 m
xbo = 3.0 m
xr = ?
xb = xr – 1.5 m
Neglecting any external forces,
the cm of the system will not
change!
xcm = mrxro + mbxbo
mr + mb = 2.0 m
A dog weighing 10.8 lb is standing on a flatboat so
that he is 21.4 ft from the shore. He walks the 8.50
ft length of the flatboat and then halts. The boat
weighs 46.4 lb, and you can assume the water to be
frictionless. How far is the dog from the shore
after walking? (Assume the boat is perpendicular to
the shore and realize that you have two possible
answers depending upon whether the dog walks
toward the shore or away).
Richard, mass 78.4 kg, and Judy, who is less
massive, sit at opposite ends of a canoe, 2.93 m
apart, on placid water. The canoe is 31.6 kg. The
two changes seats and Richard notices that the
canoe has moved 41.2 cm relative to a sunken log.
Richard then decides to calculate Judy’s mass.
(Richard is a loser who will die alone playing a
video game in his mother’s basement which he
will call his mancave.) What is Judy’s mass?
R
J
∆xc = .412 m
55.2 kg
An 84.4 kg man is standing at the rear of a 425 kg
iceboat that is moving at 4.16 m/s across the ice
that is frictionless. He decides to walk to the front
of the 18.2 m boat and does so at a speed of 2.08
m/s relative to the boat. How far does the boat
move across the ice while he is walking?