8.4 Conservation of Linear Momentum
As stated before, linear momentum is a vector quantity because velocity is a
vector quantity.
DEF: Linear momentum = P (for progress ☺) = m·v
Linear momentum can also be construed to mean the amount of effort to bring
a particle to rest, or, un-oomph your particle. ☺
A plausible manipulation of N2 allows us to rewrite N2 as:
dp = d (mv) = m dv = ma
dt
dt
dt
Since Fnet = ma, Fnet = dp
dt
Q: What does that really mean?
A: The Fnet acting on a particle = the time rate of change of the particle’s
linear momentum.
This gives the following:
Q: What do you get if you differentiate both sides of eq 8-14?
A: dPsys = M dvcm = Ma
dt
dt
If Ma = Fnet then Σ Fnet, ext = Fnet, ext = dPsys
dt
Q: What happens to the net rate of change of the total momentum when the
net external force acting on a system of particles is zero?
A: It remains at zero, and the total momentum remains constant. Graphically,
that means it’s linear.
This is the Law of Conservation of Momentum:
See Ex 8-6, p 229
During repair of the Hubble Space Telescope, an astronaut replaces a solar
panel whose frame is bent. Pushing the detached panel away into space, she is
propelled in the opposite direction. The astronaut’s mass is 60 kg and the
panel’s mass is 80 kg. The astronaut is at rest relative to her spaceship when
she shoves away the panel, and she shoves it a 0.3 m/s relative to the spaceship.
What is her subsequent velocity relative to the space ship? (During this
operation, assume the astronaut is tethered to the ship; for our calculation,
assume the tether remains slack.)
The velocity of the astronaut can be found from the velocity of the panel using
conservation of momentum. Choose the direction of the motion of the panel to
be positive.
Apply conservation of momentum to find the velocity of the astronaut. Since
the total momentum is initially zero. It remains zero:
Pp + Pa = mpvp + mava = 0
Solve for the astronaut’s velocity:
va = -(mp/ma)vp
= -(80kg/60kg)(.3 m/s)
va = -0.4 m/s
See Example 8-7, p 229
A runaway 14,000 kg railroad car is rolling horizontally at 4 m/s toward a
switchyard. As it passes by a grain elevator, 2000 kg of grain are suddenly
dropped into the car. How long does it take the car to cover the 500 m distance
from the elevator to the switchyard? Assume that the grain falls straight down
and that slowing due to rolling friction and air drag are negligible.
We find the travel time that we seek from the distance traveled and the speed
of the car. Consider the car and the grain as our system. There are no
horizontal external forces acting on this system, so the horizontal component of
the momentum is conserved. The final speed of the grain-filled car is found
from its final momentum, which equals the car’s initial momentum. (The grain
initially has no horizontal momentum.) Let mc and mg be the masses of the car
and grain, respectively.
The time for the car to travel from the elevator to the yard is the distance to
the yard d divided by the car’s speed vf following the grain dump:
Apply conservation of momentum to relate the final velocity vf to the initial
velocity vi. Be careful, only the horizontal component of the system’s
momentum is conserved:
(mc + mg)vf = mcvi + mg(0)
Solve for vf:
Substitute the result for vf into step 1 and solve for the time:
Q: How does a rocket move forward through space?
A: By throwing its exhaust out backward.
See Example 8-8, p 231
A 40 kg skateboarder on a 3 kg skateboard is training with two 5 kg weights.
Beginning from rest, she throws the weights horizontally, one at a time from
her board. The speed of each weight is 7 m/s relative to her and the board after
it is thrown. Assume the board rolls without friction.
(a) How fast is she propelled in the opposite direction after throwing the first
weight?
(b) after throwing the second weight?
There are no external forces acting horizontally, so horizontally, momentum is
conserved.
a) Assume: m = mass of 1st weight = mass of 2nd weight = 5 kg
M = mass of boarder + board = 43 kg
She moves in the positive x-direction, so
Velocity of the weight relative to her, vw,s = - 7 m/s
Vsg1 = Velocity of her wrt the ground
vwg1 = velocity of the weight wrt the ground
Psys0 = momentum of the system before the 1st throw
Psys1 = momentum of the system between the 1st and 2nd throw, etc.
Psys1 = Psys0
(M + m) Vsg1 = m vwg1 = 0
The velocity of the thrown weight wrt the ground = the velocity of the weight
wrt the skateboarder + the velocity of the skateboarder wrt the ground
vwg1 = vws1 + Vsg1
Substitute for vwg1 and you get Vsg1 = - ___m___ vws1 = - ____5 kg____ (- 7 m/s)
M+2m
43 kg + 10 kg
= 0.66 m/s = 0.7 m/s
b) Do the same thing as you did in part (a) for the second throw
Psys2 = Psys1
M Vsg2 + mvw’g2 = (M + m) Vsg1
vw’g2 = vw’s2 + Vsg2
M Vsg2 + m(vw’s2 + Vsg2) = (M + m) Vsg1
Solve for Vsg2 and you get 0.66 m/s – 5 kg / 48 kg (-7 m/s) = 1.39 m/s = 1 m/s
Try Ex 8-9, p 232 now ☺
8.5 Kinetic Energy of a System
As we discussed earlier, if the net external force of a system is zero, the total
momentum of the system is constant.
Q: Can the total mechanical energy of a system vary?
A: Yep. If the internal forces are not conservative, they can change the total
mechanical energy of the system.
Just like you’d expect, the KE of the system is the sum of all of the KE’s of the
particles that constitute the system.
Assume:
vi = velocity of an individual particle
vcm = velocity of the center of mass
ui = velocity of the particle relative to the center of mass
Then KE = Σ ½ mi (vi·vi) = Σ ½ mi (vcm + ui) · (vcm + ui)
= Σ ½ mi (v2cm + 2 vcm · ui + ui2)
= Σ ½ mi v2cm + vcm · Σ miui + Σ ½ miu2i
NOTE: You can factor out a 2 from the middle term because the vcm does not
change from particle to particle. Also, Σ miui is the total momentum of the
system relative to the center of mass = Mucm = 0, because, relative to the center
of mass, ucm is zero. So, Mucm has to be zero, as well. ☺
M = Total mass, Krel = KE of the particles relative to the c.o.m. Only relative
KE can change in an isolated system.
8.6 Collisions
Q: What is a collision?
DEF: Collision = two objects that approach one another and interact strongly
for a very short time.
In a collision, in the physics sense, the strength of the interaction between the
forces of the objects involved far outweighs the effects of any external forces,
for that brief time period.
Q: Technically speaking, then, do like sides of magnets collide? Does the
moon collide with the earth?
A: Yep. Yep. ☺
DEF: Elastic collision = KE is conserved = “perfect” bounce
DEF: Inelastic collision = KE is not conserved = crunch – (separate)
DEF: Perfectly inelastic collision = KE is not conserved and all of the KE
relative to the c.o.m. is converted to thermal or internal energy of the system,
and the objects stick together = crunch – stick – travel together
Impulse and Average Force
The area under the Fx-vs-t curve is the x-component of the impulse, Ix
Fave = the average force for the time interval, ∆t. The rectangular area,
Fave,x ∆t, is the same as the area under the Fx-vs-t curve.
Q: What is impulse, again?
The magnitude of the impulse is the area under the Fx-vs-t curve.
Q: What are the units for impulse?
A: Gotta be N·s
This relates to N2 via the following:
The following theorems also apply:
See Ex 8-10, p 234-235.
With an expert karate blow, you shatter a concrete block. Consider your fist
to have a mass of 0.70 kg, to be moving at 5.0 m/s as it strikes the block, and to
stop within 6.0 mm of the point of contact.
a) What impulse does the block exert on your fist?
b) What is the approximate collision time and the average force the block
exerts on your fist?
a) I = ∆p = pf - pi
pf = 0 and pi = mv, so
pi = mv = 0 - (0.70 kg)(- 5.0 m/s) = 3.5 kg · m/s (or in the j direction)
b)
∆t = ∆y / vave = - 0.0060 m / ½ v = - 0.0060 m / - 2.5 m/s = 0.0024 s
Fave = I / ∆t = 3.5 kg · m/s / 0.0024 s = 1.5 x 103 N in the j direction
NOTE: That’s a big force! ☺
Q: What does impulse cause?
A: A change in momentum
TRY Ex 8-11, p 235-236. (Note the assumption that they make!)
Collisions in One Dimension (Head-on Collisions)
m1v1f + m2v2f = m1v1i + m2v2i
What you actually substitute in depends on the type of collision you have. If
the objects stick together, then their final velocity gets applied to both. If they
don’t stick, you keep the equation as is. You also have to know whether or not
they are initially traveling in the same direction or not.
Perfectly Inelastic Head-On Collisions
In a perfectly inelastic head-on collision (pi-hoc), the particles stick together
after the collision. This changes the equation to be:
(m1 + m2)vcm = m1v1i + m2v2i
Try Ex’s 8-15 p 240 now. ☺ Make sure you read the remarks before you
attempt the problem.
Elastic Head-On Collisions
Kf = Ki
This is really only possible in a perfect physics world. Still, we use it as a
model. If two objects collide, then v2i – v1i must be negative. Thus, their speed
of approach has to be – (v2i – v1i). Once they collide, their speed of recession is
v2f – v1f. NOTE: This eq applies only to elastic collisions!
TRY Ex’s 8-16 – 8-17, p 242-243. ☺
Q: What happens when a huge object collides with an itty-bitty one?
A: Relatively good news for the huge object; not so good news for the itty-bitty
one. The huge object barely changes its velocity, which means the itty-bitty
object has to change its velocity by 2vi so that the speed of recession = the speed
of approach.
The Coefficient of Restitution (Optional Section)
In reality, most collisions are not “perfectly” anything, but are rather,
somewhere in between.
DEF: Coefficient of restitution = e = a measure of the elasticity of a collision.
Collisions in 3-D
Perfectly Inelastic Collisions in 3-D
Total initial P = Σ Pi for each object involved. Because the objects stick
together, and because P is conserved, we get m1v1i + m2v2i = (m1 + m2)vf
This is the one we worked with in 10th grade. If this is our equation, then all
three velocity vectors, and thus the collision, are in the same plane. From the
def of c.o.m., we also know that vf = vcom.
See Ex 8-18, p 244-245
A guy is at the wheel of a 1200-kg car traveling east through an intersection
when a 3000-kg truck traveling north crashes into his car. The car and truck
stick together after impact. The truck driver claims the other guy was at fault
because he was speeding. The sign on the road says the speed limit is 80 km/h.
The speedometer on the truck is stuck at 50 km/h. The wreck skidded from
the zone of impact no less than 590 N of E. Who is at fault?
This is exactly the type of work done by Sgt Tim Robbins, State Crash Expert.
If he ever has free time, he’ll come and talk to us. For now, it’s up to us to
“solve” the mystery!
mcvc + mtvt = (mc + mt)vf
Since one driver was heading E and the other N, the respective equations look
like:
mcvc + 0 = (mc + mt)vf cos θ
0 + mtvt = (mc + mt)vf sin θ
Eliminating vf gives mtvt = sin θ = tan θ
mcvc cos θ
Solve this for vc and you get vc = (3000 kg)(50 km/h)
(1200 kg) tan 590
= 75.1 km / h for the car
This means the car wasn’t speeding.
NOTE: If at all possible, the court system would still want eye-witness
accounts! Jeez!
Elastic Collisions in 3-D (Optional Section)
Q: What is an impact parameter?
Take 2 minutes and read p 245-246. ☺
Psys = m1v1i = m1v1f + m2v2f
Figure 8-39
Figure 8-40
Just as a final note (finally!!), equation 8-34, which says v1i2 = v1f2 + v2f2 is
really just an application of the Pythagorean Theorem for the right triangle
formed by the vectors v1f, v2f, and v1i.