newton’s 2 law can be expressed as...
TRANSCRIPT
Chapter 8: Momentum • Momentum is defined as (for non-relativistic speeds)
• Newton’s 2nd law can be expressed as
– In fact, this is the preferred version of Newton’s 2nd law, since it is valid even for relativistic speeds.
• Impulse: The net change in the momentum due to an interaction is given by the impulse:
�p = m�v
�F =d�p
dt
~J =
Z t2
t1
~F (t) dt =
Z t2
t1
d~p
dtdt =
Z t2
t1
d~p = �~p
Clicker Question
An egg is dropped from rest and falls a distance h and then hits the concrete ground. Another egg (same mass as the first) falls a distance h and then lands on a pillow. The impulse exerted on the egg by the round is ________ the impulse exerted on the egg by the pillow. a) equal to b) greater than c) less than
A 2.0 kg object moving to the right with speed 0.50 m/s experiences the force shown. What are the object’s speed and direction after the force ends? A. 0.50 m/s left. B. At rest. C. 0.50 m/s right. D. 1.0 m/s right. E. 2.0 m/s right.
QuickCheck 9.2
Slide 9-‐31
A 2.0 kg object moving to the right with speed 0.50 m/s experiences the force shown. What are the object’s speed and direction after the force ends? A. 0.50 m/s left. B. At rest. C. 0.50 m/s right. D. 1.0 m/s right. E. 2.0 m/s right.
QuickCheck 9.3
Slide 9-‐33
Compare impulse and work
– Changes in momentum depend on the time over which the net force acts, but changes in kinetic energy depend on the distance over which the net force acts.
A light plasHc cart and a heavy steel cart are both pushed with the same force for 1.0 s, starHng from rest. ALer the force is removed, the momentum of the light plasHc cart is ________ that of the heavy steel cart.
QuickCheck 9.5
A. greater than
B. equal to
C. less than
D. Can’t say. It depends on how big the force is.
Slide 9-‐39
Clicker Question
Conservation of Momentum • Consider two particles interacting with each other
– No external forces acting on the system – Example: two astronauts pushing on each other – Newton’s 3rd law applies
�FB on A = ��FA on B
d�pA
dt= �d�pB
dt
d
dt(�pA + �pB) = 0
�P = �pA + �pB = constant
A mosquito and a truck have a head-on collision. Splat! Which has a larger change of momentum?
A. The mosquito.
B. The truck.
C. They have the same change of momentum.
D. Can’t say without knowing their initial velocities.
QuickCheck 9.8
Slide 9-‐60
Conservation of Momentum • This result generalizes for the case where there are N
interacting particles.
�P =N�
i
�pi = constant
�F1 =N�
i=2
�Fi1
(valid if there are no external forces acHng on system)
Clicker Question
Chalkboard Question Two friends are initially moving with the same speed of
5 m/s on individual skateboards. Boe, with a mass of 30 kg, pushes on Gilbert, who has a mass of 70 kg. After the push, Gilbert is moving 10 m/s. How fast, and in what direction is Boe moving after the push?
Clicker Question
Block A on the left has mass 1.00 kg. Block B on the right
has mass 3.00 kg. The blocks are forced together,
compressing the spring. Then the system is released from
rest on a level, frictionless surface. After the blocks are
released, the kinetic energy (KE) of block A is
A. 1/9 the KE of block B B. 1/3 the KE of block B
C. 3 times the KE of block B D. 9 times the KE of block B
E. the same as the KE of block B
Q8.7
Collisions • During collisions, it is very complicated to express
the force as a function of position (or time) • Since most collisions are short events, often we don’t
care about the motion during the collision; only interested in how motion changed. Momentum conservation is perfect for this analysis!
1D Perfectly Inelastic Collisions • Consider a perfectly inelastic head-on collision,
where the two objects stick together (so they have the same final velocity):
• Ignoring friction during the collision, momentum is conserved
M1v1 �M2v2 = (M1 + M2)v�
v� =M1v1 �M2v2
M1 + M2
Clicker Question
Two objects with different masses collide and s#ck to each other. Compared to before the collision, the system of two objects a,er the collision has
A. the same total momentum and the same total kineHc energy.
B. the same total momentum but less total kineHc energy.
C. less total momentum but the same total kineHc energy.
D. less total momentum and less total kineHc energy.
E. not enough informaHon given to decide
A B
QuickCheck 9.9
The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is
A. 0 m/s.
B. 1 m/s.
C. 2 m/s.
D. 3 m/s.
E. There’s not enough information to tell.
Slide 9-‐70
Chalkboard Question A gun fires a bullet horizontally, which hits a chair and
lodges into the wood of the chair. As a result, the chair skids a distance of 4 cm. The mass of the bullet is 10 g, the mass of the chair is 20 kg, and coefficient of kinetic friction is 0.2. What was the initial speed of the bullet?
Chalkboard Question Ballistic Pendulum Determine the speed of an incoming bullet by
measuring the maximum angle of the pendulum.
L = 1 m θmax = 15° mW = 2 kg mB = 20 g
Clicker Question
Challenge Question An inverted garbage can of mass M is suspended in air by water from a geyser (moving vertically upward). The water shoots up from the ground with an initial speed v(y = 0) = v0 at a constant rate dm/dt. Find the height h at which the ``floor” of the garbage levitates above the ground
Chalkboard Question A freight car of mass M runs under a sand dispenser, which is releasing sand vertically into the car at a rate
dm/dt. If the car is moving with a velocity v, what horizontal force is required to keep the velocity constant?
2D Perfectly Inelastic Collisions • For 2D (and 3D) collisions, need to remember that
momentum is a vector, and each of its components are conserved.
�P = �P �
Px = P �x
Py = P �y
Example 8.9: What is angle and final speed?
Two balls, labeled 1 and 2, collide. The iniHal momenta of the two balls and the final momentum of ball 1 are shown in the figure.
p1i
p1f
p2i
What is the x-‐component of the final momentum of ball 2? A: -‐2 B: -‐1 C: 0 D: +1 E: +2
Clicker Question
Elastic Collisions • For elastic collisions, the kinetic energy of the system
is also conserved.
• Consider 1D elastic collision with body B initially at rest.
�K = 0��P = 0
12mAv2
A =12mAv�2
A +12mBv�2
B
mAvA = mAv�A + mBv�
B
• The two equations can be solved for the two unknowns (the final velocities), yielding:
– If the two masses are the same, what happens? – If mB >> mA, what happens? – What is maximum speed of object B after collision?
v�A =
mA �mB
mA + mBvA
v�B =
2mA
mA + mBvA
JITT Question • Initially, the red ball is chasing after the blue ball with a speed
of 3 m/s while the speed of the blue ball is 2 m/s. After the collision, the red ball rebounds at 2 m/s and the blue ball speeds away at 3 m/s. What was the change in velocity experienced by the red ball during the collision? What was the change in velocity experienced by the blue ball during the collision? If the mass of the red ball is the same as the mass of the blue ball, can the collision we are describing occur in the real world?
Gravitational Slingshot • Notice that these equations don’t depend on the
nature of the force (as long as the collision is elastic), so they don’t have to “touch.”
• One common example is a gravitational slingshot between a satellite and a planet.
– What is the final speed of the satellite? • Use velocity transformation such that initial velocity of
planet is zero. This yields |Δv| = 2vB
JITT Question • Consider the two-dimension collision shown below.
The two objects have the same mass. Is this collision possible? Is it elastic? Explain your reasoning.
2D Elastic Collision • For two-dimensional collisions, need to (like before)
conserve momentum in both the x and y directions. Thus there are three equations to solve:
12mAv2
A =12mAv�2
A +12mBv�2
B
mAvAx = mAv�Ax + mBv�
Bx
mAvAy = mAv�Ay + mBv�
By
turns out that θ+Φ=90° if the masses are the same.
p0A = pA
2
4mA cos ✓ ±
qm2
B +m2A sin
2 ✓
mA +mB
3
5
An open cart is rolling to the left on a horizontal surface. A package slides down a chute and lands in the cart. Which quantities have the same value just before and just after the package lands in the cart?
A. the horizontal component of total momentum
B. the vertical component of total momentum
C. the total kinetic energy
D. two of A., B., and C.
E. all of A., B., and C.
Q8.8
Skateboard Propulsion and Rockets • Consider a person of mass M standing on a
skateboard, initially stationary. He throws a rock of mass m to the right. What is his final speed if the relative speed after the throw between the person and the rock is u?
Mv01 = mv02 = m(u� v01)
) v01 =m
M +mu
Skateboard Propulsion and Rockets • What if the skateboarder was already moving to the
left with an initial speed v1? What would her velocity be after throwing the rock? – No need to redo this problem. We already have the answer! – In a frame of reference in which the skateboarder was
initially stationary, her velocity changed by an amount
– This change in velocity occurs in any inertial frame of reference, so we are done!
�v1 = v�1 � v1 =
m
M + mu
Skateboarder Propulsion and Rockets • Now suppose the skateboarder is throwing many
rocks in succession (or has a machine gun).
• What is the acceleration of the skateboarder?
– If the skateboarder is throwing rocks at a rate of dN/dt, and each rock has a mass m<<M, then during a short dime interval Δt, the change in velocity is
where dm/dt is the mass rate of rocks/fuel being ejected. Thus the acceleration is a =
�v
�t=
u
M
dm
dt
�v ⇡ m
M(t)udN
dt�t =
u
M(t)
dm
dt�t
�v1 = v�1 � v1 =
m
M + mu
Rockets • Rearranging,
– Note that the mass M includes the mass of the rocks (or fuel for rocket) still onboard, so it decreases with time.
– Fuel leaving rocket exerts a thrust force on the rocket. This force was derived simply by using conservation of momentum!
Ma = udm
dt= Fthrust
Atlas 5 Rocket • Mass: About 400,000 kg (plus payload,
roughly 5,000-10,000 kg) • Height: 58 meters • Thrust of main engine (1st stage): 4,152
kN (not counting solid-engine boosters which contribute about 1,100 kN each) – The burn rate is roughly R=1000 kg/s and
the exhausts speed is roughly u=4000 m/s – Fuel is refined kerosene and liquid oxygen – For New Horizons mission (Pluto), total
thrust was 9,000 kN. Acceleration initially 8 m/s2. Reached 16 km/s within minutes!
Rockets • For a rocket taking off from Earth, need
to also include the force of gravity:
Ma = udm
dt�Mg
Velocity vs Time of Rocket Launch • To determine the velocity as a function of time, the
differential equation law must be solved:
Mdv
dt= u
dm
dt�Mg
Mdv
dt= �u
dM
dt�Mg
dv
dt= � u
M
dM
dt� g
� v
0dv = �
� M
M0
u
MdM �
� t
0g dt
(rocket is losing mass as exhaust is expelled)
v = �u ln�
M0 �Rt
M0
⇥� gt
Motion of Center of Mass • Center of mass (CM) is defined as • Total momentum of the system: • Consider several interacting
particles subject to an external force
– Motion of the CM depends solely on mass Mtot Fext (regardless on where on object it acts and internal forces)
• If no external force, Vcm = constant
�Rcm =1
Mtot
�
i
mi�ri
�P =�
i
mi�vi = Mtot�Vcm
d~P
dt= M
tot
~Acm
=X
i
~Fneti = ~F
ext
Clicker Question: Dumbbell I
1) case (a)
2) case (b)
3) no difference
4) it depends on the rota9onal iner9a of the dumbbell
A force is applied to a dumbbell for a certain period of time, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed ?
Examples of Center of Mass • A planet orbiting a star
– If we place the origin in the middle of the star, then
– For Sun and Jupiter
– Note, Sun must move around center-of-mass too!
xCM =1
Mtot
�
i
Mixi =M2
M1 + M2x2
xCM =1.9� 1027
2.0� 1030 + 1.9� 10277.8� 1011 m = 7.4� 108 m = 1.1R�
Planet Detection • The fact that stars have to orbit, or wobble as planets
orbit them has lead to a method of planet detection.
• What is velocity of Sun due to Jupiter?
A yellow block and a red rod are joined together. Each object is of uniform density. The center of mass of the combined object is at the position shown by the black “X.”
Which has the greater mass, the yellow block or the red rod?
A. the yellow block
B. the red rod
C. they both have the same mass
D. not enough informaHon given to decide
Q8.9
Examples of Center of Mass • Person standing on plank on cliff. Does he fall?
– If xcm > d, the person will fall!
xCM =1
M1 + M2
�M1
L
2+ M2L
⇥
CM for Continuous Mass Distributions • For continuous mass distributions, the summations
turn into integrals:
• For one-dimensional mass distributions, so the center of mass is given by
• For a uniform rod of length L, with the left end on the
origin,
�RCM =1M
�
i
mi�ri �1M
⇥�r dm
dm = � dx
xcm =1M
�� x dx
xcm =1M
� L
0� x dx =
�L2
2M=
L
2
• Now suppose the rod has a non-uniform density given by .
Then the CM is given by But the mass is equal to So,
�(x) = a x
xcm =1M
� L
0�(x) x dx =
1M
� L
0ax2 dx =
13M
aL3
M =� L
0�(x) dx =
12aL2
xcm =23L
A baseball bat is cut in half at its center of mass. Which end is heavier?
A. The handle end (leL end). B. The hi_ng end (right end). C. The two ends weigh the same.
QuickCheck 12.3
Slide 12-‐45
Center of Mass of a Solid Object
Divide a solid object into many small cells of mass Δm. As Δm → 0 and is replaced by dm, the sums become
Before these can be integrated: § dm must be replaced by
expressions using dx and dy. § Integration limits must be established.
Slide 12-‐41