newton’s 2 law can be expressed as...

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