Review of Lecture 9

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Review of Lecture 9
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Review of Lecture 9
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Collisions

• We certainly have an intuitive feel for what the
  word collision means, but what is the specific
  definition given to a collision in physics?
• A collision is an isolated event in which two or
  more bodies (the colliding bodies) exert
  relatively strong forces on each other for a
  relatively short time

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Collisions

• An interesting point to note is that the
  definition doesnt necessarily require the bodies
  to actually make contact a near miss (near
  enough so that there are relatively strong
  forces involved) will suffice

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Collisions

• To analyze a collision we have to pay attention
  to the 3 parts of a collision
• Before
• During
• After

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Collisions

• By using knowledge of various conservation laws,
  and looking at the system of particles before,
  during and after the collision, physicists can
  deduce much about what is going on in the system

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Impulse Linear Momentum

• Lets start by examining a head-on collision
  between just two bodies of different masses
• We know that there will be a pair of 3rd law
  forces involved the force from body 1 on body 2
  and vice versa the force from body 2 on body 1
• And of course we know that these forces will be
  equal in magnitude but opposite in direction
• The forces are therefore F(t) and -F(t)

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Impulse Linear Momentum

• These forces will change the linear momentum of
  the bodies the amount of change will depend on
  average value of the force as well as how long
  the force acts on the body (e.g., ?t)

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Impulse Linear Momentum

• To state this qualitatively, lets go back to
  Newtons 2nd law in the form of

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Impulse Linear Momentum

• Lets see what is happening to object R (on the
  right)

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Impulse Linear Momentum

• We can rearrange the previous equation a bit to
  getwhere F(t) is a time varying force as
  shown in the following graph

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Impulse Linear Momentum

• If we now integrate over the interval ?t we will
  get
• The left side evaluates to pf pi which is
  just the change in linear momentum

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Impulse Linear Momentum

• The right side is a measure of the strength as
  well as the duration of the collision force and
  is defined as the impulse J of the collision
  (not to be confused with the SI units of joules)

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Impulse Linear Momentum

• We can see from equation 10-3 that the impulse is
  simply the area under the force curve during the
  interval in question
• We can also see that
• This is called the impulse linear momentum
  theorem

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Impulse Linear Momentum

• We can see that the impulse is just the area
  under the time-varying force curve over the time
  interval in question
• If we take the average value of the force
  instead, we will get the same result for the
  impulse

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Impulse Linear Momentum

• We can see that in the plot to the right

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Impulse Linear Momentum

• Checkpoint 1

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Series of Collisions

• What happens when there are multiple collisions
  to consider?
• Lets imagine an object that is securely bolted
  down so it cant move and is continuously pelted
  with a steady stream of projectiles

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Series of Collisions

• Each projectile has a mass of m and is moving at
  velocity v along the x axis (so we will drop the
  vector arrows and remember that we are talking
  about what is happening only along the x axis)
• The linear momentum of each projectile is
  therefore mv lets also suppose that n
  projectiles arrive in an interval of ?t

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Series of Collisions

• As each projectile hits (and is absorbed by) the
  mass, the change in the projectiles linear
  momentum is ?p thus the total change in linear
  momentum during the interval ?t is n ?p

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Series of Collisions

• The total impulse on the target is the same
  magnitude but opposite in direction to the change
  in linear momentum thus
• But we also know that

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Series of Collisions

• This leads us to
• So now we have the average force in terms of the
  rate at which projectiles collide with the target
  (n/?t) and each projectiles change in velocity
  (?v)

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Series of Collisions

• In the time interval ?t, we also know that an
  amount of mass ?m nm collides with the target
• Thus our equation for the average force finally
  turns out to be

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Series of Collisions

• Assuming that each projectile stops upon impact,
  we know that
• In this case the average force is

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Series of Collisions

• Suppose instead that each projectile rebounds
  (bounces directly back) upon impact in this
  case we have
• In this case the average force is

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Series of Collisions

• Checkpoint 2

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Momentum Kinetic Energyin Collisions

• If we have a system where two bodies collide,
  then there must be some kinetic energy (and
  therefore linear momentum) present
• During the collision, the kinetic energy and
  linear momentum of each body is changed by the
  impulse from the other body

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Momentum Kinetic Energyin Collisions

• We define two different kinds of collisions
• An elastic collision is one where the total
  kinetic energy of the system is conserved
• An inelastic collision is one where the total
  kinetic energy is not conserved
• A collision is not necessarily completely elastic
  or completely inelastic most are in fact
  somewhere in between
• However a completely inelastic collision is one
  where the two bodies end up sticking together

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Momentum Kinetic Energyin Collisions

• For any kind of collision however, linear
  momentum must be preserved
• This is because in a closed, isolated system, the
  total linear moment cannot change without the
  presence of an external force (and the forces
  involved in a collision are internal to the
  system not external)

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Momentum Kinetic Energyin Collisions

• This does not mean that the linear momentum of
  the various colliding bodies cannot change
• The linear momentum of each body involved in the
  collision may indeed change, but the total linear
  momentum of the system cannot change

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Inelastic Collisions inOne Dimension

• Here we have the before and after pictures of
  an inelastic collision in one dimension

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Inelastic Collisions inOne Dimension

• From the rule that total linear momentum must be
  preserved, we know that
• Given that this is, by definition, a
  1-dimensional problem we can drop the arrows

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Inelastic Collisions inOne Dimension

• Knowing that p mv, we can also rewrite our
  equation (10-16) as
• Thus if we know the masses and initial velocities
  and one of the final velocities then we can
  compute the other final velocity

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Completely Inelastic Collisionin One Dimension

• Here we again have the before and after
  pictures but this time we have a completely
  inelastic collision
• The body m2 happens to be at rest initially (it
  is therefore called the target)

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Completely Inelastic Collisionin One Dimension

• In light of the new situation, we can rewrite our
  previous equation (10-16) asorwhere V is
  the velocity of the stuck-together masses after
  the collision

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Completely Inelastic Collisionin One Dimension

• Note that because the ratiothe value of V
  must always be less than v1,i

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Velocity of theCenter of Mass

• In a closed, isolated system we know that the
  velocity of the center of mass cannot change
  why?
• Only external forces can change the velocity of
  the center of mass and, by definition, an
  isolated system has no forces acting on it

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Velocity of theCenter of Mass

• As a result, we know that the total momentum of
  our system is the same as the momentum of the
  center of mass thus
• Because total linear momentum is conserved during
  a collision, we can also state that

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Velocity of theCenter of Mass

• Substituting the above into the previous equation
  gives us
• But we can see that the right side of the
  equation is constant, therefore we have proved
  that Vcom is constant in a closed, isolated
  system

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Velocity of theCenter of Mass

• The figure on the right shows a series of freeze
  frames which tracks the velocity of the center
  of mass in a completely inelastic collision

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Velocity of theCenter of Mass

• The target (m2) initially has no momentum, so the
  total momentum of the system is just that of the
  projectile (m1)

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Velocity of theCenter of Mass

• After the collision, the two bodies stick
  together, so their combined velocity is

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Velocity of theCenter of Mass

• Checkpoint 3
• Sample Problem 10-2

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Sample Problem 10-2

• For many years ballistic pendulums were used to
  determine the velocity v of a rifle bullet
• Let M 5.4 kg,m 9.5 g andh 6.3 cm

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Sample Problem 10-2

• This simple mechanism depended on conservation of
  linear momentum as well as conservation of
  mechanical energy in order to allow an accurate
  determination of the bullets speed

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Sample Problem 10-2

• We begin by recognizing the fact that while
  mechanical energy is not conserved during the
  bullet/block collision, linear momentum is
  conserved
• Why is this true?

We are assuming here that no pieces of the
block go flying off when the bullet hits. If they
did, we would have to account for their momentum
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Sample Problem 10-2

• Using conservation of linear momentum we
  havewhere V is the velocity of the block of
  wood immediately after the collision, v is the
  velocity of the bullet before the collision, and
  m and M are the masses of the bullet and block
  respectively

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Sample Problem 10-2

• Thats nice, but since we dont know v we cant
  calculate V
• However we do know h (the height the block rose
  due to the collision)
• We also know that the height the block rises (h)
  is related to the kinetic energy just after the
  collision mechanical energy is conserved once
  the collision is over

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Sample Problem 10-2

• So we have
• Substituting V from the previous equation we get

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Sample Problem 10-2

• Plugging in the values we get
• We will use a ballistic pendulum in one of our
  labs (no rifles allowed however)

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Elastic Collisions inOne Dimension

• Now lets look at what happens in an elastic
  collision (even though we know that most
  collisions are inelastic to some degree)
• We know from the definition that an elastic
  collision is one where the total kinetic energy
  of the system is conserved thus
• Total kinetic energy before the collision Total
  kinetic energy after the collision

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Elastic Collisions inOne Dimension

• Having said that, it is not true that the kinetic
  energy of each body involved in the collision
  must remain the same before vs. after the
  collision
• In fact (depending on the nature of the
  collision) the final kinetic energies of the
  participating bodies may be quite different from
  their initial kinetic energies

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Elastic Collisions inOne Dimension

• Lets take billiard balls as a reasonable
  approximation of an elastic collision
• We know that in a head-on collision, the kinetic
  energy of the cue ball may be nearly entirely
  transmitted to the target ball
• We also know that if the collision is not
  head-on, then a more complicated interaction
  takes place as both balls end up with velocities
  in different directions

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Elastic Collisions inOne Dimension

• The figure below shows two particles that are
  involved in an elastic collision we see the
  before and after pictures

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Elastic Collisions inOne Dimension

• The target initially was at rest, however after
  the collision both objects have some velocity in
  the positive x direction

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Elastic Collisions inOne Dimension

• Assuming the system is closed and isolated, we
  know that linear momentum is conserved thus we
  get
• And because kinetic energy is conserved, we also
  know that

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Elastic Collisions inOne Dimension

• Assuming that we know the masses and the initial
  velocities, we then have two equations and two
  unknowns (v1,f and v2,f)
• Rearranging the first equation we get

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Elastic Collisions inOne Dimension

• And rearranging the 2nd equation a little give us

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Elastic Collisions inOne Dimension

• Plugging in the value for m2v2,f from the 1st
  equation into the 2nd equation and turning the
  crank a while results in

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Elastic Collisions inOne Dimension

• Now lets look at these results for a minute
• We can see that v2,f must always be positive
• But the same cannot be said for v1,f the final
  velocity of body 1 may be positive or negative
  depending on the relative sizes of masses m1 and
  m2

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Elastic Collisions inOne Dimension

• Case 1 m1 m2 Equal masses
• In this case you can see that v1,f will be zero
• What happens is that all of the kinetic energy of
  the first body is transferred to the 2nd body
  (note that v2,f v1,i which must be true if KE
  is conserved)

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Elastic Collisions inOne Dimension

• Case 2 m1 ltlt m2 Massive target (golf ball
  hits a cannon ball)
• In this case you can see that v1,f will be
  negative (the body rebounds from the collision)
  and v1,f ? - v1,i
• And v2,f ? (2m1/m2)v1,i which will be a very
  small number

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Elastic Collisions inOne Dimension

• Case 3 m1 gtgt m2 Massive projectile (cannon
  ball hits a golf ball)
• In this case you can see that v1,f will be
  positive (the projectile continues forward) and
  v1,f ? v1,i
• And v2,f ? 2v1,i (the target leaves at twice the
  initial velocity of the projectile)

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Elastic Collisions inOne Dimension

• Why does the target in the 3rd case leave at
  twice the initial velocity of the projectile?
• Lets go back to the opposite case where the
  light projectile hit the massive target
• In that case the velocity of the projectile
  changed from v to v for a ?v -2v (for a net
  change in velocity of 2v)

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Elastic Collisions inOne Dimension

• Here we have the opposite case the massive
  projectile hits the light target and we get a
  corresponding ?v 2v this time from an initial
  velocity of zero to 2v (again for a net change in
  velocity of 2v)

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Elastic Collisions with aMoving Target

• In the previous examples we assumed that the
  initial velocity of the target was zero but
  that of course doesnt always have to be the case
• Now lets look at what happens if the target is
  moving when the collision occurs

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Elastic Collisions with aMoving Target

• In the figure below we have two bodies that are
  moving towards each other for a head-on collision

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Elastic Collisions with aMoving Target

• Conservation of linear momentum tells us
  thatand from conservation of KE we get

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Elastic Collisions with aMoving Target

• We rearrange the 1st equation asand the 2nd
  as

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Elastic Collisions with aMoving Target

• Do the substitution and turn the crank for a
  while you will eventually get

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Elastic Collisions with aMoving Target

• Note that these equations are symmetric with
  respect to the subscripts they can be swapped
  and the same equations will result

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Elastic Collisions with aMoving Target

• You should also convince yourself that if you set
  v2,i 0, you will get back the equations you saw
  previously

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Elastic Collisions with aMoving Target

• Checkpoint 4
• Sample Problem 10-4

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Sample Problem 10-4

• Sphere 1 with mass m1 30 g is pulled back to a
  height h1 8.0 cm and released from rest

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Sample Problem 10-4

• Sphere 1 then has an elastic collision with
  sphere 2 (mass m2 75 g)
• What is the velocity v1,f of sphere 1 immediately
  after the collision?

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Sample Problem 10-4

• We will attack this as two problems
• The descent of sphere 1, and
• The collision with sphere 2
• We need to know the initial collision velocity of
  sphere 1 but we know how to do that using
  energy considerations

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Sample Problem 10-4

• We know that the potential energy of sphere 1
  when released will equal the kinetic energy at
  the bottom of the swing (via conservation of
  kinetic energy)
• Using that relationship we get

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Sample Problem 10-4

• Solving for v1,i we get
• Now that we know v1,i we can move on to the 2nd
  part of the problem an elastic collision
  between unequal masses with the target mass
  (sphere 2) initially at rest

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Sample Problem 10-4

• We assume that our system is isolated and closed,
  which allows us to use equation 10-30 to find the
  velocity of sphere 1 just after the collision

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Sample Problem 10-4

• Plugging in the values we get
• So from this we can see that sphere 1 moves to
  the left after the collision (e.g., it rebounds)
  which makes sense given the higher mass of
  sphere 2

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Collisions in Two Dimensions

• So far we have dealt with situations where the
  two objects collide head-on
• When that is not the case, the final velocities
  will not be along the initial centerline of the
  masses

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Collisions in Two Dimensions

• The same conservation rules still apply however
  so we know that linear momentum must be conserved
  thus
• If the collision is also elastic (which wont
  always be the case of course) we also have

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Collisions in Two Dimensions

• Because this is a 2-dimensional collision, we
  know that we will have to break the momentum
  equation down into its vector components
• So for the x axis we have

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Collisions in Two Dimensions

• Similarly, for the y axis we have
• Note that we really are taking the absolute value
  of ?1 here, so we need to account for that by
  flipping the sign of this term

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Collisions in Two Dimensions

• Again because we are dealing with an elastic
  collision, we can deal with just the speeds of
  the particles in the kinetic energy equation
  which results in

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Collisions in Two Dimensions

• So at this point we have three equations in seven
  variables 2 masses (m1 and m2), 3 speeds (v1,i,
  v1,f, v2,f), and 2 angles (?1 and ?2)
• If we know any four of these variables, then we
  can solve for the other three

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Collisions in Two Dimensions

• Checkpoint 5
• Sample Problem 10-5

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Sample Problem 10-5

• Here we have two skaters Alfred(mA 83 kg
  moving with speed vA 6.2 km/h east) and Barbara
  (mB 55 kg moving at vB 8.7 km/h north)

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Sample Problem 10-5

• Alfred and Barbara experience an inelastic
  collision, e.g., they stick together after impact
• What is the velocity of the couple after they
  collide?

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Sample Problem 10-5

• We know that Pi Pf so we can express that
  as
• Solving for V we get

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Sample Problem 10-5

• If you have a vector-capable calculator, you can
  now solve directly for V
• For those of us who dont have such a calculator,
  we can go back to the first equation and break it
  down into components along the two axis knowing
  that momentum must be conserved along each of the
  axis

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Sample Problem 10-5

• So for the x axis we getand correspondingly
  for the y axis we get

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Sample Problem 10-5

• We now have two equations and two unknowns, so
  after the appropriate algebra we get

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Sample Problem 10-5

• We can now take that value for ? and plug it back
  into our formula for the y axis (rearranged a
  bit) and get

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Sample Problem 10-5

• What is the velocity of the center of mass of the
  two skaters before as well as after the collision?

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Sample Problem 10-5

• The velocity vcom of the COM after the collision
  is the value we just calculated 4.9 km/h at an
  angle ? 40
• The velocity vcom of the COM before the
  collision is exactly the same (since there are no
  external forces)

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Next Class

• Homework Problems Chapter 106, 8, 26, 36, 38,
  48,
• Read sections Chapter 11