Physics 221 Chapter 9

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Physics 221Chapter 9
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On a Collision Course ...

• Linear Momentum mass x velocity
• p mv
• Conservation of Momentum In a collision,
  momentum BEFORE the collision equals the momentum
  AFTER the collision

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Problem 1 . . . The Odd Couple I

• A model railroad car (mass 100 g) collides and
  locks onto a similar stationary car. The coupled
  cars move as a unit on a straight and
  frictionless track. The speed of the moving cars
  is most nearly
• A. 1 m/s
• B. 2 m/s
• C. 3 m/s
• D. 4 m/s

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Solution 1 . . . The Odd Couple I

• Momentum BEFORE Momentum AFTER
• 100 x 4 0 200 x v
• v 2 m/s

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Problem 2 . . . The Odd Couple II

• Is the K.E. BEFORE K.E. AFTER? In other words,
  is the K.E. conserved in this collision?

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Solution 2 . . . The Odd Couple II

• K.E. BEFORE the collision (1/2)(0.1)(16) 0
  0.8J
• K.E. AFTER the collision (1/2)(0.2)(4) 0
  0.4J
• Half of the K.E. has mysteriously disappeared!
• BUT this does not mean that ENERGY is not
  conserved! K.E. was transformed (converted) into
  other forms of energy.

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Problem 3 . . . Big Bang!

• A shell explodes into two unequal fragments. One
  fragment has twice the mass of the other. The
  smaller fragment moves in the NE direction with a
  speed of 6 m/s. The velocity (speed and
  direction) of the other fragment is most nearly
• A. 3 m/s SW
• B. 3 m/s NE
• C. 4 m/s NE
• D. 4 m/s SW

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Solution 3 . . . Big Bang!

• M x 6 2 M x V
• V 3 m/s in the SW direction.

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Close collisions of the second kind ...

• There are TWO types of collisions
• I. Momentum conserved and K.E. also conserved.
  This type is called an ELASTIC collision.
  Elastic collisions are non-sticky as in
  billiard balls or steel ball-bearings.
• II. Momentum conserved BUT K.E. NOT conserved.
  This type is called an INELASTIC collision.
  Inelastic collisions are sticky as in coupled
  railroad cars and putty.

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Problem 4 . . . Elastic I

• Two hockey pucks collide elastically on ice. P2
  is at rest and P1 strikes it head-on with a
  speed of 3 m/s.
• A. P1 stops and and P2 moves forward at 3 m/s
• B. P1 and P2 move forward at 1.5 m/s
• C. P1 and P2 move in opposite directions at 1.5
  m/s
• D. P1 moves forward at 1 m/s and P2 moves forward
  at 2 m/s

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Solution 4 . . . Elastic I

• A. P1 stops and P2 moves forward at 3 m/s.
• Please verify that
• 1. Momentum is conserved
• 2. K.E. is also conserved (elastic)

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Problem5 . . . Elastic II

• Two hockey pucks collide elastically on ice. P2
  has twice the mass of P1 and is at rest and P1
  strikes it head-on with a speed of 3 m/s.
  Calculate their velocities after the collision.

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Solution 5 . . . Elastic II

• P1 moves backward at 1 m/s and P2 moves forward
  at 2 m/s.
• HINT
• 1. Take 5 sheets of scrap paper ( Trust me!!!)
• 2. Write the equation for momentum conservation
• 3. Write the equation for K.E. conservation
• 4. Solve two equations for two unknowns.

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Tell me more about momentum!

• O.K. good boys and girls. Here is everything you
  always wanted to know about momentum but were
  afraid to ask!
• In the beginning there was F ma
• So F m(vf - vi) / t
• If F 0 then mvf mvi
• Another interesting observation If F is NOT zero
  then momentum WILL change and change in momentum
  equals F x t. This is called IMPULSE

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Problem 6 . . . Tiger in the woods

• A golf ball has a mass of 48 g. The force
  exerted by the club vs. time is a sharp spike
  that peaks at 180 N for 0.01 s and then drops
  back to zero as the ball leaves the club head at
  high speed. It is estimated that the average
  force is 90 N over a short time (?t) of 0.04 s.
  The speed of the ball is
• A. 35 m/s
• B. 75 m/s
• C. 95 m/s
• D. 135 m/s

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Solution 6 . . . Impulse to ride the tiger

• Impulse F ?t 90 x 0.04 3.6 Ns
• Impulse change in momentum
• 3.6 mv -0
• 3.6 0.048 x v
• v 75 m/s

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Problem 7 . . . Teeter- Totter

• FB weighs 180 pounds and sits at the 20 cm. mark.
  Where should SP (120 pounds) sit in order to
  balance the teeter-totter at the playground?

0 20
50
100
180
120
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Solution 7 . . . See-Saw

• 30 x 180 what x 120
• what 45 or 95 cm. mark

0 20
50
100
180
120
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Problem 8 . . . Center of Mass

• The C.M. is a weighted average position of a
  distribution of masses where the system can be
  balanced.
• Where is the C.M. ?

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Solution 8 . . . Center of Mass

• (100)( x -20) (300)(80 - x)
• x (100) (20) (300)( 80) / (100 300)
• x 65

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A formula for C.M.

• C.M. M1 X1 M2 X2
• M1 M2

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Problem 9 . . . more C. M.

• Include the mass of the meter stick (60 g). Where
  is the new C.M. ?

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Solution 9 . . . more C. M.

• C.M. M1 X1 M2 X2 M3 X3
• M1 M2 M3
• C.M. (100)(20) (60)(50) (300)(80)
• 100 60 300
• C.M. 63

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Problem 10 . . . The Hole

• A circular hole of radius R/2 is drilled in a
  disc of mass M and radius R as shown in the
  figure. The center of mass is closest to
• A. (R/6 , 0)
• B. (R/4 , 0)
• C. (R/3 , 0)
• D. (R/2 , 0)

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Circular plate with hole

• If the mass of the missing piece is 100 then the
  mass of the remaining piece is 300 for a total of
  400. In other words, if M is the mass of the
  original disc (no hole) then the mass of the disc
  with the hole is 3M/4 and the mass of the
  circular piece removed is M/4.
• NOTE Non-Linear thinking! Disc with HALF the
  radius has only a quarter of the mass!

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Solution 10A model for the hole problem

• (100)(R/2) (300)(x)
• x R/6

x