Center of Mass and Linear Momentum

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CHAPTER-9

• Center of Mass and Linear Momentum

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Ch 9-2 The Center of Mass

• Center of mass (com) of a system of particles is
  required to describe the position and motion of
  the system
• The com of a system of particles is the point
  that moves as if the entire mass of the system
  were concentrated there and all external forces
  were applied there

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Ch 9-2 The Center of Mass

• com is defined with reference to origin of an
  axis
• For Fig. b
• Xcom(m1x1m2x2)/(m1m2)
• (m1x1m2x2)/M
• where M m1m2 ?mi
• xcom ?mixi / M ?mixi/?mi
• ycom?miyi /M
• zcom?miyi /M
• rcom xcom i ycom j zcom k
• ?miri / M

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Ch 9-2 The Center of Masssolid bodies

• For a solid body having continuous distribution
  of matter, the particle becomes differential mass
  element dm
• xcom(1/M) ?x dm ycom(1/M) ?y dm
• zcom(1/M) ?z dm
• M is mass of the object and its density ? are
  related to its volume through ? M/V dm/dV
• Then xcom(1/V) ?x dV ycom(1/V) ?y dV
• zcom(1/V) ?z dV

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Ch-9 Check Point 1

• (a) origin
• (b) fourth quadrant
• (c) on y axis below origin
• (d) origin
• (e) third quadrant
• (f) origin

• The figure shows a uniform square plate from
  which four identical squares at the corners will
  be removed
• a) where is com of plate originally?
• b) where it is after removal of square 1
• c) after removal of square 1 and 2
• d) after removal of square 1 and 3
• e) after removal of square 1, 2 and 3
• f) All four square
• Answers in term of Quadrant, axes or points

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Ch 9-3 Newtons Second Law for a System of
Particles

• For a system of particles with com defined by
• rcom ?miri / M , Newtons Second law FnetM
  acom
• Mrcom ?miri differentiating w.r.t time
• Mvcom ?mivi differentiating w.r.t time
• Macom ?miai ?Fi Fnet
• Fnet-x Macom-x Fnet-y Macom-y Fnet-z Macom-z
  

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Ch 9-4,5 Linear Momentum

• Linear momentum of a particle p- a vector
  quantity p mv (linear momentum of a particle)
• Newtons second law of motion
• Time rate change of the momentum of a particle is
  equal to the net force acting on the particle and
  is in the direction of force
• Fnet d/dt (p) d/dt (mv) m dv/dt ma
• Linear momentum P of a system of particles is
  vector sum of individuals particles linear
  momenta p
• P ?pi ?mivi Mvcom (System of particles)
• Fnet d/dt (P) Macom (System of particles)

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Ch-9 Check Point 3

• The figure gives the magnitude of the linear
  momentum versus time t for a particle moving
  along an axis. A force directed along the axis
  acts on the particle.
• Rank the four regions indicated according to the
  magnitude of the force, greatest first
• b) In which region particle is slowing

• F dp/dt
• Region 1 largest slope
• 2 Zero slope
• 3 Negative slope
• 4 Zero slopes
• Ans
• 1, 3, then 2 and 4 tie (zero force)
• (b) 3

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Ch 9-6 Collision and Impulse

• Momentum p of any point- like object can be
  changed by application of an external force
• Single collision of a moving particle-like object
  (projectile) with another body ( target)
• Ball (Projectile-R) bat (target-L) system
• Change in momentum of ball in time dt , dpF(t)
  dt.
• Net change ? dp ? F(t) dt.
• Impulse J ? F(t) dt. Favg ?t
• Change in momentum ?P Pf-Pi J

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Ch 9-7 Conservation of Linear Momentum

• If Fnet-external 0 then J ? F(t) dt 0
• ?P Pf-Pi J0 then Pf Pi
• Law of conservation of linear momentum
• If no net external force acts on a system of
  particles, the total linear momentum of the
  system of particles cannot change.
• Pfx Pix Pfy Piy

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Ch-9 Check Point 4

• A paratrooper whose chute fails to open lands in
  snow, he slightly hurt. Had he landed on bare
  ground, the stoppong time would be 10 times
  shorter and the collision lethal. Does the
  presence of snow increases, decreases or leaves
  unchanged the values of
• (a) the paratrooper change in momentum
• (b) the impulse stopping the paratrooper
• C) the force stopping the paratrooper

• 4.
• (a)?p m(vf-vi) unchanged
• (b) J ?p unchanged
• (c) J?F.dt ?t increase, F decrease

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Ch-9 Check Point 5

• The figure shows an overhead view of a ball
  bouncing from a vertical wall without any change
  in its speed. Consider the change ?p in the
  balls linear momentum
• a) Is ?px positive, negative, or zero
• b) Is ?px positive, negative, or zero
• b) What is direction of ?p?

• ?pxpxf-pxi
• 0
• ?pypyf-pyipyf-(-pyi)
• positive
• Direction of ?P towards y-axis

• q

q
x
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Ch-9 Check Point 6

• An initially stationary device lying on a
  frictionless floor explodes into two pieces,
  which then slides across the floor. One piece
  slides in the positive direction of an x axis.
• a) What is the sum of the momenta of the two
  pieces after the explosion?
• b) Can the second piece move at an angle to the
  x-axis?
• c) What is the direction of the momentum of the
  second piece?

• a) Pi Pf 0
• b) No because
• Pf 0P1fxP2
• Then P2 -P1fx
• c) Negative x-axis

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Ch 9-8 Momentum and Kinetic Energy in collisions

• Elastic collision Momentum and kinetic energy of
  the system is conserved
• then Pf Pi and Kf Ki
• Inelastic collision Momentum of the system is
  conserved but kinetic energy of the system is not
  conserved
• then Pf Pi and Kf ? Ki
• Completely inelastic collision Momentum of the
  system is conserved but kinetic energy of the
  system is not conserved. After the collision the
  colliding bodies stick together and moves as a
  one body.

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Ch-9-9 Inelastic Collision in one Dimension

• One-dimensional inelastic collision
• For a two body system
• Total momentum Pi before collision Total
  momentum Pf after collision
• p1ip2ip1fp2f
• m1v1im2v2im1v1fm2v2f
• One-dimensional completely inelastic collision
  (v2i0)
• m1v1i(m1m2)V
• m1v1i/(m1m2)
• Hence V?v1i m1 ? (m1m2)

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Ch-9-9 Velocity of the Center of Mass

• For One-dimensional completely inelastic
  collision of a two body system
• P(m1m2)vcom
• PiPf and m1v1i(m1m2)V
• Pi(m1m2)vcom m1v1i and Pf(m1m2)vcom
  (m1m2)V
• vcom m1v1i /(m1m2) V
• Vcom has a constant speed

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Sample-Problem-9-8Ballistic Pedulum

• One dimensional completely inelastic collision
• Conservation of linear momentum
• mv (mM)V Vmv/(mM)
• Conservation of mechanical energy
• (KPE)initial (KPE)final
• (mM)V2/2 (mM)gh
• V ?(2gh) Vmv/(mM)
• v (mM)?(2gh)/m

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Ch-9-10 Elastic Collision in One Dimensions

• Elastic collision Momentum and kinetic energy of
  the system is conserved
• then Pf Pi and Kf Ki
• Two classes
• Stationary target (v2i0)
• m1v1im1v1fm2v2f
• m1v1i2/2m1v1f2/2m2v2f2/2
• v1fv1i(m1-m2)/(m1m2)
• v2f2v1im1/(m1m2)

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Ch-9-10 Elastic Collision in One Dimensions

• Stationary target (v2i0)
• v1fv1i(m1-m2)/(m1m2)
• v2f2v1im1/(m1m2)
• Three cases
• Equal masses m1m2
• v1f0 and v2fv1i
• Massive target m2gtgtm1
• v1f-v1i and v2f ? (2m1/m2)v1i
• Massive projectile m1gtgtm2
• v1f ? v1i and v2f? 2v1i

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Ch-9-10 Elastic Collision in One Dimensions

• Moving target
• m1v1i m2v2i m1v1fm2v2f
• m1v1i2/2 m2v2i2/2 m1v1f2/2m2v2f2/2
• v1fv1i (m1-m2)/(m1m2) 2m2v2i/(m1m2)
• v2f2v1i m1/(m1m2)
• v2i (m2-m1) / (m1m2)

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Ch-9 Check Point 10

• What is the final linear momentum of the target
  if the initial linear momentum of the projectile
  is 6 kg.m/s and final linear momentum of the
  projectile is
• a) 2 kg.m/s
• b) -2 kg.m/s
• c) what is the final kinetic energy of the target
  if the initial and final kinetic energies of the
  projectile are , respectively 5 J and 2 J?

• PiPf
• Pf2Pi-Pf1
• 6-2 4 kg.m/s
• 6-(-2)8 kg.m/s
• c) Ki1 Kf1Kf2
• Then
• Kf2Ki1-Kf1
• 5 23 J

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Ch-9-11 Elastic Collision in Two Dimensions

• Solve equations
• P1i P2i P1f P2f
• K1i K2i K1f K2f
• Along x-axis
• m1v1i m1v1f cos?1m2v2f cos?2
• Along y-axis
• 0 - m1v1f sin?1m2v2f sin?2
• m1v1i2/2 m1v1f2/2m2v2f2/2