Physics 111: Lecture 15 Todays Agenda

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Physics 111 Lecture 15Todays Agenda

• Elastic collisions in one dimension
• Center of mass reference frame
• Colliding carts problem
• Some interesting properties of elastic collisions
• Killer bouncing balls

2
Momentum Conservation Review

• The concept of momentum conservation is one of
  the most fundamental principles in physics.
• This is a component (vector) equation.
• We can apply it to any direction in which there
  is no external force applied.
• You will see that we often have momentum
  conservation even when kinetic energy is not
  conserved.

3
Comment on Energy Conservation

• We have seen that the total kinetic energy of a
  system undergoing an inelastic collision is not
  conserved.
• Energy is lost
• Heat (bomb)
• Bending of metal (crashing cars)
• Kinetic energy is not conserved since work is
  done during the collision!
• Momentum along a certain direction is conserved
  when there are no external forces acting in this
  direction.
• In general, momentum conservation is easier to
  satisfy than energy conservation.

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Lecture 15, Act 1Collisions

• A box sliding on a frictionless surface collides
  and sticks to a second identical box which is
  initially at rest.
• What is the ratio of initial to final kinetic
  energy of the system?

(a) 1 (b) (c) 2

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Lecture 15, Act 1Solution

• No external forces in the x direction, so PX is
  constant.

v
m
m
m
m
v / 2
x
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Lecture 15, Act 1Solution

• Compute kinetic energies

v
m
m
m
m
v / 2
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Lecture 15, Act 1Another solution

• We can write

• P is the same before and after the collision.

• The mass of the moving object has doubled, hence
  thekinetic energy must be half.

m
m
m
m
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Lecture 15, Act 1Another Question

• Is it possible for two blocks to collide
  inelastically in such a way that the kinetic
  energy after the collision is zero?

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Lecture 15, Act 1 Another Question

• Is it possible for two blocks to collide
  inelastically in such a way that the kinetic
  energy after the collision is zero?

YES If the CM is not moving!
CM
CM
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Elastic Collisions

• Elastic means that kinetic energy is conserved as
  well as momentum.
• This gives us more constraints
• We can solve more complicated problems!!
• Billiards (2-D collision)
• The colliding objectshave separate motionsafter
  the collision as well as before.
• Start with a simpler 1-D problem

Initial
Final
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Elastic Collision in 1-D
m2
m1
initial
v1,i
v2,i
x
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Elastic Collision in 1-D
m1
m2
before
v1,i
v2,i
x
after
v2,f
v1,f
Suppose we know v1,i and v2,i We need to solve
for v1,f and v2,f
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Elastic Collision in 1-D
Airtrack
Collision balls

• However, solving this can sometimes get a little
  bit tedious since it involves a quadratic
  equation!!
• A simpler approach is to introduce the Center
  of Mass Reference Frame

m1v1,i m2v2,i m1v1,f m2v2,f
1/2 m1v21,i 1/2 m2v22,i 1/2 m1v21,f 1/2
m2v22,f
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CM Reference Frame

• We have shown that the total momentum of a system
  is the velocity of the CM times the total mass
• PNET MVCM.
• We have also discussed reference frames that are
  related by a constant velocity vector (i.e.
  relative motion).
• Now consider putting yourself in a reference
  frame in which the CM is at rest. We call this
  the CM reference frame.
• In the CM reference frame, VCM 0 (by
  definition) and therefore PNET 0.

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Lecture 15, Act 2Force and Momentum

• Two men, one heavier than the other, are standing
  at the center of two identical heavy planks. The
  planks are at rest on a frozen (frictionless)
  lake.
• The men start running on their planks at the same
  speed.
• Which man is moving faster with respect to the
  ice?

(a) heavy (b) light (c)
same
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Lecture 15, Act 2Conceptual Solution

• The external force in the x direction is zero
  (frictionless)
• The CM of the systems cant move!

X
X
X
X
x
CM
CM
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Lecture 15, Act 2Conceptual Solution

• The external force in the x direction is zero
  (frictionless)
• The CM of the systems cant move!
• The men will reach the end of their planks at the
  same time, but lighter man will be further from
  the CM at this time.
• The lighter man moves faster with respect to the
  ice!

X
X
X
X
CM
CM
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Lecture 15, Act 2Algebraic Solution

• Consider one of the runner-plank systems
• There is no external force acting in the
  x-direction
• Momentum is conserved in the x-direction!
• The initial total momentum is zero, hence it must
  remain so.
• We are observing the runner in the CM reference
  frame!

x
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Lecture 15, Act 2Algebraic Solution

• The speed of the runner with respect to the plank
  is V vR vP (same for both runners).

MvP mvR (momentum conservation)
m
vR
vP
So vR is greater if m is smaller.
M
x
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Example 1 Using CM Reference Frame

• A glider of mass m1 0.2 kg slides on a
  frictionless track with initial velocity v1,i
  1.5 m/s. It hits a stationary glider of mass m2
  0.8 kg. A spring attached to the first glider
  compresses and relaxes during the collision, but
  there is no friction (i.e. energy is conserved).
  What are the final velocities?

Video of CM frame
m2
v1,i
v2,i 0
m1
VCM
CM
m2
m1
x
v2,f
m2
m1
v1,f
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Example 1...

• Four step procedure
• First figure out the velocity of the CM, VCM.
• VCM (m1v1,i m2v2,i), but
  v2,i 0 so
• VCM v1,i
• So VCM 1/5 (1.5 m/s) 0.3 m/s

Step 1
(for v2,i 0 only)
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Example 1...

• If the velocity of the CM in the lab reference
  frame is VCM, and the velocity of some particle
  in the lab reference frame is v, then the
  velocity of the particle in the CM reference
  frame is v
• v v - VCM (where v, v, VCM are vectors)

v
VCM
v
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Example 1...

• Calculate the initial velocities in the CM
  reference frame (all velocities are in the x
  direction)

Step 2
v1,i v1,i - VCM 1.5 m/s - 0.3 m/s 1.2 m/s
v2,i v2,i - VCM 0 m/s - 0.3 m/s -0.3 m/s
v1,i 1.2 m/s
v2,i -0.3 m/s
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Example 1 continued...
Movie

• Now consider the collision viewed from a frame
  moving with the CM velocity VCM. ( jargon in
  the CM frame)

m2
v1,i
v2,i
m1
m2
m1
x
m2
v2,f
m1
v1,f
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Energy in Elastic Collisions

• Use energy conservation to relate initial and
  final velocities.
• The total kinetic energy in the CM frame before
  and after the collision is the same
• But the total momentum is zero
• So

(and the same for particle 2)
Therefore, in 1-D v1,f -v 1,i
v2,f -v2,i
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Example 1...
Step 3
v1,f -v 1,i v2,f -v2,i
m2
v1,i
v2,i
m1
m2
m1
x
v2,f - v2,i .3 m/s
v1,f - v1,i -1.2m/s
m2
m1
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Example 1...
v v - VCM
Step 4

• So now we can calculate the final velocities in
  the lab reference frame, using

v v VCM
v1,f v1,f VCM -1.2 m/s 0.3 m/s -0.9
m/s
v2,f v2,f VCM 0.3 m/s 0.3 m/s 0.6
m/s
Four easy steps! No need to solve a quadratic
equation!!
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Lecture 15, Act 3 Moving Between Reference Frames

• Two identical cars approach each other on a
  straight road. The red car has a velocity of 40
  mi/hr to the left and the green car has a
  velocity of 80 mi/hr to the right.
• What are the velocities of the cars in the CM
  reference frame?

(a) VRED - 20 mi/hr (b) VRED - 20
mi/hr (c) VRED - 60 mi/hr VGREEN
20 mi/hr VGREEN 100 mi/hr VGREEN
60 mi/hr
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Lecture 15, Act 3 Moving Between Reference Frames

• The velocity of the CM is

• So VGREEN,CM 80 mi/hr - 20 mi/hr 60 mi/hr

• So VRED,CM - 40 mi/hr - 20 mi/hr - 60 mi/hr

x
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Lecture 15, Act 3 Aside

• As a safety innovation, Volvo designs a car with
  a spring attached to the front so that a head on
  collision will be elastic. If the two cars have
  this safety innovation, what will their final
  velocities in the lab reference frame be after
  they collide?

80mi/hr
- 40mi/hr
x
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Lecture 15, Act 3 Aside Solution
vGREEN,i 60 mi/hr vRED,i -60 mi/hr
vGREEN,f -v GREEN,i vRED,f -vRED,i
vGREEN,f -60 mi/hr vRED,f 60
mi/hr
v v VCM
vGREEN,f -60 mi/hr 20 mi/hr - 40 mi/hr
vRED,f 60 mi/hr 20 mi/hr 80 mi/hr
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Summary Using CM Reference Frame
(m1v1,i m2v2,i)
VCM
Step 1

• Determine velocity of CM
• Calculate initial velocities
  in CM reference frame
• Determine final velocities in
  CM reference frame
• Calculate final velocities in
  lab reference frame

Step 2
v v - VCM
Step 3
vf -vi
Step 4
v v VCM
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Interesting Fact
v1,i
v2,i

• We just showed that in the CM reference frame
  the speed of an object is the same before and
  after the collision, although the direction
  changes.
• The relative speed of the blocks is therefore
  equal and opposite before and after the
  collision. (v1,i - v2,i) - (v1,f - v2,f)
• But since the measurement of a difference of
  speeds does not depend on the reference frame, we
  can say that the relative speed of the blocks is
  therefore equal and opposite before and after the
  collision, in any reference frame.
• Rate of approach rate of recession

v2,f -v2,i
v1,f -v1,i
This is really cool and useful too!
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Basketball Demo.
Drop 2 balls

• Carefully place a small rubber ball (mass m) on
  top of a much bigger basketball (mass M). Drop
  these from some height. The height reached by the
  small ball after they bounce is 9 times the
  original height!! (Assumes M gtgt m and all bounces
  are elastic).
• Understand this using the speed of approach
  speed of recession property we just proved.

3v
m
v
v
v
v
M
v
(a)
(b)
(c)
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More comments on energy

• Consider the total kinetic energy of the system
  in the lab reference frame

but
so
(same for v2)
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More comments on energy...

• Consider the total kinetic energy of the system
  in the LAB reference frame

KREL
KCM
So ELAB KREL KCM
KCM is the kinetic energy of the center of mass.
KREL is the kinetic energy due to relative
motion in the CM frame.
This is true in general, not just in 1-D
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More comments on energy...

• ELAB KREL KCM
• Does total energy depend on the reference frame??
• YOU BET!

KREL is independent of the reference frame,
but KCM depends on the reference frame (and 0
in CM reference frame).
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Recap of todays lecture

• Elastic collisions in one dimension (Text
  8-6)
• Center of mass reference frame (Text 8-7)
• Colliding carts problem
• Some interesting properties of elastic collisions
  
• Killer bouncing balls
• Look at textbook problems Chapter 11 63, 67, 71