Transcript SPH4U: Lecture 15 Today’s Agenda

SPH4U: Lecture 15
Today’s Agenda

Elastic collisions in one dimension
 Analytic solution (this is algebra, not rocket science!)

Center of mass reference frame
 Colliding carts problem
 Two dimensional collision problems (scattering)
 Solving elastic collision problems using COM and inertial reference
frame transformations

Some interesting properties of elastic collisions
 Center of mass energy and energy of relative motion
Momentum Conservation: Review
FEXT 



dP
dt
dP
0
dt
FEXT  0
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.
Comment on Energy Conservation

We have seen that the total kinetic energy of a system undergoing
an inelastic collision is not conserved.
 Mechanical Energy is lost: (remember what this is??)
» Heat (bomb)
» Bending of metal (crashing cars)

Kinetic energy is not conserved since dissipative work is done
during an inelastic collision! (here, KE equals mechanical energy)

Total momentum, PT, along a certain direction is conserved when
there are no external forces acting in this direction.
 F = ma = dPT/dt says this has to be true!! (Newton’s Laws)
 In general, momentum conservation is easier to satisfy than
mechanical energy conservation.
 Remember: in the absence of external forces, total energy
(including heat…) of a system is always conserved even when
mechanical energy is not conserved.
 How much do two objects that inelastically collide heat up?
Lecture 15, Act 1
Collisions


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) 2
(c) 2
Lecture 15, Act 1
Solution

No external forces in the x direction, so PX is constant.
PI  mv
v
PF  2m  
2
v
m
m
m
m
v/2
x
Lecture 15, Act 1
Solution

Compute kinetic energies:
KI 
1
mv 2
2
v
m
m
2
1
1
v
K F  2 m   K I
2
2
 2
m
KI
2
KF
m
v/2
Lecture 15, Act 1
Another solution

1
P2
2
We can write K  mv 
2
2m

P is the same before and after the collision.

The mass of the moving object has doubled, hence the
kinetic energy must be half.
KI
2
KF
m
m
m
m
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?
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
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 objects
have separate motions
after the collision as
well as before.
Initial
Final
 all 3D collision problems can be solved in 2 dimensions
by using center of mass inertial reference frame

Start with a simpler 1-D problem
Elastic Collision in 1-D
what has to happen
Why is this elastic?
m2
m1
initial
v1,i
v2,i
Maybe, it depends…
x
m1
final
m2
v2,f
v1,f
Kinetic energy  potential energy  kinetic energy
The spring is conservative
Elastic Collision in 1-D
the spring is conservative
m1
Conserve PX: (no external forces!)
before
m2
v1,i v2,i
x
m1v1,i + m2v2,i = m1v1,f + m2v2,f
after
Conserve Kinetic Energy: (it’s elastic!)
1/
2
v1,f
m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f
Suppose we know v1,i and v2,i
We need to solve for v1,f and v2,f
Should be no problem
2 equations & 2 unknowns!
v2,f
Elastic Collision in 1-D

However, solving this can sometimes get a little
bit tedious since it involves a quadratic
equation!!
m1v1,i + m2v2,i = m1v1,f + m2v2,f
1/
2

m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f
A simpler approach is to introduce the
Center of Mass Reference Frame

First, describe the solution to the problem using
algebra.
 Useful analysis and useful formulae
momentum
energy
Elastic Collision in 1-D
special case: equal masses

If the masses of the two objects are equal the
algebra is not too bad. Let’s see what we get…
m1v1,i + m2v2,i = m1v1,f + m2v2,f
1/
2

m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f
Divide through by m = m1 = m2
v1,i + v2,i = v1,f + v2,f
v21,i + v22,i = v21,f + v22,f
momentum
energy
Elastic Collision in 1-D
special case: equal masses

Now just rearrange equations to bring v1,i and v1,f to left hand
side, and v2,i and v2,f to rhs
v1,i - v1,f = v2,f - v2,i
v21,i - v21,f = v22,f - v22,i
momentum
energy
(v1,i - v1,f )(v1,i + v1,f ) = (v2,f - v2,i )(v2,f + v2,i )

Divide through energy equation by momentum equation which
gives
v1,i + v1,f = v2,f + v2,i
v1,i - v1,f = v2,f - v2,i

v2,f = v1,i
v1,f = v2,i
Particles just trade velocities in 1-D elastic collision of
equal mass objects (let equation talk to you…)
Elastic Collision in 1-D
general case: unequal masses

Conserve linear momentum and mechanical energy, but now
the masses are different:
m1(v1,i - v1,f ) = m2(v2,f - v2,I )
m1(v21,i - v21,f ) = m2(v22,f - v22,I )
momentum
energy
m1(v1,i - v1,f )(v1,i + v1,f ) =m2 (v2,f - v2,i )(v2,f + v2,i )

Divide through energy equation by momentum equation which
gives
v1,i + v1,f = v2,f + v2,i
m1(v1,i - v1,f ) = m2(v2,f - v2,I )

Now solving these two linear equations is only a bit more
complicated
Elastic Collision in 1-D
general case: unequal masses

Algebra just gave us the following equations based on
conservation of momentum and mechanical energy:
v1,i + v1,f = v2,f + v2,i
m1(v1,i - v1,f ) = m2(v2,f - v2,I )

Now just solve for final velocities, v1,f and v2,f
in terms of v1,i and v2,i
v1, f 
v2, f 
2m2
m  m2
v2,i  1
v1,i
m1  m2
m1  m2
2m1
m  m1
v1,i  2
v2,i
m1  m2
m1  m2
When m1 = m2
v2,f = v1,i
v1,f = v2,i
Another way to solve elastic collision problems:
CM Reference Frame

We have shown that the total momentum of a system of
particles 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.they’re in 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.
 This is a cool mathematical tool that makes the algebra
solving this much simpler

(it doesn’t change the physical situation)
Lecture 15, Act 2
Force 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 somewhere in Ontario.
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
Lecture 15, Act 2
Conceptual Solution

The external force in the x direction is zero (frictionless):
 The CM of the systems can’t move!
 Aha! this is the key!!
X
X
X
X
CM
CM
x
Lecture 15, Act 2
Conceptual Solution



The external force in the x direction is zero (frictionless):
 The CM of the systems can’t 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.
 His motion doesn’t count as much, since he is less massive
The lighter man moves faster with respect to the ice!
X
X
X
X
CM
CM
Lecture 15, Act 2
Algebraic 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!

Let the mass of the runner be m and the plank be M.

Let the speed of the runner and the plank with respect
to the ice be vR and vP respectively.
m
vR
vP
M
x
Lecture 15, Act 2
Algebraic Solution

The velocity of the runner with respect to the plank
is V = vR - vP (same for both runners).
MvP = - mvR
(momentum conservation, it’s zero!)
Plugging vP = vR - V into this
we find:
M
vR  V
m M
So vR is greater if m is smaller.
m
vR
vP
M
x
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?
m1
m2
v1,i
v2,i = 0
VCM
+ = CM
m1
v1,f
m1
m2
x
m2
v2,f
Example 1...

Four step procedure
 First
figure out the velocity of the CM, VCM.


1
 (m1v1,i + m2v2,i), but v2,i = 0 in this case so
 m1  m 2 
» VCM = 
 m1 
VCM = 
 v1,i
m

m
 1
2

(for v2,i = 0 only)
So VCM = 1/5 (1.5 m/s) = 0.3 m/s
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* where:
v* = v - VCM
(where v*, v, VCM are vectors)
This is the “lab”
frame velocity
v
VCM
v*
If you were traveling along with the CM, you
would see the velocity of the mass to be less
than in the lab frame in this case
This is the CM
frame velocity
Example 1...
 Calculate
the initial velocities in the CM reference frame
(all velocities are in the x direction):
v*1,i = v1,i - VCM = 1.5 m/s - 0.3 m/s = 1.2 m/s
v*2,i = v2,i - VCM = 0 m/s - 0.3 m/s = -0.3 m/s
v*1,i = 1.2 m/s
v*2,i = -0.3 m/s
Movie
Example 1 continued...

Now consider the collision viewed from a frame moving
with the CM velocity VCM. ( jargon: “in the CM frame”)
m1
m1
v*1,f m1
m2
v*1,i
v*2,i
m2
x
m2
v*2,f
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, it’s elastic!! (look how we write this…)
1
1
1
1
m12 v *12,i 
m22 v * 22,i 
m12 v *12,f 
m22 v * 22,f
2 m1
2 m2
2 m1
2 m2

But the total momentum is zero, both initial and final:
m v *   m v * 
2

1
So:
1,i
2
2
2 ,i
Likewise for final v’s
 1
1  2 2  1
1  2 2

m1 v *1,i  
m1 v *1,f


 2 m1 2 m2 
 2 m1 2 m2 
v *12,i  v *12,f
Therefore, in 1-D:
(and the same for particle 2)
v*1,f = -v* 1,i
v*2,f = -v*2,i
Example 1...
Calculate the final velocities in the CM
frame:
v*1,f = -v* 1,i
m1
v*2,f = -v*2,i
m2
v*1,i
m1
m2
x
v*2,f = - v*2,i =.3 m/s
v*1,f = - v*1,i = -1.2m/s
m1
v*2,i
m2
v* = v - VCM
v*f = -v*,i
Example 1...
 Now
we can calculate the final velocities in the lab reference
frame, using:
v = v* + VCM
v1,f = v*1,f + VCM = -1.2 m/s + 0.3 m/s = -0.9 m/s
v2,f = v*2,f + VCM = 0.3 m/s + 0.3 m/s = 0.6 m/s
v1,f = -0.9 m/s
v2,f = 0.6 m/s
Four easy steps! No need to solve a quadratic equation!!
Especially important in 2D
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
(a)
are the velocities of the cars in the CM reference frame?
VRED = - 20 mi/hr
VGREEN = + 20 mi/hr
(b) VRED = - 20 mi/hr
VGREEN = +100 mi/hr
(c) VRED = - 60 mi/hr
VGREEN = + 60 mi/hr
Lecture 15, Act 3
Moving Between Reference Frames

The velocity of the CM is:


1

  m1v1  m2v2 
 m1  m2 
VCM 
m  80  m  40
mi / hr
2m
= 20 mi / hr

So VGREEN,CM = 80 mi/hr - 20 mi/hr = 60 mi/hr

So VRED,CM = - 40 mi/hr - 20 mi/hr = - 60 mi/hr

The CM velocities are equal and opposite since PNET = 0 !!
20mi/hr
 40mi/hr
80mi/hr
CM
x
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
20mi/hr
 40mi/hr
CM
x
Lecture 15, Act 3
Aside Solution
v*GREEN,i = 60 mi/hr
v*RED,i = -60 mi/hr
v*GREEN,f = -v* GREEN,i
v*RED,f = -v*RED,i
v*GREEN,f = -60 mi/hr
v*RED,f = 60 mi/hr
v´ = v* + VCM
v´GREEN,f = -60 mi/hr + 20 mi/hr = - 40 mi/hr
v´RED,f = 60 mi/hr + 20 mi/hr = 80 mi/hr
Summary: Using CM Reference Frame
 (m1v1,i + m2v2,i)
: Determine velocity of CM VCM = 
m1  m2

: Calculate initial velocities in CM
reference frame
v* = v - VCM
: Determine final velocities in CM
reference frame
v*f = -v*i
: Calculate final velocities in lab
reference frame
v = v* + VCM



Interesting Fact

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.
v*1,i
v*2,i
v*1,f = -v*1,i v*2,f = -v*2,i

The relative speed of the blocks is therefore equal and opposite
before and after the collision.
(v*1,i - v*2,i) = - (v*1,f - v*2,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
This is really cool and useful too!
Recap of lecture
Elastic Collision –a collision in which the total kinetic
energy after the collision equals the total kinetic energy
before the collision.
If m2 is initially at rest, then you can use: conservation of energy
and conservation of momentum, or:
 m  m2 
v1 f   1
 v1i
 mT 
 2m1 
v2 f  
 v1i
 mT 


v1i  v2i   v1 f  v2 f  v2 f  v1 f
Example
A ball with mass 2.5x10-2 kg moving at 2.3 m/s collides with a
stationary 2.0x10-2 kg ball. If the collision is elastic, determine the
velocity of each ball after the collision.
Solution
A ball with mass 2.5x10-2 kg moving at 2.3 m/s collides with a
stationary 2.0x10-2 kg ball. If the collision is elastic, determine the
velocity of each ball after the collision.
K.E.
Momentum
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
m1v1i  m2v2i  m1v1 f  m2v2 f
m2
v1 f  v1i 
v2 f
m1
Given
m1  2.5 102 kg
m2  2.0 102 kg
v1i  2.3
v2i  0
m
s
m
s
Re-arranging the
Momentum equation with
v2i=0, we obtain:
m 2.0 102 kg
v1 f  2.3 
v2 f
2
s 2.5 10 kg
m
 2.3  0.80v2 f
s
Substitute this into the K.E. equation:
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
m

2
2.5

10
kg
2.3

  s    2.5 102 kg  v12f   2.0 102 kg  v22 f
Substitute in:
m2
13.2kg 2   2.5kg  v12f   2.0kg  v22 f
s
m
v1 f  2.3  0.80v2 f
s
2
m2
m


13.2kg 2   2.5kg   v1 f  2.3  0.80v2 f    2.0kg  v22 f
s
s


0  9.2kg
Note: Both
balls end up
moving in the
same direction.
v2 f
m
 0.80v2 f
s
m
m

 2.3  0.80  2.6 
s
s

m
 0.26
s
v1 f  2.3
m
0
s
v2 f  2.6
m
v2 f  3.6kgv22 f
s
m
s
Solution by CM
A ball with mass 2.5x10-2 kg moving at 2.3 m/s collides with a stationary
2.0x10-2 kg ball. If the collision is elastic, determine the velocity of each ball
after the collision.
VCM 
Given
m1  2.5 102 kg
 1.2778
m2  2.0 102 kg
v1i  2.3
v2i  0
m
s
m
s
1

m

 m
2
2
2.3
2.5

10
kg

0
2.0

10
kg










2.5 102 kg  2.0 102 kg  
s
 s

m
s
m
m
V  2.3  1.2778
s
s
m
 1.0222
s
*
1i
m
m
 1.2778
s
s
m
 1.2778
s
V2*i  0
m
V  1.0222
s
V2*f  1.2778
m
s
m
m
V1i  1.0222  1.2778
s
s
m
 0.26
s
V2 f  1.2778
m
m
 1.2778
s
s
*
1f
 2.6
m
s