sph4u: lecture 15 today’s agenda l elastic collisions in one dimension analytic solution (this is...
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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
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.
FP
EXTddt
ddtP 0 FEXT 0
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 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
2
Lecture 15, Act 1Solution
No external forces in the x direction, so PX is constant.
v m m
2
v2mPF m m v / 2
x
mvPI
Lecture 15, Act 1Solution
Compute kinetic energies:
v m m
m m v / 2
2
Imv
2
1K
I
2
K2
1
2
vm2
2
1
FK
2I
F
K
K
Lecture 15, Act 1Another solution
We can write
m m
m m
K12
mv2 P2m
2
P is the same before and after the collision.
The mass of the moving object has doubled, hence thekinetic energy must be half.
2I
F
K
K
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?
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 motionsafter the collision as well as before.
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
Initial Final
Elastic Collision in 1-Dwhat has to happen
v1,i v2,i
initial
x
m1m2
v1,fv2,f
finalm1
m2
Why is this elastic?
Kinetic energy potential energy kinetic energyThe spring is conservative
Maybe, it depends…
Elastic Collision in 1-Dthe spring is conservative
x
Conserve PX: (no external forces!)
m1v1,i + m2v2,i = m1v1,f + m2v2,f
Conserve Kinetic Energy: (it’s elastic!)
1/2 m1v21,i + 1/2 m2v2
2,i = 1/2 m1v21,f + 1/2 m2v2
2,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!
v1,i v2,i
before
m1 m2
v1,fv2,f
after
Elastic Collision in 1-D
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
First, describe the solution to the problem using algebra. Useful analysis and useful formulae
m1v1,i + m2v2,i = m1v1,f + m2v2,f
1/2 m1v21,i + 1/2 m2v2
2,i = 1/2 m1v21,f + 1/2 m2v2
2,f
momentum
energy
Elastic Collision in 1-Dspecial case: equal masses
If the masses of the two objects are equal the algebra is not too bad. Let’s see what we get…
Divide through by m = m1 = m2
m1v1,i + m2v2,i = m1v1,f + m2v2,f
1/2 m1v21,i + 1/2 m2v2
2,i = 1/2 m1v21,f + 1/2 m2v2
2,f
momentum
energy
v1,i + v2,i = v1,f + v2,f
v21,i + v2
2,i = v21,f + v2
2,f
Elastic Collision in 1-Dspecial 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
Divide through energy equation by momentum equation which gives
Particles just trade velocities in 1-D elastic collision of equal mass objects (let equation talk to you…)
momentum
energy
v1,i + v1,f = v2,f + v2,i
v1,i - v1,f = v2,f - v2,i
v21,i - v2
1,f = v22,f - v2
2,i
(v1,i - v1,f )(v1,i + v1,f ) = (v2,f - v2,i )(v2,f + v2,i )
v2,f = v1,i v1,f = v2,i v1,i - v1,f = v2,f - v2,i
Elastic Collision in 1-Dgeneral case: unequal masses
Conserve linear momentum and mechanical energy, but now the masses are different:
Divide through energy equation by momentum equation which gives
Now solving these two linear equations is only a bit more complicated
momentum
energy
v1,i + v1,f = v2,f + v2,i
m1(v1,i - v1,f ) = m2(v2,f - v2,I )
m1(v21,i - v2
1,f ) = m2(v22,f - v2
2,I )
m1(v1,i - v1,f )(v1,i + v1,f ) =m2 (v2,f - v2,i )(v2,f + v2,i )
m1(v1,i - v1,f ) = m2(v2,f - v2,I )
Elastic Collision in 1-Dgeneral case: unequal masses
Algebra just gave us the following equations based on conservation of momentum and mechanical energy:
Now just solve for final velocities, v1,f and v2,f in terms of v1,i and v2,i
2 1 21, 2, 1,
1 2 1 2
2f i i
m m mv v v
m m m m
v1,i + v1,f = v2,f + v2,i
m1(v1,i - v1,f ) = m2(v2,f - v2,I )
1 2 12, 1, 2,
1 2 1 2
2f i i
m m mv v v
m m m m
v2,f = v1,i v1,f = v2,i
When m1 = m2
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 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 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 2Conceptual 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
X
CM CM
Lecture 15, Act 2Conceptual 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 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
Let the mass of the runner be m and the plank be M.
m
M
Let the speed of the runner and the plank with respect to the ice be vR and vP respectively.
vR
vP
Lecture 15, Act 2Algebraic Solution
The velocity of the runner with respect to the plank is V = vR - vP (same for both runners).
x
m
M
vR
vP
MvP = - mvR (momentum conservation, it’s zero!)
Plugging vP = vR - V into thiswe find:
v VM
m MR
So vR is greater if m is smaller.
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?
x
m1m2 v1,i v2,i = 0
VCM
m1m2
m1v1,f v2,fm2
+ = CM
Example 1...
Four step procedure
First figure out the velocity of the CM, VCM.
» VCM = (m1v1,i + m2v2,i), but v2,i = 0 in this case so
VCM = v1,i
So VCM = 1/5 (1.5 m/s) = 0.3 m/s
1
1 2m m
m
m m1
1 2
(for v2,i = 0 only)
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)
VCM
v
v*
This is the “lab” frame velocity
This is the CM frame velocityIf 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
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
Example 1 continued...
Now consider the collision viewed from a frame moving with the CM velocity VCM. ( jargon: “in the CM frame”)
m1m2 v*1,i v*2,i
xm2 m1
m1v*1,f v*2,f
m2
Movie
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…)
But the total momentum is zero, both initial and final:
So:
2f,2
22
2
2f,1
21
1
2i,2
22
2
2i,1
21
1
*vmm21
*vmm21
*vmm21
*vmm21
2f,1
21
21
2i,1
21
21
*vmm21
m21
*vmm21
m21
2f,1
2i,1 *v*v (and the same for particle 2)
Therefore, in 1-D: v*1,f = -v* 1,i v*2,f = -v*2,i
2i,22
2
i,11 *vm*vm Likewise for final v’s
Example 1...
v*1,i v*2,i
x
m1m2
m1m2
v*1,f = - v*1,i = -1.2m/s v*2,f = - v*2,i =.3 m/s
m1m2
v*1,f = -v* 1,i v*2,f = -v*2,i
Calculate the final velocities in the CM frame:
Example 1...
Now we can calculate the final velocities in the lab reference frame, using:
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
v = v* + VCM
v* = v - VCM
Four easy steps! No need to solve a quadratic equation!!Especially important in 2D
v*f = -v*,i
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
Lecture 15, Act 3 Moving Between Reference Frames
The velocity of the CM is:
x
Vm m
mhrCM 80 40
2 mi /
= 20 mi / hr
20mi/hr
CM
80mi/hr - 40mi/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 !!
1 1 2 21 2
1m v m v
m m
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?
Lecture 15, Act 3 Aside
20mi/hr
CM
80mi/hr - 40mi/hr
x
Lecture 15, Act 3 Aside Solution
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
v*GREEN,i = 60 mi/hr
v*RED,i = -60 mi/hr
Summary: Using CM Reference Frame
: Determine velocity of CM
: Calculate initial velocities in CM reference frame
: Determine final velocities in CMreference frame
: Calculate final velocities in lab reference frame
VCM =
21 mm
v* = v - VCM
v*f = -v*i
v = v* + VCM
(m1v1,i + m2v2,i)
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.
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
v*1,i v*2,i
v*1,f = -v*1,iv*2,f = -v*2,i
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.
1 21 1f i
T
m mv v
m
12 1
2f i
T
mv v
m
1 2 1 2 2 1i i f f ffv v v v v v
If m2 is initially at rest, then you can use: conservation of energy and conservation of momentum, or:
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.
2 2 2 21 1 2 2 1 1 2 2
1 1 1 1
2 2 2 2ii f fm v m v m v m v
1 1 2 2 1 1 2 2ii f fm v m v m v m v
K.E.
Momentum
21
22
1
2
2.5 10
2.0 10
2.3
0
i
i
m kg
m kg
mv
sm
vs
21 1 2
1f i f
mv v v
m
2
1 22
2
2.0 102.3
2.5 10
2.3 0.80
f f
f
m kgv v
s kg
mv
s
Given
Re-arranging the Momentum equation with v2i=0, we obtain:
Substitute this into the K.E. equation:
1 22.3 0.80f f
mv v
s
2 2 2 2 21 22.5 10 2.3 2.5 10 2.0 10f f
mkg kg v kg v
s
2 2 2 21 1 2 2 1 1 2 2
1 1 1 1
2 2 2 2ii f fm v m v m v m v
2
2 21 22
13.2 2.5 2.0f f
mkg kg v kg v
s
Substitute in:
22
21 2 22
13.2 2.5 2.3 0.80 2.0f f f
m mkg kg v v kg v
s s
22 20 9.2 3.6f f
mkg v kgv
s
2
2
0
2.6
f
f
m
sm
s
v
v
1 22.3 0.80
2.3 0.80 2.6
0.26
f f
m
s
mv v
sm
s
m
s
Note: Both balls end up moving in the same direction.
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.
21
22
1
2
2.5 10
2.0 10
2.3
0
i
i
m kg
m kg
mv
sm
vs
Given 2 22 2
12.3 2.5 10 0 2.0 10
2.5 10 2.0 10
1.2778
CM
m mkg kg
kg kg s s
m
V
s
*1 2.3 1.2778
1.0222
i
m m
s s
s
V
m
*2 0 1.2778
1.2778
i
m m
s s
s
V
m
*1 1.0222fV
m
s
*2 1.2778fV
m
s
1 1.0222 1.2778
0.26
i
m mV
s sm
s
2 1.2778 1.2778
2.6
f
m m
sm
s
Vs