Physics 111: Lecture 15, Pg 1
Physics 111: Lecture 15
Today’s Agenda
Elastic collisions in one dimension
Center of mass reference frame Colliding carts problem
Some interesting properties of elastic collisionsKiller bouncing balls
Physics 111: Lecture 15, Pg 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.
F PEXT
ddt
ddtP 0 FEXT 0
Physics 111: Lecture 15, Pg 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.
Physics 111: Lecture 15, Pg 4
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
Physics 111: Lecture 15, Pg 5
Lecture 15, Act 1Solution
No external forces in the x direction, so PX is constant.
mvK I v
m m
2v2mPF m m v / 2
x
Physics 111: Lecture 15, Pg 6
Lecture 15, Act 1Solution
Compute kinetic energies:
v m m
m m v / 2
2I mv
21K
I
2
K21
2vm2
21
FK
2KK
F
I
Physics 111: Lecture 15, Pg 7
Lecture 15, Act 1Another solution
We can write
m m
m m
K 12
mv 2 P2m
2
P is the same before and after the collision. The mass of the moving object has doubled, hence the
kinetic energy must be half.
2KK
F
I
Physics 111: Lecture 15, Pg 8
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?
Physics 111: Lecture 15, Pg 9
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
Physics 111: Lecture 15, Pg 10
Elastic Collisions
Elastic means that kinetic energy is conserved as well as momentum.
This gives us more constraintsWe can solve more complicated problems!!Billiards (2-D collision)The colliding objects
have separate motionsafter the collision as well as before.
Start with a simpler 1-D problem
Initial Final
Physics 111: Lecture 15, Pg 11
Elastic Collision in 1-D
v1,i v2,i
initial
x
m1m2
v1,fv2,f
finalm1
m2
Physics 111: Lecture 15, Pg 12
Elastic Collision in 1-D
v1,i v2,i
v1,fv2,f
before
after
x
m1 m2
Conserve PX:
m1v1,i + m2v2,i = m1v1,f + m2v2,f
Conserve Kinetic Energy:
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!
Physics 111: Lecture 15, Pg 13
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
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
Airtrack
Collisionballs
Physics 111: Lecture 15, Pg 14
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.
Physics 111: Lecture 15, Pg 15
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
Physics 111: Lecture 15, Pg 16
Lecture 15, Act 2Conceptual Solution
The external force in the x direction is zero (frictionless):The CM of the systems can’t move!
x
X
X
X
X
CM CM
Physics 111: Lecture 15, Pg 17
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.
The lighter man moves faster with respect to the ice!
X
X
X
X
CM CM
Physics 111: Lecture 15, Pg 18
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
Physics 111: Lecture 15, Pg 19
Lecture 15, Act 2Algebraic Solution
The speed 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)
Plugging vP = V - vR into thiswe find:
v V Mm MR
So vR is greater if m is smaller.
Physics 111: Lecture 15, Pg 20
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?
m1m2 v1,i v2,i = 0
xm1
VCM
m2
m1v1,f v2,fm2
+ = CM
Videoof
CM frame
Physics 111: Lecture 15, Pg 21
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
1
1 2m m
mm m
1
1 2
(for v2,i = 0 only)
Physics 111: Lecture 15, Pg 22
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)
VCM
v
v*
Physics 111: Lecture 15, Pg 23
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
Physics 111: Lecture 15, Pg 24
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
Physics 111: Lecture 15, Pg 25
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:
2f,2
22
2
2f,1
21
1
2i,2
22
2
2i,1
21
1
*vmm21*vm
m21*vm
m21*vm
m21
2f,1
2i,1 *v*v
2i,222
i,11 *vm*vm
2f,1
21
21
2i,1
21
21
*vmm21
m21*vm
m21
m21
(and the same for particle 2)
Therefore, in 1-D: v*1,f = -v* 1,i v*2,f = -v*2,i
Physics 111: Lecture 15, Pg 26
Example 1...
v*1,i v*2,i
v*1,f = - v*1,i = -1.2m/s
x
v*2,f = - v*2,i =.3 m/s
m1m2
m1
m1
m2
m2
v*1,f = -v* 1,i v*2,f = -v*2,i
Physics 111: Lecture 15, Pg 27
Example 1...
So 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!!
Physics 111: Lecture 15, Pg 28
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
Physics 111: Lecture 15, Pg 29
Lecture 15, Act 3 Moving Between Reference Frames
The velocity of the CM is:
x
V m mm
hrCM 80 402
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 !!
Physics 111: Lecture 15, Pg 30
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
Physics 111: Lecture 15, Pg 31
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
Physics 111: Lecture 15, Pg 32
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)
Physics 111: Lecture 15, Pg 33
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!
Physics 111: Lecture 15, Pg 34
Basketball Demo. 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 >> m and all bounces are elastic).Understand this using the “speed of approach = speed of
recession” property we just proved.
v
v
v
v
3v
v
(a) (b) (c)
m
M
Drop 2 balls
Physics 111: Lecture 15, Pg 35
More comments on energy
Consider the total kinetic energy of the system in the lab reference frame:
E m v m vLAB 12
121 1
22 2
2 but
so
(same for v2)
= KREL = KCM= PNET,CM = 0
CM2
CM1
VvVv v*1
v*2
1CM21
2CM11
21 2Vv Vvv v* v*
2211CM2
CM21222
211LAB mmVmm
21m
21m
21E Vv* v* v* v*
Physics 111: Lecture 15, Pg 36
More comments on energy...
Consider the total kinetic energy of the system in the LAB reference frame:
= KREL = KCM
So ELAB = KREL + KCM
KREL is the kinetic energy due to “relative” motion in the CM frame.
KCM is the kinetic energy of the center of mass.
This is true in general, not just in 1-D
2CM21
222
211LAB Vmm
21m
21m
21E v* v*
Physics 111: Lecture 15, Pg 37
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).
Physics 111: Lecture 15, Pg 38
Recap of today’s 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