Midterm Exam #2
§  Midterm is Wednesday, October 21 (80 minutes!!!)
Ø  Units 6-10: Friction – Center of Mass
Ø  Things to bring: Pencil(s), calculator UofU ID card
Ø  Bring your U of U ID card: you will not receive a grade unless you
show it upon turning in your exam
Ø  No makeups!!! (Unless you are unavailable due to University/military
business or documented medical emergency – come talk to me!)
Ø  Cheating: You will receive a zero and referred to the University
administration – may result in an “E” for the course and expulsion
§  Exam Location: SFEBB 1110 (large lecture hall on first floor of
Spencer Fox Eccles Business Building)
Ø  Same location as Exam #1
Mechanics Lecture 12, Slide 1
Post-Exam Review
§  Instead of discussion sessions on W/TH, you will do a post-exam
review:
Ø  Will get into groups and write solutions for one of the exam
problems together
Ø  Each group that turns in a well-formatted solution will receive 4
bonus points (does not need to be correct)
Ø  We may post some of your solutions on Canvas
§  Sections 031 and 032 (M/W discussions) will meet in CRCC 215
immediately after the exam instead of normal discussion rooms
§  Section 033 will meet in normal discussion room on TH
Mechanics Lecture 12, Slide 2
Announcements
§  Monday/Tuesday Discussion sessions:
Ø  Review practice exam
Ø  Better if you take practice exam under exam conditions before you
look at solutions.
§  Extra Help Lab hours week:
Ø  Monday: 10A - 11:30A (Jon Miller)
Ø  Tuesday: 11A - noon (Jordan); 3P - 4:30P (Jon)
Ø  Wednesday: 10A - 11:30A (Steve); 11:30A - 1P (Jordan)
Ø  Thursday: no hours
Ø  Friday: normal hours (don’t forget about homework on Sunday)
Mechanics Lecture 12, Slide 3
Lecture 12: Elastic Collisions
Today’s Concepts:
a) Elastic Collisions
b) Center-of-Mass Reference Frame
Mechanics Lecture 12, Slide 4
Unit 12 Learning Objectives
If you master this unit, you should:
§  be able to determine whether a collision is elastic or inelastic
and understand the differences.
§  Be able to “transform” between the “CM” and “laboratory”
reference frames: mathematically relate motion in one
reference frame to motion in the other.
§  Be able to “solve” inelastic and elastic 2-body collisions in 1D
or 2D: find final velocities (includes direction) given initial
velocities, masses, etc.
Mechanics Lecture 12, Slide 5
Center of Mass & Collisions so far:
!
!
Fnet ,ext = M tot Acm
!
dPtot
=
dt
The CM behaves just like a point par<cle !
If F
net
, ext
=
0 then momentum is constant (usually the case during a collision) 

dPtot
then = 0 ⇒ Ptot = const.
dt
!
!
Ptot = M totVcm
If you are in a reference frame moving along with the CM then the total momentum you measure is 0. Mechanics Lecture 12, Slide 6
CheckPoint: Identical Box Collision
A box sliding on a fric<onless surface collides and s<cks to a second iden<cal box which is ini<ally at rest. Compare the ini<al and final kine<c energies of the system. A) Kinitial > Kfinal
B) Kinitial = Kfinal
initial
C) Kinitial < Kfinal
final
About 60% of you got this correct the first time.
Mechanics Lecture 12, Slide 7
CheckPoint: Discuss & Re-vote
A) Kini*al > Kfinal B) Kini*al = Kfinal A
B
C
D
initial
C) Kini*al < Kfinal final
A) Due to this being an inellas<c collision the kine<c energy will be lost as work will be done on deforma<on of the two objects. B) The kine<c energy would be equal before and aGer the collision. The ini<al K = 1/2mv2 and the final K = 1/2*2m(v/2)2. These equa<ons are equal. C) Set up equa<on mvi = 2mvf and find that vf = vi/2 Mechanics Lecture 12, Slide 8
Momentum & Kinetic Energy
1 2
K = mv
2
2 2
mv
=
2m
2
p
=
2m
since !
!
p = mv
This is oGen a handy way to figure out the kine<c energy before and aGer a collision since Ptot is constant if Fnet,ext = 0. initial
K tot,initial
K tot, final
final
Ptot2
=
2M
Ptot2
=
2(2M )
These are the same! Mechanics Lecture 12, Slide 9
Review: Inelastic Collisions
Inelas<c collisions: Momentum is constant, but mechanical energy is not. ~CM
F~net,ext = 0 =) P~tot = Mtot V
E = WN C,net 6= 0 =) Ef inal 6= Einitial
Mechanics Lecture 12, Slide 10
CheckPoint: Unknown Box Collision
A green block of mass m slides to the right on a fric<onless floor and collides elas<cally with a red block of mass M which is ini<ally at rest. AGer the collision the green block is at rest and the red block is moving to the right. How does M compare to m? A) m > M B) m = M C) m < M M
m
m
Before Collision M
AGer Collision Mechanics Lecture 12, Slide 11
Elastic collisions: momentum & energy
constant
M
m
M
m
A) m > M
B) m = M
Before Collision
After Collision
C) m < M
A) If the first block does not have less mass than the second block than the first block would keep moving with second block when it hit it instead of coming to rest. B) If m were greater than M it would con<nue moving, if m were less than M it would bounce back. C) if mass m had equal or greater mass it would move with mass M instead of stopping Mechanics Lecture 12, Slide 12
Review: Inelastic vs. Elastic Collisions
Inelas<c collisions: Momentum is constant, but mechanical energy is not. ~CM
F~net,ext = 0 =) P~tot = Mtot V
E = WN C,net 6= 0 =) Ef inal 6= Einitial
Elas<c collisions: Momentum and mechanical energy are both constant. ~CM
F~net,ext = 0 =) P~tot = Mtot V
E = WN C,net = 0 =) Ef inal = Einitial
Mechanics Lecture 12, Slide 13
Newton’s Cradle
A great example of elas<c collisions: The only way to understand this is for both P and E to be constant (just before and just aGer collision) Mechanics Lecture 12, Slide 14
Solving 2-body Collision Problems
1.  There are 6 variables: m1, m2, v1i, v2i, v1f, v2f Ø  If we know 4 of the 6, then we need 2 equa<ons to solve for remaining 2 unknowns Ø  One equa<on comes from conserva<on of momentum: P~tot,i = P~tot,f =) m1~v1i + m2~v2i = m1~v1f + m2~v2f
2.  If collision is inelas<c: Eqn. 1
Ø  Then Ktot before and aGer are not the same, but… Ø  OGen v1 = v2 aGer collision (if they s<ck together) Eqn. 2
3.  If collision is elas<c: Ø  Then Ktot before and aGer are the same: Eqn. 2
1
1
1
1
2
2
2
2
m1 v1i + m1 v2i = m1 v1f + m1 v2f
2
2
2
2
Mechanics Lecture 12, Slide 15
The difficulty with elastic collisions…
Conserva<on of momentum: m1 v1i + m2 v2i = m1 v1f + m2 v2f
=) v2f =
m1 (v1i
v1f ) + m2 v2i
m2
Conserva<on of energy: 2
2
2
2
m1 v1i + m2 v2i = m1 v1f
+ m2 v2f

m1 (v1i
2
= m1 v1f + m2
v1f ) + m2 v2i
m2
2
YIKES!!! Can be tedious!!! Use Center of Mass frame! Mechanics Lecture 12, Slide 16
Center-of-Mass Frame
!
VCM
!
!
!
m1v1 + m2 v2 + ...
Ptot
=
=
M tot
m1 + m2 + ...
In the CM reference frame, VCM = 0 In the CM reference frame, Ptot = 0 Mechanics Lecture 12, Slide 17
Center of Mass Frame & Elastic Collisions
The speed of each object is the same before and aGer an elas<c collision as viewed in the CM frame: Before:
v*1,i
m1
m2
v*2,i
m2
v*2, f = -v*2,i
v*1 = v*2 = 0
During:
m1
m2
After:
v*1, f = -v*1,i m1
Mechanics Lecture 12, Slide 18
Example: Using CM Reference Frame
A glider of mass m1 = 0.2 kg slides on a fric<onless track with ini<al velocity v1i = 1.5 m/s. It hits a sta<onary glider of mass m2 = 0.8 kg. A spring abached to the first glider compresses and relaxes during the collision, but there is no fric<on (i.e., energy is conserved). What are the final veloci<es? v1i
+
m1
m2 v2i = 0
vCM
+ : CM
m1
v1f
m1
x
+ m2
+
m2
v2f
Mechanics Lecture 12, Slide 19
Example
Four step procedure: Step
1:
First figure out the velocity of the CM, VCM. ⎛ 1 ⎞
VCM = ⎜⎜ ⎟⎟ (m1v1,i + m2v2,i), but v2,i = 0 so ⎝ m1 + m2 ⎠
⎛ m1 ⎞
VCM = ⎜⎜ ⎟⎟ v1,i (for v2,i = 0 only) ⎝ m1 + m2 ⎠
So VCM = 1/5 (1.5 m/s) = 0.3 m/s Mechanics Lecture 12, Slide 20
Example
Now consider the collision viewed from a frame moving with the CM velocity VCM. v*1i
m1
m1
v*1,f
m1
m2
v*2i
m2
v*2,f
m2
Mechanics Lecture 12, Slide 21
Example
Step 2: Calculate the ini<al veloci<es in the CM reference frame (all veloci<es are in the x direc<on): v
v = v* +VCM
VCM
v* = v - VCM
v*
v*1i = v1i – VCM = 1.5 m/s - 0.3 m/s = 1.2 m/s
v*2i = v2i - VCM = 0 m/s - 0.3 m/s = -0.3 m/s
v*1i = 1.2 m/s
v*2i = -0.3 m/s
Mechanics Lecture 12, Slide 22
Example
Step 3: Use the fact that the speed of each block is the same before and aGer the collision in the CM frame. v*1f = -v*1i
v*1i
m1
m1
v*1f = -v*1i = -1.2 m/s
v*1f
m1
v*2f = -v*2i
m2
v*2i
x
m2
v*2f = -v*2i = 0.3 m/s
m2
v*2f
Mechanics Lecture 12, Slide 23
Example
Step 4: Calculate the final veloci<es back in the lab reference frame: v
v = v* +VCM
VCM
v*
v1f = v*1f + VCM = -1.2 m/s + 0.3 m/s = -0.9 m/s
v2f = v*2f + VCM = 0.3 m/s + 0.3 m/s = 0.6 m/s
v1f = -0.9 m/s
v2f = 0.6 m/s
Four easy steps! No need to solve a quadra<c equa<on! Mechanics Lecture 12, Slide 24
Review: Elastic Collisions in the CM Frame
Step 1: Figure out the velocity of the CM: ✓
◆
m1
VCM =
v1i
for v 2i
=
0 only! m1 + m2
Step 2: Find the ini<al veloci<es of each object in the CM frame: ⇤
⇤
v2i
= v2i VCM
1D: v1i = v1i VCM
⇤
⇤
~CM
~CM
~v2i
= ~v2i V
3D: ~v1i = ~v1i V
Step 3: The speed of each block is the same before and aGer the collision in the CM frame: ⇤
⇤
⇤
⇤
v1f
= v1i
v2f
= v2i
Step 4: Transform back to the Lab frame: ⇤
⇤
~CM
~CM
~v1f = ~v1f
+V
~v2f = ~v2f
+V
Mechanics Lecture 12, Slide 25
CheckPoint: Double Mass Half Speed 1
Two blocks on a horizontal fric<onless track head toward each other as shown. One block has twice the mass and half the velocity of the other. The velocity of the center of mass of this system before the collision is. A) Toward the leG B) Toward the right C) zero 2v
v
m
2m
Before Collision
Mechanics Lecture 12, Slide 26
CheckPoint Results
The velocity of the center of mass of this system before the collision is A) Toward the leG B) Toward the right C) zero v
2v
2m
m
P
j
VCM = P
mj vj
j
mj
This is the CM frame! m(2v) + 2m( v)
=
=0
3m
Mechanics Lecture 12, Slide 27
CheckPoint: Double Mass Half Speed 2
Two blocks on a horizontal fric<onless track head toward each other as shown. One block has twice the mass and half the velocity of the other. Suppose the blocks collide elas<cally. Picking the posi<ve direc<on to the right, what is the velocity of the bigger block aGer the collision takes place? A) +2v B) +v C) 0 D) -­‐v E) -­‐2v This is the CM frame
2v
v
m
2m
+
Before Collision
Mechanics Lecture 12, Slide 28
CheckPoint Results: Double Mass Half Speed 2
Suppose the blocks collide elas<cally. Picking the posi<ve direc<on to the right, what is the velocity of the bigger block aGer the collision takes place? A) +2v B) +v C) 0 D) -­‐v E) -­‐2v 2v
m
v
2m
+
Before Collision
This is the CM frame
Mechanics Lecture 12, Slide 29