1. Section 9-4: Conservation of Energy & Momentum in Collisions

2. HOLDS for ALL collisions!
Given some information, using conservation laws, we can determine A LOT about collisions without knowing the collision forces!
To analyze ALL collisions use:
Rule # 1
Momentum is ALWAYS (!!!) conserved in a collision!
 mAvA + mBvB = mA(vA) + mB(vB)
HOLDS for ALL collisions!

3.  mAvA + mBvB = mA(vA) + mB(vB)
Note!!!
A VERY Special (idealized!) Case:
2 very hard objects (~billiard balls) collide ( “Elastic Collision”)
To analyze Elastic Collisions:
Rule # 1 Still holds, of course!
 mAvA + mBvB = mA(vA) + mB(vB)
Rule # 2 Holds For Elastic Collisions ONLY (!!)
The total kinetic energy (K) is conserved!!
(K)before = (K)after
 (½)mA(vA)2 + (½) mB(vB)2
= (½)mA(vA)2 + (½)mB(vB)2

4. ELASTIC COLLISIONS ONLY!!
Total Kinetic energy (K) is conserved for
ELASTIC COLLISIONS ONLY!!
Inelastic Collisions
 Collisions which are NOT elastic.
Is Kinetic Energy conserved in Inelastic Collisions? NO!!!!
Is momentum conserved in Inelastic Collisions? YES!!!!! by Rule # 1: Momentum is ALWAYS conserved in a collision!

5. 2 masses colliding elastically:
Special case: Head-on Elastic Collisions
Can analyze in 1 dimension
Possible types of head-on collisions 
2 masses colliding elastically:
We know the masses & the initial speeds. Both momentum & kinetic energy are conserved, so we have 2 equations. Doing algebra, we can solve for the 2 unknown final speeds.

6. 1 dimensional collisions: Some possible types:
Special case: Head-on Elastic Collisions.
1 dimensional collisions: Some possible types:
before
collision  or 
after
collision or
Note that vA, vB, (vA), (vB) are 1 dimensional vectors!

7. Sect. 9-5: Elastic Collisions in 1 Dimension
Special Case: Head-on Elastic Collisions.
Momentum is conserved (ALWAYS!)
Pbefore = Pafter or mAvA + mBvB = mAvA + mBvB
vA, vB, vA, vB are one dimensional vectors!
Kinetic Energy is conserved (ELASTIC!)
(K)before = (K)after
(½)mA(vA)2 + (½)mB(vB)2
= (½)mA(vA)2 + (½)mB(vB)2
2 equations, 6 quantities: vA,vB,vA, vB, mA, mB
 Clearly, we must be given 4 out of 6 to solve problems! Solve with CAREFUL algebra!!

8. Elastic head-on (1d) collisions only!!
mAvA + mBvB = mAvA + mBvB (1)
(½)mA(vA)2 + (½)mB(vB)2 = (½)mA(vA)2 + (½)mB(vB)2 (2)
Now, some algebra with (1) & (2), the results of
which will help to simplify problem solving:
Rewrite (1) as: mA(vA - vA) = mB(vB - vB) (a)
Rewrite (2) as: mA[(vA)2 - (vA)2] = mB[(vB)2 - (vB)2] (b)
Divide (b) by (a):
 vA + vA = vB + vB or
vA - vB = vB  - vA  = - (vA - vB) (3)
Relative velocity before = - Relative velocity after
Elastic head-on (1d) collisions only!!

9. vA - vB = vB - vA = - (vA - vB) (3)
Summary: 1d Elastic Collisions:
Rather than directly use momentum conservation + K conservation, it is often convenient to use:
Momentum conservation
mAvA + mBvB = mAvA + mBvB (1)
along with
vA - vB = vB - vA = - (vA - vB) (3)
(1) & (3) are equivalent to momentum conservation + Kinetic Energy conservation, since (3) was derived from these conservation laws! 
use
these!

10. Example 9-7: Pool (Billiards)
Ball A
Ball B
mA = mB = m, vA = v, vB = 0, vA = ?, vB = ?
Momentum Conservation: mv +m(0) = mvA + mvB
Masses cancel  v = vA + vB (I)
Relative velocity results for elastic head on collisions:
v - 0 = vB - vA (II)
Solve (I) & (II) simultaneously for vA & vB :
 vA = 0, vB = v
Ball A comes to rest. Ball B moves with original velocity of ball A.
Before:
 v
v = 0
Ball B
Ball A
After:
v = 0
 v

11. Example 9-8: Unequal masses, target at rest
A common practical situation is for a moving object (mA) to strike a second object (mB, the “target”) at rest (vB = 0).
Assume that mA & mB, & are unequal, that the collision is elastic & that it occurs along a line (head-on).
a. Derive equations for vB & vA in terms of the initial velocity vA of mass mA & the masses mA and mB.
b. Calculate vB & vA if mA is very large compared to mB (mA >> mB).
c. Calculate vB & vA if mA is very small compared to mB (mA << mB).
Solution: a. Both momentum and kinetic energy are conserved. The rest is algebra.
b. In this case, vA′ doesn’t change much, and vB′ = 2 vA.
c. In this case, vB’ remains zero, and mass A reverses its direction at the same speed.

12. Example 9-9: Nuclear Collision
A proton (p) of mass mp = 1.01 u (atomic mass units), traveling with speed vp = 3.60  104 m/s has an elastic head-on collision with a helium (He) nucleus (mHe = 4.00 u) initially at rest. Calculate the velocities of the proton & He nucleus after the collision.

13. Section 9-6: Inelastic Collisions
 Collisions which
Do NOT Conserve Kinetic Energy!
Some initial kinetic energy is lost to thermal or potential energy. Kinetic energy may also be gained in explosions (there is addition of chemical or nuclear energy).
A Completely Inelastic Collision
is one in which the objects
stick together afterward
so there is only one final velocity.

14. To analyze ALL collisions: Rule # 1:  m1v1 + m2v2 = m1v1 + m2v2
Given some information, using conservation laws, we can determine a LOT about collisions without knowing forces of collision.
To analyze ALL collisions:
Rule # 1:
Momentum is ALWAYS (!!!) conserved in a collision!
 m1v1 + m2v2 = m1v1 + m2v2
HOLDS for ALL collisions!

15. Total Kinetic energy (KE) is conserved for ELASTIC COLLISIONS ONLY!!
Inelastic Collisions  Collisions which are NOT elastic.
Is KE conserved for Inelastic Collisions? NO!!!!
Is momentum conserved for Inelastic Collisions?
YES!! (Rule # 1: Momentum is ALWAYS conserved in a collision!).
Special Case: Completely Inelastic Collisions  Inelastic collisions in which the 2 objects collide & stick together.
KE IS NOT CONSERVED FOR THESE!!

16. Example 9-10: Railroad cars again
Same rail cars as Ex Car A, mass mA = 10,000 kg, traveling at speed vA = 24 m/s strikes car B (same mass), initially at rest (vB = 0). Cars lock together after collision. Ex. 9-3: Find speed v after collision.
Before
Collision
After
Collision
Figure 9-5.
Solution: Momentum is conserved; after the collision the cars have the same momentum. Therefore their common speed is 12.0 m/s.
Ex. 9-3 Solution: vA = 0, (vA) = (vB) = v Use Momentum Conservation:
mAvA+mBvB = (mA + m2B)v  v = [(mAvA)/(mA + mB)] = 12 m/s
Ex. 9-10: Cars lock together after collision. Find amount of initial KE transformed to thermal or other energy forms:
Initially: Ki = (½)mA(vA)2 = 2.88  106 J
Finally: Kf = (½)(mA+ mB)(v)2 = 1.44  106 J! (50% loss!)

17. Example 9-11: Ballistic pendulum
The ballistic pendulum is a device used to measure speeds of projectiles, such as a bullet.
A projectile, mass m, is fired into a large block, mass M, which is suspended like a pendulum. After the collision, pendulum & projectile swing up to a maximum height h.
Find the relation between the initial horizontal speed of the projectile, v & the maximum height h.
Figure 9-16.
Solution: This has two parts. First, there is the inelastic collision between the bullet and the block; we need to find the speed of the block. Then, the bullet+block combination rises to some maximum height; here we can use conservation of mechanical energy to find the height, which depends on the speed.

18. Ex. 9-11 & Probs. 42 & 43 (Inelastic Collisions)
ℓ - h 
Before
After a ℓ ℓ
v = 0 a a a a a a
Momentum Conservation mv = (m + M)v´
Mechanical Energy (½)(m +M)(v´)2 = (m + M)gh
Conservation
 v = [1 +(M/m)](2gh)½

19. Problem of the bullet it before hits the block. Multi-step problem!
A bullet, m = kg hits
& is embedded in a block,
M = 1.35 kg. Friction
coefficient between block &
surface: μk = Moves d = 9.5 m before stopping. Find v
of the bullet it before hits the block. Multi-step problem!
1. Find V using Work-Energy Principle with friction.
2. Find v using momentum conservation. But, to find V, first we need to
3. Find the frictional force!
Ffr = μkFN = μk(M+m)g

20. 1. Friction force:
Ffr = μkFN = μk(M+m)g
2. The Work- Energy
Principle: Wfr = -Ffrd
= K = 0 – (½)(M+m)V2 OR: -Ffrd = - (½)(M+m)V2
μk(M+m)gd = (½)(M+m)V2 (masses cancel!)
Stops in distance d = 9.5 m
 V = 6.82 m/s
3. Momentum conservation:
mv + 0 = (M+m) V
 v = (M+m)V/m = 375 m/s (bullet speed)

21. Summary: Collisions Basic Physical Principles:
Conservation of Momentum: Rule # 1: Momentum is ALWAYS conserved in a collision!
Conservation of Kinetic Energy:
Rule # 2: K is conserved for elastic collisions ONLY !!
Combine Rules #1 & #2 & get relative velocity before = - relative velocity after.
As intermediate step, might use Conservation of Mechanical Energy (K + U)!!
v1 – v2 = v2 – v1

22. Elastic Collisions in 2D qualitative here, quantitative in the text
Physical Principles: The same as in 1D
1. Conservation of VECTOR momentum:
P1x + P2x = P1x + P2x
P1y + P2y = P1y + P2y
2. Conservation of K
(½)m1(v1)2 + (½)m2(v2)2 = (½)m1(v1)2 + (½)m2(v2)2