1. Section 7-4: Conservation of Energy & Momentum in Collisions

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

3. Total Kinetic Energy (KE) is conserved!!
Note!!
Ideal Very Special Case: 2 very hard objects (like billiard balls) collide. An “Elastic Collision”
To analyze Elastic Collisions:
Rule # 1 Still holds!
 mAvA + mBvB = mAvA + mBvB
Rule # 2
For Elastic Collisions ONLY (!!)
Total Kinetic Energy (KE) is conserved!!
(KE)before = (KE)after
 (½)mA(vA)2 + (½) mB(vB)2 = (½)mA(vA)2 + (½)mB(vB)2

4. ELASTIC COLLISIONS ONLY!!
Total Kinetic energy (KE) is conserved for
ELASTIC COLLISIONS ONLY!!
Inelastic Collisions
 Collisions which are AREN’T elastic.
Is KE conserved for Inelastic Collisions?
NO!!!!!!
Is momentum conserved for Inelastic Collisions?
YES!!
(By Rule # 1: Momentum is ALWAYS conserved in a collision!)

5. Special case: Head-on Elastic Collisions
Can analyze in 1 dimension
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. Special case: Head-on Elastic Collisions.
1 dimensional collisions: Some possible types:
before
collision  or 
after
collision or
vA, vB, (vA), (vB), are 1 dimensional vectors!

7. Sect. 7-5: Elastic Collisions in 1 Dimension
Special case: Head-on Elastic Collisions.
Momentum is conserved (ALWAYS!)
Pbefore = Pafter
mAvA + mBvB = mAvA + mBvB
vA, vB, vA, vB are one dimensional vectors!
Kinetic Energy is conserved (ELASTIC!)
(KE)before = (KE)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. vA - vB = vB  - vA  = - (vA - vB) (3)
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. Momentum conservation: mAvA + mBvB = mAvA + mBvB (1) along with:
Summary: 1d Elastic collisions: Rather than directly use momentum conservation + KE conservation, 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 7-7: 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 collision:
v - 0 = vB - vA (II)
Solve (I) & (II) simultaneously for vA & vB :
 vA = 0, vB = v
Ball 1: to rest. Ball 2 moves with original velocity of ball 1
Before:
 v
v = 0
Ball 2
Ball 1
Before:
 v
v = 0

11. Example: Unequal Masses, Target at Rest
A very common practical situation is for a moving object (mA) to strike a second object (mB, the “target”) at rest (vB = 0). Assume the objects have unequal masses, and that the collision is elastic and occurs along a line (head-on). (a) Derive equations for vB and vA in terms of the initial velocity vA of mass mA and the masses mA and mB. (b) Determine the final velocities if the moving object is much more massive than the target (mA >> mB). (c) Determine the final velocities if the moving object is much less massive than the target (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 7-8: Nuclear Collision
A proton (p) of mass 1.01 u (unified atomic mass units) traveling with a speed of 3.60 x 104 m/s has an elastic head-on collision with a helium (He) nucleus (mHe = 4.00 u) initially at rest. What are the velocities of the proton and helium nucleus after the collision? Assume the collision takes place in nearly empty space.

13. Section 7-6: Inelastic Collisions
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. 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!!

15. Example 7-9: 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. 7-3: 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. 7-3 Solution: vA = 0, (vA) = (vB) = v Use Momentum Conservation:
mAvA+mBvB = (mA + m2B)v  v = [(mAvA)/(mA + mB)] = 12 m/s
Ex. 7-9: Cars lock together after collision. Find amount of initial KE transformed to thermal or other energy forms:
Initially: KEi = (½)mA(vA)2 = 2.88  106 J
Finally: KEf = (½)(mA+ mB)(v)2 = 1.44  106 J ! (50% loss!)

16. Example 7-10: 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.

17. Ex. 7-10 & Probs. 32 & 33 (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

18. Problem 71 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

19. 1. Friction force:
Ffr = μkFN = μk(M+m)g
2. The Work- Energy
Principle: Wfr = -Ffrd
= KE = 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)

20. Summary: Collisions Basic Physical Principles:
Conservation of Momentum: Rule # 1: Momentum is ALWAYS conserved in a collision!
Conservation of Kinetic Energy:
Rule # 2: KE 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 (KE + PE)!!
vA – vB = vB – vA

21. 7-7 Collisions in Two or Three Dimensions
Conservation of energy & momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy.
Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities.

22. Elastic Collisions in 2D qualitative here, quantitative in the text
Physical Principles: The same as in 1D
1. Conservation of VECTOR momentum:
PAx + PBx = PAx + PBx
PAy + PBy = PAy + PBy
2. Conservation of KE
(½)mA(vA)2 + (½)mB(vB)2 = (½)mA(vA)2 + (½)mB(vB)2