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

Section 7-4: Conservation of Energy &
Momentum in Collisions
• 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!
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
• 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!)
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.
• 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!
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!!
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:
– Rewrite (2) as:
mA(vA - vA) = mB(vB - vB)
(a)
mA[(vA)2 - (vA)2] = mB[(vB)2 - (vB)2] (b)
– Divide (b) by (a):

vA + vA = vB + vB
vA - vB = vB  - vA  = - (vA - vB)
or
(3)
Relative velocity before= - Relative velocity after
Elastic head-on (1d) collisions only!!
• Summary: 1d Elastic collisions: Rather than
directly use momentum conservation + KE
conservation, often convenient to use:
Momentum conservation:
mAvA + mBvB = mAvA + mBvB (1)

use
along
with:
these!

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!
Example 7-7: Pool (Billiards)
Ball A
Before:
Ball B
 v
v=0
mA = mB = m, vA = v, vB = 0, vA = ?, vB = ?
Momentum Conservation: mv +m(0)=mvA + mvB
Masses cancel 
v = v A  + v B
(I)
• Relative velocity results for elastic head on collision:
v - 0 = vB - vA
(II)
Solve (I) & (II) simultaneously for vA & vB :

vA = 0,
v B = v
Ball 1: to rest. Ball 2 moves with original velocity of ball 1
Before:
Ball 2
Ball 1
v=0
 v
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).
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 headon 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.
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.
• 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!!
Example 7-9: Railroad cars again
Same rail cars as Ex. 7-3. 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
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!)
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.
Ex. 7-10 & Probs. 32 & 33 (Inelastic Collisions)
Before
ℓ
ℓ



ℓ-h


a


After
a
v=0
aa
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
Problem 71
A bullet, m = 0.025 kg hits
& is embedded in a block,
M = 1.35 kg. Friction
coefficient between block &
surface: μk = 0.25. 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
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)
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. vA – vB = vB – vA
• As intermediate step, might use Conservation of
Mechanical Energy (KE + PE)!!
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.
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