Conservation of Momentum
& Energy 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!
 m1v1i + m2v2i = m1v1f + m2v2f
HOLDS for ALL collisions!
Note!!
• An Ideal Very Special Case: 2 very hard objects
(like billiard balls) collide. An “Elastic Collision”
• To analyze Elastic Collisions:
Rule # 1 Still holds!
 m1v1i + m2v2i = m1v1f + m2v2f
Rule # 2
• For Elastic Collisions ONLY (!!)
Total Kinetic Energy (KE) is conserved!!
(KE)before = (KE)after
 (½)m1(v1i)2 + (½) m2(v2i)2 =
(½)m1(v1f)2 + (½)m2(v2f)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!
Types of Collisions
Momentum is conserved in all collisions
• Inelastic collisions: rubber ball and hard ball
– Kinetic energy is not conserved
– Perfectly inelastic collisions occur when the
objects stick together
• Elastic collisions: billiard ball
– both momentum and kinetic energy are conserved
• Actual collisions
– Most collisions fall between elastic and perfectly
inelastic collisions
Collisions Summary
• In an elastic collision, both momentum and kinetic
energy are conserved
• In a non-perfect inelastic collision, momentum is
conserved but kinetic energy is not. Moreover, the
objects do not stick together
• In a perfectly inelastic collision, momentum is conserved,
kinetic energy is not, and the two objects stick together
after the collision, so their final velocities are the same
• Elastic and perfectly inelastic collisions are limiting
cases, most actual collisions fall in between these
two types
• Momentum is conserved in all collisions
Special case: Head-on Elastic Collisions
These can be analyzed in 1
dimension. Some Types of
head on collisions:
For 2 masses colliding
elastically: Often we know
the masses & the initial
velocities. 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
after 
collision
or
or
vA, vB, (vA), (vB), are 1 dimensional vectors!
Elastic Collisions in 1 Dimension
• Special case: Head-on Elastic Collisions.
• Momentum is conserved (ALWAYS!)
Pbefore = Pafter
m1v1i + m2v2i = m1v1f + m2v2f
v1i, v2i, v1f, v2f are one dimensional vectors!
• Kinetic Energy is conserved (ELASTIC!)
(KE)before = (KE)after
(½)m1(v1i)2 + (½) m2(v2i)2 =
(½)m1(v1f)2 + (½)m2(v2f)2
• 2 equations, 6 quantities: v1i,v2i,v1f, v2f, m1, m2
 Clearly, we must be given 4 out of 6 to solve problems!
Solve with CAREFUL algebra!!
m1v1i + m2v2i = m1v1f + m2v2f
(1)
(½)m1(v1i)2 + (½) m2(v2i)2 =
(½)m1(v1f)2 + (½)m2(v2f)2 (2)
• Now, some algebra with (1) & (2), the results of
which will help to simplify problem solving:
• Rewrite (1) as:
m1(v1i – v1f) = m2(v2f - v2i)
(a)
• Rewrite (2) as: m1[(v1i)2 - (v1f)2] = mB[(v2f)2 - (v2i)2 (b)
• Divide (b) by (a):

v1i + v1f = v2i + v2f
v1i – v2i = v2f – v1f = - (v1f – v2f)
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:
m1v1i + m2v2i = m1v1f + m2v2f (1)
Along with:
v1i – v2i = v2f – v1f = - (v1f – v2f) (3)
• (1) & (3) are equivalent to momentum
conservation + Kinetic Energy
conservation, since (3) was derived from
these conservation laws!
Example: Pool (Billiards)
Ball 1
Before:
Ball 2
 v
v=0
m1 = m2 = m, v1i = v, v2i = 0, v1f = ?, v2f = ?
Momentum Conservation: mv +m(0)=mv1f + mv2f
Masses cancel 
v = v1f + v2f
(I)
• Relative velocity results for elastic head on collision:
v - 0 = v2f - v1f
(II)
Solve (I) & (II) simultaneously for v1f & v2f :

v1f = 0,
v2f = 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
(m1) to strike a second object (m2, the “target”) at rest (v2 =
0). Assume the objects have unequal masses, and that the
collision is elastic and occurs along a line (head-on).
(a) Derive equations for v2f & v1f in terms of the initial
velocity v1i of mass m1 & the masses m1 & m2.
(b) Determine the final velocities if the moving object
is much more massive than the target (m1 >> m2).
(c) Determine the final velocities if the moving object
is much less massive than the target (m1 << m2).
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: Railroad cars again
Same rail cars as Ex. 7-3. Car A, mass m1 = 10,000 kg, traveling at speed
v1i = 24 m/s strikes car B (same mass), initially at rest (v2i = 0). Cars lock together
after collision. Ex. 7-3: Find speed v after collision.
Before
Collision
After
Collision
Ex. 7-3 Solution: v2i = 0, v1f = v2f = v Use Momentum Conservation:
m1v1i+m2v2i = (m1 + m2)v  v = [(m1v1i)/(m1 + m2)] = 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)m1(v1i)2 = 2.88  106 J
Finally: KEf = (½)(m1+ m2)(v)2 = 1.44  106 J ! (50% loss!)
More about Perfectly Inelastic
Collisions
• When two objects stick together
after the collision, they have
undergone a perfectly inelastic
collision
• Conservation of momentum
m1v1i  m2 v2i  (m1  m2 )v f
m1v1i  m2 v2i
vf 
m1  m2
• Kinetic energy is NOT conserved
An SUV Versus a Compact
An SUV with mass 1.80103 kg is travelling
eastbound at +15.0 m/s, while a compact car with
mass 9.00102 kg is travelling westbound at -15.0
m/s. The cars collide head-on, becoming entangled.
(a) Find the speed of the
entangled cars after the
collision.
(b) Find the change in the
velocity of each car.
(c) Find the change in the kinetic
energy of the system
consisting of both cars.

An SUV Versus a Compact
(a)
Find the speed of the entangled m1  1.80 103 kg, v1i  15m / s
cars after the collision.
m  9.00 102 kg, v  15m / s
2
pi  p f
m1v1i  m2 v2i  (m1  m2 )v f
m1v1i  m2 v2i
vf 
m1  m2
v f  5.00m / s
2i
An SUV Versus a Compact
(b) Find
the change in the
velocity of each car.
v f  5.00m / s
m1  1.80 103 kg, v1i  15m / s
m2  9.00 102 kg, v2i  15m / s
v1  v f  v1i  10.0m / s
v2  v f  v2i  20.0m / s
m1v1  m1 (v f  v1i )  1.8 104 kg  m / s
m2 v2  m2 (v f  v2i )  1.8 104 kg  m / s
m1v1  m2 v2  0
An SUV Versus a Compact
(c)
Find the change in the kinetic
energy of the system
consisting of both cars.
m1  1.80 103 kg, v1i  15m / s
m2  9.00 102 kg, v2i  15m / s
v f  5.00m / s
1
1
2
KEi  m1v1i  m2 v22i  3.04 105 J
2
2
1
1
2
KE f  m1v1 f  m2 v22 f  3.38 10 4 J
2
2
KE  KE f  KEi  2.70 105 J
More About Elastic Collisions
• Both momentum and kinetic energy are
conserved
m1v1i  m2v2i  m1v1 f  m2 v2 f
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
• Typically have two unknowns
• Momentum is a vector quantity
– Direction is important
– Be sure to have the correct signs
• Solve the equations simultaneously
Elastic Collisions
• A simpler equation can be used in place of the KE equation
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
m1 (v12i  v12f )  m2 (v22 f  v22i )
v  v  ( v  v )
m1 (v11i i v1 f )( v21ii  v1 f )  m21(fv2 f  v22i )(f v2 f  v2i )
m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1 (v1i  v1 f )  m2 (v2 f  v2i )
v1i  v1 f  v2 f  v2i
m1v1i  m2 v2i  m1v1 f  m2 v2 f
Summary of Types of Collisions
• In an elastic collision, both momentum and kinetic energy are
conserved
v1i  v1 f  v2 f  v2i
m1v1i  m2 v2i  m1v1 f  m2 v2 f
• In an inelastic collision, momentum is conserved but kinetic
energy is not
m1v1i  m2 v2i  m1v1 f  m2 v2 f
• In a perfectly inelastic collision, momentum is conserved,
kinetic energy is not, and the two objects stick together after
the collision, so their final velocities are the same
m1v1i  m2 v2i  (m1  m2 )v f
Example: 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.
Example: 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
Example
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)!!
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