Chapter 9B - Conservation of
Momentum
AA PowerPoint
PowerPoint Presentation
Presentation by
by
Paul
Paul E.
E. Tippens,
Tippens, Emeritus
Emeritus Professor
Professor
Southern
Southern Polytechnic
Polytechnic State
State University
University
©
2007
NASA
Momentum is
conserved in this
rocket launch.
The velocity of
the rocket and
its payload is
determined by
the mass and
velocity of the
expelled gases.
Photo: NASA
Objectives: After completing this
module, you should be able to:
• State the law of conservation of momentum
and apply it to the solution of problems.
• Distinguish by definition and example
between elastic and inelastic collisions.
• Predict the velocities of two colliding bodies
when given the coefficients of restitution,
masses, and initial velocities.
A Collision of Two Masses
When two masses m1 and m2 collide, we will use
the symbol u to describe velocities before collision.
Before
m1
u1
u2
m2
The symbol v will describe velocities after collision.
v1
After
m1
m2
v2
A Collision of Two Blocks
Before
m1
“u”= Before
After v1
m1
u1
u2
Collision
m1
m2B
m2
“v” = After
m2
v2
Conservation of Energy
m1
u2
u1
m2
The kinetic energy before colliding is
equal to the kinetic energy after colliding
plus the energy lost in the collision.
1
2
m u  m u  m v  m v  Loss
2
1 1
1
2
2
2 2
1
2
2
1 1
1
2
2
2 2
Example 1. A 2-kg mass moving at 4 m/s
collides with a 1-kg mass initially at rest. After
the collision, the 2-kg mass moves at 1 m/s
and the 1-kg mass moves at 3 m/s. What
energy was lost in the collision?
It’s important to draw and label a sketch with
appropriate symbols and given information.
u1 = 4 m/s
m1
u2 = 0
m2
m1 = 2 kg
m1 = 1 kg
BEFORE
v1 = 1 m/s
m1
v2 = 2 m/s
m2
m1 = 2 kg
m1 = 1 kg
AFTER
Example 1 (Continued). What energy
was lost in the collision? Energy is
conserved.
u1 = 4 m/s
m1
m2
m1 = 2 kg
BEFORE:
u2 = 0
1
2
m1 = 1 kg
v1 = 1 m/s
m1
m1 = 2 kg
v2 = 2 m/s
m2
m1 = 1 kg
m1u12  12 m2u22  12 (2 kg)(4 m/s) 2  0  16 J
2
2
2
2
1
1
1
1
m
v

m
v

(2
kg)(1
m/s)

(1
kg)(2
m/s)
 3J
AFTER
2 2
2 1 1
2
2
2
Energy Conservation: K(Before) = K(After) + Loss
Loss = 16 J – 3 J
Energy
Energy Loss
Loss =
= 15
15 JJ
Impulse and Momentum
A
-FA t
vA
uA uB
B
B
Impulse = p
FB t
Ft = mvf– mvo
Opposite but Equal F t
B
A
vB
FBt = -FAt
mBvB - mBuB = -(mAvA - mAuA)
Simplifying:
m
mBBvvBB == m
mAAuuAA ++ m
mBBuuBB
mAAvvAA ++ m
Conservation of Momentum
The total momentum AFTER a collision is equal
to the total momentum BEFORE.
m
mAAvvAA ++ m
mBBvvBB == m
mAAuuAA ++ m
mBBuuBB
A
-FAt
vA
A
uA uB
B
B
FB t
B
vB
Recall that the total energy
is also conserved:
Kinetic Energy: K = ½mv2
KKA0
+
K
=
K
+
K
+
Loss
+
K
=
K
+
K
+
Loss
B0
Af
Bf
A0
B0
Af
Bf
Example 2: A 2-kg block A and a 1-kg block
B are pushed together against a spring and
tied with a cord. When the cord breaks,
the 1-kg block moves to the right at 8 m/s.
What is the velocity of the 2 kg block?
The initial velocities are
zero, so that the total
A
B
momentum before release is
zero.
0
0
mAvA + mBvB = mAuA + mBuB
mA v A = - m B v B
vA = -
mB v B
mA
Example 2 (Continued)
2 kg
A
B
1 kg
vA2
8 m/s
A
B
0
mAvA+ mBvB = mAuA + mBuB0
mB v B
mA v A = - m B v B
vA = mA
vA = -
(1 kg)(8 m/s)
(2 kg)
vvAA == -- 44 m/s
m/s
Example 2 (Cont.): Ignoring friction, how
much energy was released by the spring?
2 kg
A
B
1 kg
Cons. of E:
4 m/s
½kx2
8 m/s
A
B
2
= ½ mAvA + ½mBvB2
½kx2 = ½(2 kg)(4 m/s)2 + ½(1 kg)(8 m/s)2
½kx2 =
16 J + 32 J = 48 J
22= 48 J
½
kx
½kx = 48 J
Elastic or Inelastic?
An elastic collision loses
no energy. The deformation on collision is fully
restored.
In an inelastic collision,
energy is lost and the
deformation may be
permanent. (Click it.)
Completely Inelastic Collisions
Collisions
Collisions where
where two
two objects
objects stick
stick together
together
and
and have
have aa common
common velocity
velocity after
after impact.
impact.
Before
After
Example 3: A 60-kg football player stands
on a frictionless lake of ice. He catches a
2-kg football and then moves at 40 cm/s.
What was the initial velocity of the football?
A
Given: uB= 0; mA= 2 kg; mB= 60 kg;
B
v A = vB = vC
Momentum:
vC = 0.4 m/s
mAvA + mBvB = mAuA + mBuB
Inelastic collision:
(mA + mB)vC = mAuA
(2 kg + 60 kg)(0.4 m/s) = (2 kg)uA
uuAA== 12.4
12.4 m/s
m/s
0
Example 3 (Cont.): How much energy
was lost in catching the football?
1
2
0
m u  m u  (mA  mB )v  Loss
2
A A
1
2
2
B B
1
2
2
C
½(2 kg)(12.4 m/s)2 = ½(62 kg)(0.4 m/s)2 + Loss
154 J = 4.96 J + Loss
Loss
Loss =
= 149
149 JJ
97% of the energy is lost in the collision!!
General: Completely Inelastic
Collisions where two objects stick together
and have a common velocity vC after impact.
Conservation of Momentum:
(mA  mB )vc  mAu A  mB uB
Conservation of Energy:
1
2
m u  m u  (mA  mB )v  Loss
2
A A
1
2
2
B B
1
2
2
c
Example 4. An 87-kg skater B collides with a 22-kg
skater A initially at rest on ice. They move together
after the collision at 2.4 m/s. Find the velocity of the
skater B before the collision.
Common speed after
colliding: 2.4 m/s.
uA = 0 uB = ?
vB= vA = vC = 2.4 m/s
mAu A  mB uB  (mA  mB )vC
87 kg
A
22 kg
B
(87 kg)uB = (87 kg + 22 kg)(2.4 m/s)
(87 kg)uB =262 kg m/s
uB = 3.01 m/s
Example 5: A 50 g bullet strikes a 1-kg
block, passes all the way through, then
lodges into the 2 kg block. Afterward,
the 1 kg block moves at 1 m/s and the 2
kg block moves at 2 m/s. What was the
entrance velocity of the bullet?
u A= ?
2 kg
1 kg
1 m/s
1 kg
2 kg
2 m/s
Find entrance velocity of
bullet: mA= 0.05 kg; uA= ?
A
50 g B
1 kg
Momentum After =
Momentum Before
0
C
2 kg
1 m/s
1 kg
2 kg
2 m/s
0
mAuA + mBuB + mCuC = mBvB + (mA+mC) vAC
(0.05 kg)uA =(1 kg)(1 m/s)+(2.05 kg)(2
m/s)
kg
(0.05 kg) uA =(5.1 kg m/s)
uuAA== 102
102 m/s
m/s
Completely Elastic Collisions
Collisions where two objects collide in such a
way that zero energy is lost in the process.
APPROXIMATIONS!
Velocity in Elastic Collisions
uA
vA
A
A
B
uB
B
vB
1. Zero energy lost.
2. Masses do not change.
3. Momentum conserved.
Equal but opposite impulses (F t) means that:
(Relative v After) = - (Relative v Before)
For elastic collisions:
vvAA -- vvBB == -- (u
(uAA -- uuBB))
Example 6: A 2-kg ball moving to the
right at 1 m/s strikes a 4-kg ball moving
left at 3 m/s. What are the velocities
after impact, assuming complete
elasticity?
1 m/s
3 m/s
A
vA 1 kg
A
B
2 kg
vB
B
vA - vB = - (uA - uB)
vA - vB = uB - uA
vA - vB = (-3 m/s) - (1 m/s)
From conservation of energy (relative v):
vvAA -- vvBB == -- 44 m/s
m/s
Example 6 (Continued)
Energy: vA - vB = - 4 m/s
Momentum also conserved:
mAvA + mBvB = mAuA + mBuB
3 m/s
1 m/s
A
vA 1 kg
A
B
2 kg v
B
B
(1 kg)vA+(2 kg)vB=(1 kg)(1 m/s)+(2 kg)(-3 m/s)
Two independent
equations to solve:
vA + 2vB = -5 m/s
vA - vB = - 4 m/s
Example 6 (Continued)
vA + 2vB = -5 m/s
vA - vB = - 4 m/s
Subtract: 0 + 3vB2 = - 1 m/s
vvBB == -- 0.333
0.333 m/s
m/s
Substitution:
vA - vB = - 4 m/s
3 m/s
1 m/s
A
B
1 kg
vA
A
2 kg v
B
B
vA2 - (-0.333 m/s) = - 4 m/s
vvAA== -3.67
-3.67 m/s
m/s
Example 7. A 0.150 kg bullet is fired at 715 m/s into a 2kg wooden block at rest. The velocity of block afterward is
40 m/s. The bullet passes through the block and emerges
with what velocity?
A
mAv A  mB vB  mAu A  mB uB
B
uB = 0
(0.150 kg)vA+ (2 kg)(40 m/s) = (0.150 kg)(715 m/s)
0.150vA+ (80 m/s) = (107 m/s)
0.150vA = 27.2 m/s)
vA = 181 m/s
27.2 m/s
vA 
0.150
Example 8a: Inelastic collision: Find vC.
2 m/s
5 kg
uB=0
7.5 kg
A
Common
vC after
A
B
vC
B
After hit: vB= vA= vC
mAu A  mB uB  (mA  mB )vC
(5 kg)(2 m/s) = (5 kg + 7.5 kg)vC
12.5 vC =10 m/s
vvCC == 0.800
0.800 m/s
m/s
In
In an
an completely
completely inelastic
inelastic collision,
collision, the
the two
two balls
balls
stick
stick together
together and
and move
move as
as one
one after
after colliding.
colliding.
Example 8. (b) Elastic collision: Find vA2 and vB2
2 m/s
Conservation of Momentum:
vB1=0
5 kg
7.5 kg
A
B
vA
vB
A
B
mAv A  mAv A  mB vB
(5 kg)(2 m/s) = (5 kg)vA2 + (7.5 kg) vB
5 vA + 7.5 vB = 10 m/s
For Elastic Collisions:
v A  vB  (u A  u B )
v A  vB  2 m/s
Continued . . .
Example 8b (Cont). Elastic collision: Find vA & vB
Solve simultaneously:
x (-5)
v A  vB  2 m/s
2 m/s
5 kg
5 vA + 7.5 v B = 10 m/s
A
vB =0
7.5 kg
A
B
vA
vB
B
5 vA + 7.5 vB = 10 m/s
-5 vA + 5 vB = +10 m/s
12.5 vB = 20 m/s
20 m/s
vB 
 1.60 m/s
12.5
vA - 1.60 m/s = -2 m/s
vvAA == -0.400
-0.400 m/s
m/s
vvBB == 1.60
1.60 m/s
m/s
General: Completely Elastic
Collisions where zero energy is lost
during a collision (an ideal case).
Conservation of Momentum:
mAv A  mB vB  mAu A  mB uB
Conservation of Energy:
1
2
m u  m u  m v  m v  Loss
2
A A
1
2
2
B B
1
2
2
A A
v A  vB  u B  u A
1
2
2
B B
Example 9: A 50 g bullet lodges into a 2-kg
block of clay hung by a string. The bullet and
clay rise together to a height of 12 cm. What
was the velocity of the 50-g mass just before
entering?
B
uA
A
B
A
The ballistic pendulum!
12 cm
Example (Continued):
Collision and Momentum:
mAuA+0= (mA+mB)vC
2.05 kg
B
uA
A
(0.05 kg)uA = (2.05 kg)vC
To find vA we need vC .
50 g
12 cm
2 kg
After collision, energy is conserved for masses.
1
2
(mA  mB )v  (mA  mB ) gh
2
C
vvCC == 22gh
gh
Example (Continued):
vC = 2gh = 2(9.8)(0.12)
After Collision: vC = 1.53 m/s
Momentum Also
Conserved:
mAuA+0= (mA+mB)vC
2.05 kg
B
uA
A
50 g
2 kg
(0.05 kg)uA = (2.05 kg)(1.53 m/s)
uuAA == 62.9
62.9 m/s
m/s
12 cm
Summary of Formulas:
Conservation of Momentum:
mAv A  mB vB  mAu A  mB uB
Conservation of Energy:
1
2
m u  m u  m v  m v  Loss
2
A A
1
2
2
B B
For
For elastic
elastic only:
only:
1
2
2
A A
1
2
2
B B
v A  vB  u B  u A
CONCLUSION: Chapter 9B
Conservation of Momentum