Page 1
Phys101 Lectures 14, 15, 16
Momentum and Collisions
Key points:
• Momentum and impulse
• Condition for conservation of momentum and why
• How to solve collision problems
• Centre of mass
Momentum
is a vector:
It’s a quantity that represents the amount and direction of motion.
It was once called “the quantity of motion”. Now we know that kinetic energy is another quantity of motion.
Newton’s 2nd Law
where is called the impulse of force . (area under the F-t curve)
Impulse-momentum principle : The net impulse on an object is equal to the change in momentum of the object.
dt
p
d
dt
v
d
m
a
m
F
p
p
p
dt
F
f i t t f i
t
F
dt
F
J
ave t t f i
p
J
F
Example 9-1: Force of a tennis serve.
For a top player, a tennis ball may leave the
racket on the serve with a speed of 55 m/s
(about 120 mi/h). If the ball has a mass of 0.060 kg and is in contact with the racket for about 4 ms (4 x 10-3 s), estimate the average
force on the ball. [Solution]
)
(
800
004
.
0
55
06
.
0
0
N
t
mv
t
p
F
p
t
F
J
ave ave
x
A 0.140-kg baseball is dropped from rest. It has a speed of
1.20 m/s just before it hits the ground. It rebounds with a
speed of 1.00 m/s. The ball is in contact with the ground
for 0.0140 s. What is the average force exerted by the
ground on the ball during that time?
A) 2.00 N
B) 10.00 N
C) 22.0 N
D) 24.0 N
E) 12.0 N
i-clicker question 14-1
x
)
N
(
.
)
.
(
.
.
t
)
v
v
(
m
t
p
F
p
t
F
p
t
F
ix fx x x x x22
014
0
20
1
00
1
140
0
MasteringPhysics Assignment #6
Impulse-Momentum Theorem, Part I
• Consider a particle of mass m=5.00kg moving along the x axis. It is
at rest at t=0.00 sec. A varying force F(t)=6.00t2-4.00t+3.00 is acting on the particle between t=0.00 sec and t=5.00sec. Find the speed v of the particle at t=5.00 sec.
x
) S N ( ) ( ) ( ) ( ) ( ) ( ) ( t t t dt t t dt F J dt F J t t x x t t
215 0 3 0 2 0 2 5 3 5 2 5 2 3 2 2 3 4 6 2 3 2 3 5 0 2 3 5 0 2 2 1 2 1 f i f i fp
m
(
v
v
)
m
v
p
p
J
)
s
/
m
(
m
J
v
fx x43
5
215
Many-body System
Two students pulling each other on ice. Internal forces: and .
External forces:
Newton’s law:
Adding the two equations
Since the internal forces always cancel,
1
F
12 2
21F
dt
p
d
F
F
g
m
,
dt
p
d
F
F
g
m
1 N1 12 1 2 N2 21 2
12F
F
21 1 1 2 1g
,
m
g
,
F
Nand
F
Nm
is
the
total
momentum
of
the
system
.
2 1 21 12 2 2 1 1P
dt
P
d
F
F
dt
p
d
dt
p
d
F
F
F
g
m
F
g
m
int ext N N
120
21F
,
F
intF
momentum)
total
the
change
can
forces
external
(only
dt
P
d
F
ext
DemoConservation of Momentum
This is the law of conservation of linear
momentum:
When the net external force on a system of objects is zero,
the total momentum of the system remains constant.
Note 1: If one of the components of the net external
force is zero, the corresponding component of the total
momentum of the system is conserved (even though the
total momentum vector may or may not be conserved).
Note 2: For a one-object system, the condition for
momentum conservation is that the net force acting on
the object is zero.
t
tan
cons
P
.,
e
.
i
,
dt
P
d
,
F
ext
0
0
When
i-clicker question 14-2
The condition for mechanical energy to be conserved is
(A) It’s a closed system.
(B) The net external force is zero.
(C) No nonconservative work.
(D) The momentum is never conserved.
(E) The momentum is always conserved.
Example 9-4: Rifle recoil. Calculate the recoil
velocity of a 5.0-kg rifle that shoots a 0.020-kg bullet at a speed of 620 m/s. [Solution] X-component of external Force is zero;
Thus the x-component of the total momentum is conserved:
)
s
/
m
(
.
v
)
s
/
m
(
.
.
m
v
m
v
v
m
v
m
P
P
R R Bx B Rx Rx R Bx B xi xf5
2
5
2
5
620
02
0
0
0
Collisions
Momentum is conserved in all
collisions. Why?
Because the impulse of external forces can be ignored (much smaller than internal impulse).
Collisions in which kinetic energy
is conserved as well are called
elastic collisions, and those in which it is not are called inelastic.
Elastic Collisions in One Dimension
Here we have two objects
colliding
elastically
. We
know the masses and the
initial speeds.
Since both
momentum
and
kinetic energy
are
conserved, we can write
two
equations. This
allows us to solve for the
two
unknown final
Example 9-7: Billiard ball A of mass m moving with speed vA collides head-on with ball B of equal mass. What are the speeds of the two balls after the collision, assuming it is elastic? Assume ball B is initially at rest (vB = 0). Conservation of momentum: x
comp)
-(x
B A A
m
v
m
v
mv
Conservation of kinetic energy:
2 2 2
2
1
2
1
2
1
B A Am
v
m
v
mv
(2)
(1)
2 2 2 B A A B A Av
v
v
v
v
v
B Av
v
(2)
:
0
2
(1)
2 A B Av
v
v
0
:
1
Solution
A A Bv
v
v
0
:
2
Solution
Solution 2 should be rejected because it means no collision.
Inelastic Collisions
With
inelastic
collisions, some of the initial
kinetic energy is lost to
thermal
or
potential
energy. Kinetic energy may also be gained
during
explosions
, as there is the 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.
B A
v
Example 9-3: Railroad cars collide
A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the
collision, what is their common speed immediately after the collision? B A A
m
v
m
v
mv
2
A B Av
v
v
B Av
v
Conservation of momentum (x-comp.):
Hit and stick (perfectly inelastic):
Example 9-11: Ballistic pendulum.
The ballistic pendulum is a device used to measure the speed of a projectile, such as a
bullet. The projectile, of mass m, is fired into
a large block of mass M, which is
suspended like a pendulum. As a result of the collision, the pendulum and projectile together swing up to a maximum height h. Determine the relationship between the initial horizontal speed of the projectile, v, and the maximum height h.
[Solution] Two events:
1. Hit and stick;
2. Swing .
v
)
M
m
(
mv
gh
)
M
m
(
v
)
M
m
(
2
2
1
gh
m
M
m
v
gh
)
M
m
(
v
M
m
m
v
,
2
2
1
:
Eliminate
2 2
Collisions in Two or Three Dimensions
Example 9-13: Proton-proton collision.
A proton traveling with speed 8.2 x 105 m/s collides elastically with
a stationary proton in a hydrogen target. One of the protons is observed to be scattered at a 60°angle. At what angle will the
second proton be observed, and what will be the velocities of the two protons after the collision?
Elastic collision:
(1)
2
1
2
1
2
1
2 2 2 2 2 2 B A A B A Av
v
v
.,
e
.
i
v
m
v
m
mv
m
m
m
A
B
(3)
0
(2)
B B A A B B A A A B A A B A A
sin
v
sin
v
:
y
cos
v
cos
v
v
:
x
,
v
v
v
.,
e
.
i
,
v
m
v
m
v
m
Conservation of momentum:(3).
(2),
(1),
:
equations
3
.
:
unknowns
3
v
A,
v
B
,
B
i-clicker question 14-3
A car with a mass of 1200 kg and a speed of 12 m/s heading north approaches am intersection. At the same time, a minivan with a mass of 1300 kg and speed of 24 m/s heading east is also
approaching the intersection. The car and the minivan collide and
stick together. Consider the total momentum and the total kinetic energy of the two vehicles before and after the collision.
A. Both the total momentum and total kinetic energy are conserved.
B. The total momentum is conserved but the total kinetic energy is not conserved.
C. Neither the total momentum nor the total kinetic energy is conserved.
D. The total kinetic energy is conserved but the total momentum is not conserved.
E. The change in total momentum equals the change in total kinetic energy.
What is the velocity of the wrecked vehicles just after the collision?
x (E) y (N) Conservation of momentum: R V C
p
P
p
Hit and stick:
v
C
v
V
v
R
)
s
/
m
(
.
m
m
P
v
)
s
/
m
kg
(
v
m
v
m
P
C V R R C C V V R7
13
1200
1300
34400
34400
2 2
Cp
Vp
RP
65
1 1 C C V V C Vv
m
v
m
tan
p
p
tan
Center of Mass (
CM
)
In (a), the diver’s motion is pure
translation
; in (b)
it is
translation
plus
rotation
.
There is one point that moves in the
same
path a
particle would
take if subjected
to the same force
as the diver. This
point is called the
center of mass
Center of Mass (
CM
)
The
general
motion of an object can be
considered as the
sum
of the
translational
motion of the
CM
, plus
rotational
,
vibrational
,
Center of Mass (CM)
For two particles, the
center of mass
lies closer to the
one with the most mass:
where M is the
total
mass.
In general,
,
m
m
y
m
y
m
y
,
m
m
x
m
x
m
x
B A B B A A CM B A B B A A CM
Exercise 9-15: Three particles in 2-D.
Three particles, each of mass
2.50 kg, are located at the
corners of a right triangle
whose sides are 2.00 m and 1.50 m long, as shown. Locate the center of mass. [Solution]
)
m
(
.
.
.
m
m
m
x
x
x
m
m
m
x
m
x
m
x
m
x
C B A C B A C B A C C B B A A CM33
1
0
2
0
2
0
3
1
)
(
3
1
)
m
(
.
.
y
y
y
m
m
m
y
m
y
m
y
m
y
C B A C B A C C B B A A CM50
0
5
1
0
0
3
1
3
1
Example 9-17: CM of L-shaped flat object.
Determine the CM of the uniform thin L-shaped object shown.
[Solution] The object consists of two
rectangular parts: A and B, whose
centres of mass are (xA, yA) and (xB, yB).
m
.
y
,
m
.
x
m
.
y
,
m
.
x
B B A A74
0
96
1
10
0
03
1
t
.
)
.
.
(
t
m
B
1
48
0
20
0
296
)
thickness
t
,
density
t
.
)
.
.
(
t
m
A
(
412
0
20
0
06
2
)
m
(
.
.
.
)
.
(
.
.
.
m
m
y
m
y
m
y
)
m
(
.
.
.
.
.
.
.
m
m
x
m
x
m
x
B A B B A A CM B A B B A A CM25
0
296
0
412
0
74
0
296
0
10
0
412
0
42
1
296
0
412
0
96
1
296
0
03
1
412
0
Center of Mass and Translational Motion
The total momentum of a system of particles is equal to the product
of the total mass and the velocity of the center of mass.
The sum of all the forces acting on a system is equal to the total mass of the system multiplied by the acceleration of the center of mass:
Therefore, the center of mass of a system of particles
(or objects) with total mass M moves like a single
particle of mass M acted upon by the same net
external force.
C
Total
