Outline  Center of Mass, Momentum,
and Collisions


 Center of Mass
 The center of mass of an extended system is the point whose
dynamics typifies the system as a whole when it is treated
as a particle.
 Sample Problem. Find the location of the center of mass
of the system of three particles shown in Fig. 1 below. Each
particle has the same mass m. The coordinates (x,y,z) of particle
1 are (0,L,0) of particle 2 are (0,0,0) and of particle 3
(L,0,0).
 X_{CM } = (1/3m)[0(m) + 0(m) + L(m)] = L/3
Y_{CM } = (1/3m)[L(m) + 0(m) + 0(m)] = L/3
Z_{CM } = (1/3m)[0(m) + 0(m) + 0(m)] = 0
 Symmetry arguments show that the center of mass of a uniform
rectangular plate, a uniform spherical shell, a uniform sphere
or a sphere with a spherically symmetric distribution of mass
is located at the geometric center of the object.
 Two Methods to find center of mass
Addition
method: In Fig. 2(a) above, I show a thin sheet
of metal in the shape of an L. In Fig. 2(b) above, I
have indicated the center of mass of each segment. For segment
1, the (x,y) coordinates are (L/2, 3L/2), for segment 2 they
are (L/2, L/2), and for segment 3 are (3L/2, L/2). Note that
we do not need a Zcomponent for the thin sheet.
Since each segment has mass M/3,
X_{CM } = (1/M)[L/2(M/3) + L/2(M/3) +
3L/2(M/3)] = 5L/6
Y_{CM } = (1/M)[3L/2(M/3) + L/2(M/3) + L/2(M/3)]
= 5L/6
Subtraction
Method: In Fig. 2(c) above, the center of mass
of the square is at the center of the square. The XY coordinates
are (L, L). The center of mass of dashed square is (3L/2,
3L/2). The Xcomponent center of mass of the original figure
in 2a is the Xcomponent of the center of mass of the square
minus the Xcomponent of the "added" square. A similar
approach finds the Ycomponent of the original figure.
X_{CM } = (1/M)[L(4M/3)  3L/2(M/3)] =
5L/6
Y_{CM } = (1/M)[L(4M/3)  3L/2(M/3)] = 5L/6
 Center of Mass for Continuous Medium
 Densities:
 Linear density λ=
mass/length or mass = λ(length)
 Surface density σ=
mass/area or mass = σ(area)
 Volume density ρ=
mass/volume or mass = ρ(volume)
 The equations become:
 Applications:
Find the centers of mass of the three objects below:
(a) Were the "line" of mass uniform, the problem
would be trivial. The center of mass would be at the center
of the rod. To make the problem more interesting let λ
= λ_{o}(x/L),
where λ_{o}
= a constant and L = the length of the rod. For the element
of length dx, its mass element of mass dm = λ
dx = λ_{o}(x/L)
dx.
(b) Mass/area = σ. σ(area) = mass.
For mass of area dA =ydx (see Fig.3b above), dm = σdA
= σydx.
The area of the triangle = ab/2. Let the mass of
the triangle = M,
then σ = Mass/area
= M/(ab/2) = 2M/ab.
The last equation involves two variables. You must eliminate
one of them. The equation of the line
y = (slope)x + intercept on Yaxis
y =  (b/a)x + b
(c) The volume of a hemisphere is 2 πR^{3}/3.
The density of the hemisphere ρ
=
Mass/volume = M/(2 πR^{3}/3)
= 3M/2 πR^{3}.
In Fig. 3(c) above, the volume of an element dV = πx^{2}
dy.
The mass of the element dm =
ρdV = (3M/2 πR^{3}) πx^{2}
dy.
Notice that x^{2 }= R^{2 } y^{2}
and that dm = (3M/2 πR^{3}) π(R^{2
} y^{2}) dy.
 Sample problems in 107
Problem Set for Momentum: 2, 4, 8, 9.
 Linear Momentum p
 Linear momentum of an object p = mv, where
v is the velocity of an object of mass m. Linear momentum
is a vector quantity. p has the same direction as the
velocity v.
 Newton's second law F_{net external} = ma
can be rewritten in terms of momentum. Since a = dv/dt,
we can rewrite ma = d(mv)/dt. If m is a constant,
then you can take it outside of the derivative and m dv/dt
goes back to ma. Newton actually wrote the second law as F_{net
external} = dp/dt.
 If no net external forces act on a system dp/dt =
0 and the momentum of the system remains a constant. Linear
momentum is conserved.
 In "collision problems," momentum is always conserved,
but energy is only conserved for elastic collisions.
In Fig. 4 below, an object of mass m_{1} = 2 kg moves
to the right with a speed of v_{1i} = 5 m/s. It collides
with a mass mass m_{2} = 3 kg initially at rest. After
the collision, the two objects stick together and move with
a velocity v_{f}.
We find the velocity after the collision using conservation
of momentum. The initial momentum of the system equals the
final momentum:
p_{I } = p_{f}
2kg(5 m/s) = (2 + 3)kg v_{f, }or
v_{f} = 2 m/s
The initial kinetic energy of the system = 1/2 (2 kg)(5 m/s)^{2}
= 25 J
The final kinetic energy of the system = 1/2 (5 kg)(2 m/s)^{2}
= 10 J
While momentum is always conserved in a system, kinetic energy
is only conserved in an elastic collision. This is an inelastic
collision.
 Since momentum is a vector quantity, you may consider the
conservation of the components of momentum. If two vectors
are equal, their components are equal. If the initial momentum
p_{i} = p_{f}
the final momentum then,
(p_{i})_{x} = (p_{f})_{x}
and (p_{i})_{y} = (p_{f})_{y}
An object of mass 4m is initially at rest. It breaks into
three parts, two of mass m and one of mass 2m. After the explosion,
one of the objects of mass m moves along the negative Yaxis
with a speed of 10 m/s and the other one of mass m moves along
the positive Xaxes (Fig. 5a below).
Find the magnitude and the direction of the velocity of 2m
after the explosion. The initial momentum of the system is
zero. After the collision one of the pieces moves in the +X
direction and the other moves in the Y direction (Fig. 5b
above).
To solve this problem you must separate it into two onedimensional
problems:
(p_{i})_{x} = (p_{f})_{x}
0 = m(10 m/s) + 2mv_{x}
v_{x} =  5 m/s (the minus sign indicates that
it is to the left)
(p_{i})_{y} = (p_{f})_{y}
0 = m(10 m/s) + 2mv_{y}
v_{y} = +5 m/s (the plus sign indicates that
it is up)
v = (v_{x}^{2} + v_{y}^{2})^{1/2}
= (25 + 25)^{1/2} m/s = 5√2
m/s
Tangent of the angle that v makes with X axis = v_{y}/v_{x}
= 5/5 = 1
Angle Θwith the +Xaxis
= 135^{o
}
 Kinetic energy is conserved in an elastic collision
For the elastic collision shown in the figure above
p_{i} 
= 
p_{f} 
m_{1}v_{1i} 
= 
m_{1}v_{1f}
+ m_{2}v_{2f} (Equation
1) 
K_{i} 
= 
K_{f} 
1/2 m_{1}v^{2}_{1i} 
= 
1/2 m_{1}v^{2}_{1f}
+ 1/2 m_{2}v^{2}_{2f}
(Equation
2) 
Solving Eq. 1 and Eq. 2 for v_{1f}
and v_{f2}:
(if
they are not in the direction shown in the figure, your calculation
will give you a minus sign)
v_{1f} = [(m_{1}
 m_{2})/(m_{1} + m_{2})]v_{1i}
v_{2f} = [(2m_{1})/(m_{1} + m_{2})]v_{1i}
Special cases
 If m_{1} = m_{2},
v_{1f} = 0 and v_{2f} = v_{1i}
All momentum and energy of m_{1 }transferred to
m_{2
}
 If m_{2} >>
m_{1}, v_{1f} » v_{1i}
and v_{2f} ≈ (2m_{1}/m_{2})v_{1i}
Momentum transferred to m_{2} = m_{2}v_{2f}
= 2m_{1}v_{1i} = 2p_{1} for greatest
amount of momentum transferred
Kinetic energy transferred to m_{2} = (2m^{2}_{1}/m_{2})v^{2}_{1f},
a very small amount, because m_{1} << m_{2}
 If m_{1 }>>
m_{2}, v_{1f }» v_{ii
}
 Sample problems in 107
Problem Set for Momentum: 57.
 Impulse J
 Impulse J = d(mv)
=dp = F_{av}dt, where F_{av}
is the average force.
 Impulsemomentum theorem states
that the impulse equals the change in linear momentum. J
= mv_{f}  mv_{i}. Since impulse
and momentum are vector quantities, you may set the Xcomponent
of the impulse equal to the change in the Xcomponent of momentum
and the Ycomponent of the impulse equal to the change in
the Ycomponent of momentum.
 Sample problems in 107
Problem Set for Momentum: 1, 3, 1025.

