Outline  Center of Mass, Momentum,
and Collisions


 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. 1 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 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. 2a 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. 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 Θ with the Xaxis
= 135^{o
}
 Kinetic energy is conserved in an elastic collision.
For the elastic collision shown in the above figure:
p_{i} = p_{f}_{}

m_{1}v_{1i} =
m_{1}v_{1f} + m_{2}v_{2f}

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}

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 } are 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} = 1/2 m_{2}[(2m_{1}/m_{2})v_{1i}]^{2},
a very small amount, because m_{1} << m_{2}.
 If m_{1 }>> m_{2}, v_{1f
}≈ v_{1i.}
 Sample Problems in 104
Problem Set for Momentum: 57, 911, 1323.
 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 104
Problem Set for Momentum: 1, 3.
 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. 3 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.
 Sample Problems in 104
Problem Set for Momentum: 2, 4, and 8.

