Linear Momentum and Collisions:
Two particles interact with each other. According to the
Newton’s Third law:
F12=−
F21
,
d
dt ¿
)=0
Observe that
d
dt ¿
)=0. Consequently, this sum must be constant
over an arbitrary time interval. The total energy of an isolated
system is constant over a time interval, because energy is
conserved. Therefore,
m1
v1+m2
v2
is constant in an isolated
system. This quantity is known as linear momentum.
The linear momentum
p
of an object that can be modelled as a
particle of a mass
m
moving a velocity
v
defined to be the
product of the mass and velocity of the particle:
p ≡m
v
If a particle is moving in an arbitrary direction,
p
has three
components:
px=m vxpy=m v ypz=m v z
Examples:
1.A particle of mass 𝑚 moves with momentum of magnitude 𝑝.
(a) show that the kinetic energy of the particle is 𝐾=𝑝^2/2𝑚.
(b) Express the magnitude of the particle’s momentum in
terms of its kinetic energy and mass.
For (a): Recall:
K=1
2mv2
. We want to prove
K=p2
2m
So,
K=1
2mv2=
(
m v2
2
)
m
m=m2v2
2m=
(
mv
)
2
2m
Note:
p=mv
. Thus,
K=
(
mv
)
2
2m=p2
2m(true)
For (b) From
K=p2
2m→ p2=2mK → p=
√
2mK
2. A 3.00-kg particle has a velocity of (3.00𝑖1−4.00𝑗1 )𝑚/𝑠.(a)
Find its x and y components of momentum. (b) Find the
magnitude and direction of its momentum.
For (a)
px=m vx=
(
3kg
)
(
3m
s
)
=9
^
i kg ∙ m
s
