MOMENTUM

1. Momentum is the "quantity of motion" possessed by a body. More precisely, it is defined as:
The dimensions of momentum are
Mass of the body × Velocity of the body.
-1 and the units of momentum are kg-m/s.
2. Momentum is a vector quantity and has both magnitude and direction, v. We can
easily see that Newton's Second Law can be reexpressed in terms
MLT
p=m􀁇
of momentum. When
I wrote it down originally, it was in the form a . But since v a, this can also be
written as (new form). In words, the rate of change of momentum of a bod
m d
dt
d
dt
= =
=
F
p F
􀁇 􀁇 􀁇􀁇
􀁇 􀁇
y equals
the total force acting upon it. Of course, the old and new are exactly the same,
( ) .
3. When there are many particles, then th
d dm md m
dt dt dt
p= v= v= a=F
􀁇 􀁇 􀁇􀁇 􀁇
1 2
1 2
1 2
e total momentum is,
Thi
N
N
N
d d d d
dt dt dt dt
= + +⋅⋅⋅
= + + ⋅ ⋅ ⋅
= + + ⋅ ⋅ ⋅ =
P
P p p p
P p p p
F F F F
􀁇
􀁇 􀁇 􀁇􀁇
􀁇 􀁇 􀁇􀁇
􀁇 􀁇 􀁇􀁇
s shows that when there are several particles, the rate at which the total momentum
changes is equal to the total force. It makes sense!
4. A very important conclusion of the above is that if the sum of the total external forces
vanishes, then the total momentum is conserved, 0 = 0. This is quite
independent of what sort of forces act between the bodies - electric, gravitati
ext

F d
dt
Σ = ⇒ P
􀁇 􀁇
onal, etc. -
or how complicated these are. We shall see why this is so important from the following
examples.
5. Two balls, which can only move along a straight line, collide with each other. The
1 1 2 2 1 1 2 2
initial
momentum is and the final momentum is v v . Obviously one
ball exerts a force on the other when they collide, so its momentum changes. But, from the
fact that th
i f P=mu +mu P =m +m
1 1 2 2 1 1 2 2 ere is no external force acting on the balls, , or v v .
6. A bomb at rest explodes into two fragments. Before the explosion the total momentum is
i f P=P mu +mu =m +m
zero. So obviously it is zero after the explosion as well, . During the time that the
explosion happens, the forces acting upon the pieces are very complicated and changing
rapidly with
Pf =0
1 1 2 2 1 1 2 2
time. But when all is said and done, there are two pieces flying away with a
total zero final momentum v v . Hence v v . In other words, the
fragments fly apart with equal momentum
f P =m +m m =−m
but in opposite directions. The centre-of-mass
stays at rest. So, knowing the velocity of one fragment permits knowing the velocity of
the other fragment.
0
7. If air resistance can be ignored, then we can do some
interesting calculations with what we have learned.
So, suppose a shell is fired from a cannon with a speed
10 m/s at an angle 60 with the horizontal. At the highest
point in its path it explodes into two pieces of equal
masses. One of the pieces retraces its path to the cannon.
Let us find the velocity of the other piec
1 1 2
2 2 2 2
e immediately
after the explosion.
Solution: After the explosion: 5 (why?). But 10cos60
2
5 5 . Now use: v v 15 / .
2 2
8. When yo
x x x x
x x x x
P M P P P M
P M M P M ms
= − + = = ×
⇒ = + = ⇒ =
u hit your thumb with a hammer it hurts, doesn't it? Why? Because a large
amount of momentum has been destroyed in a short amount of time. If you wrap your
thumb with foam, it will hurt less. To understand this better, remember that force is the
rate of change of momentum: . Now define the as:
F dp dp Fdt I
dt
= ⇒ = impulse
force × time over which the
2 2
1 1
If the force changes with time between the limits , then one should define as,
. Since , therefore . In words, the change of momentum
equal
f
i
t t p
f i
t t p
I
I=∫Fdt ∫Fdt=∫dp I=p −p
force acts.
s the impulse, which is equal to the area under the curve of force versus time. Even
if you wrap your thumb in foam, the impulse is the same. But the force is definitely not!
Δt
9. Sometimes we only know the force numerically
(i.e. there is no expression like F=something).
But we still know what the integral means: it is
the area under the curve of force versus time.
The curve here is that of a hammer striking a
table. Before the hammer strikes, the force iszero, reaches a peak, and goes back to zero