Momentum
Physicists have defined a quantity called
momentum (動量) of a moving object as...
Momentum = mass  velocity
P = mv
Unit: kg m s1
A vector quantity
1
Newton’s 3rd law and Conservation of
momentum


A
FAB
B
FBA
Newton’s 3rd law
FBA = -FAB
m A v A  m Au A
mB v B  mB u B

t
t
mA v A  mAu A  mB vB  mB u B
 m A u A  mB u B  m A v A  m B v B
If there is no external force acting on a system, then the
total momentum of the system is conserved
2
Example 1
A bullet of mass 10 g traveling horizontally at a speed of
100 ms-1 embeds itself in a block of wood of mass 990 g
suspended by strings so that it can swing freely. Find
(a)
the vertical height through which the block rises, and
(b)
how much of the bullet’s energy becomes internal
energy.

Solution:
(a) Let v be the velocity of the block just after
impact.
By conservation of momentum,
(0.01)(100) = (0.01 + 0.99)v
100 ms-1
v = 1 ms-1
By conservation of energy,
½ mv2 = mgh
½ (1)2 = 10h
h = 0.05 m
h
v
3
Example 1
A bullet of mass 10 g traveling horizontally at a speed of
100 ms-1 embeds itself in a block of wood of mass 990 g
suspended by strings so that it can swing freely. Find
(a)
the vertical height through which the block rises, and
(b)
how much of the bullet’s energy becomes internal
energy.

Solution:
(b) Some of the K.E. is converted into internal
energy.
Energy required
100 ms-1
= ½ mu2 – ½ (m + M)v2
= ½ (0.01)(100)2 – ½ (0.01 + 0.99)(1)2
= 49.5 J
h
v
4
Application 1 – measuring the inertial mass




Weighing machine or beam balance only
measures the weight.
To find the mass (gravitational mass), we must
depend on the equation m = W/g.
However, g varies from place to place and we
cannot use this method to find the mass in the
outer space.
Another way to determine the mass (inertia
mass) without depending on the value of g is to
use the principle of conservation of momentum.
5
Application 1 – measuring the inertial mass

Consider two trolleys of m1 and m2 are in contact. A
spring is used to cause them to explode, moving off the
velocity v1 and v2.
m1
m2
Before explosion
v1
m1
m2
v2
After explosion

If no external force acts on them, conservation of
m2
v1

momentum gives 0 = m1v1 + m2v2 
m1
v2

Therefore, by measuring their velocities, the ratio of their
masses can be found.

If a standard trolley is used, the other mass can be found.
6
Example 2
Compare the masses of 2 objects X and Y which are all
initially at rest. They explode apart and their speeds become
0.16ms-1 and 0.96ms-1 respectively.
m1
m2
Before explosion


0.16 ms-1
m1
m2
0.96 ms-1
After explosion
By conservation of momentum,
m2 0.16 1


0 = m1(-0.16) + m2 (0.96) 
m1 0.96 6
The ratio of the masses m1: m2 = 6:1
7
Applications 2 – Rocket engine



The rocket engine pushes
out large masses of hot gas.
The hot gas is produced by
mixing the fuel (liquid
hydrogen) with liquid oxygen
in the combustion chamber
and burning the mixture
fiercely.
The thrust arises from the
large increase in momentum
of the exhaust gases.
8
Applications 2 – Jet engine



A jet engine uses the
surrounding air for its oxygen
supply.
The compressor draws in air
at the front, compresses it,
fuel is injected and the
mixture burns to produce hot
exhaust gases which escape
at high speed from the rear of
the engine.
These cause forward
propulsion.
9
Recoil of rifles
Two factors affects the recoil speed of a rifle.
1 momentum given to the bullet
2 momentum given to the gases produced by the
explosion.
gas
bullet
rifle
10
Example 3
A rife fires a bullet with velocity 900 ms-1 and the mass of
the bullet is 0.012 kg. The mass of the rifle is 4 kg and the
momentum of gas ejected is about 4 kg ms-1. Find the
velocity of the recoil of the rifle.
-1
gas (4 kg ms-1) 900 ms
v
rifle

bullet
Solution:
By conservation of momentum
0 = 0.012 x 900 + (-4) + 4v
v = -1.7 ms-1
The recoil velocity is 1.7 ms-1 backward
11
mv  mu
Rearrange terms in F =
t
 Ft = mv  mu
impulse: product of force & time during
which the force acts (vector)
Impulse = change in momentum
12
a
force / N
Force-time graph of impact
force / N
time / s
time / s
Area under F-t graph
= impulse
= change in momentum
13
Collisions
General properties
 A large force acts on each
colliding particles for a very
relatively short time.



The total momentum is conserved during a collision if
there is no external force.
In practice, if the time of collision is small enough, we
can ignore the external force and assume momentum
conservation.
For example, when a racket strikes a tennis ball, the
effect due to gravitational force (external force) is
neglected since the time of impact is very short.
14
Collisions

For another example, when a cannon fires a metal ball,
the effect due to frictional force (external force) on the
cannon is neglected since the time of explosion is very
short.
15
Different kinds of collisions

Assume no external force acts on colliding bodies.
Collisions
Momentum
conserved?
K.E. conserved?
Elastic collisions
Yes
Yes
Inelastic collisions
Yes
No
Completely inelastic
collision
Yes
No (K.E. loss
is maximum)
16
Relative velocity rule
(For elastic collisions only)

relative speed before collision = relative speed after
collision
u1
u2
v1
Before collision
v2
After collision
u1 – u2 = v2 – v1
17
Find the velocities of A and B after the elastic
collision.
5
1 kg
ms-1
3 ms-1
v1
2 kg
Before collision



v2
After collision
By conservation of momentum,
(1)(5) + (2)(3) = (1)(v1) + (2)(v2)
v1 + 2v2 = 11
--(1)
By relative velocity rule, 5 – 3 = -(v1 – v2)
v2 – v1 = 2
--(2)
∴ v1 = 2.33 ms-1 and v2 = 4.33 ms-1
18
Trolley A of mass m with velocity u collides elastically with
trolley B of mass M at rest. Find their velocities v and V
after collision.
v
V
At rest
u
A
B
A
B




By conservation of momentum, mu = mv + MV
--- (1)
By relative velocity rule, u = V – v
--- (2)
(1) + (2) x m:
2mu
2mu = (m + M)V ⇒ V 
mM
(1) – (2) x M:
mM
u
mu – Mu = (m + M)v ⇒ v 
mM
19
Trolley A of mass m with velocity u collides elastically with
trolley B of mass M at rest. Find their velocities v and V
after collision.
2mu
mM
v
mM
u
V 
mM
At rest
u
A
B
A
B
Some interesting results

stop after the collision
If m = M, trolley A will ___________________________

If m < M, trolley A will

move in the opposite direction
after the collision
________________________
remain at rest after the collision
If m << M, trolley B will __________________________
20
Collisions in two dimensions
Momentum can be resolved along different directions.
 Consider the following oblique impact. A particle of mass
m with velocity u collides with another stationary particle
of mass M obliquely as shown below.
y
Apply conservation of
Mv2
M
momentum along x-axis,
mu = mv1cos b + Mv2cos a
m
a
mu
x Apply conservation of
b
momentum along y-axis,
0 = Mv2sin a + (-mv1sin b)
mv1

21
A stationary object of mass 500 g explodes into three
fragments as shown below. Find the speed the two
larger fragments v1 and v2 after the explosion.
0.2v2
200 g
200 g
30o
0.2v1
500
45o
After explosion
g
100 g
Before explosion
-1
0.1
x
40
ms
 Along vertical direction:

0 = 0.2 v2 sin 30o – 0.1(40) sin 45o
v2 = 28.3 ms-1
Along horizontal direction:
0 = 0.2v2cos30o + 0.1(40)cos45o – 0.2v1
v1 = 38.6 ms-1
22
Right-angled fork track

If one particle collides with another identical particle
obliquely, the angle between the directions of motion of
the two is always 90o if the collision is elastic.
Significance



Alpha particles travel in a cloud chamber and have
collisions with helium atoms.
The figure above shows after collision the alpha particle
and the helium atom travel at right angle to each other.
This implies alpha particles must have the same mass
as helium atom and actually they are nuclei of helium.
23
Mathematical proof of right-angled fork
track (i.e. q = 90o)
u
a
v2
a collision between identical particles
q
v1
Along the line of centres,
mucosa  mv1 cosq  mv2  u cosa  v1 cosq  v2 (1)
Perpendicular to the line of centres,
musin a  mv1 sin q  u sin a  v1 sin q (2)
For elastic collision,
1
1
1
2
2
2
2
2
2
mu  mv1  mv 2  u  v1  v 2  (3)
2
2
2
24
(1)2 + (2)2:
u cos a  u sin a  v1 cosq  v2   v1 sin 2 q
2
2
2
2
2
2
 u 2  v1 cos2 q  2v1v2 cosq  v2  v1 sin 2 q
2
2
2
 u 2  v1  2v1v2 cosq  v2 (4)
2
2
Sub.(4) into (3):
v1  2v1v2 cosq  v2  v1  v2
2
2
2
2
 2v1v2 cosq  0
 cosq  0
 q  90
The two particles must move at right angle to each other.
25
Elastic collision between a smooth ball
and a fixed surface

Momentum along the fixed surface is unaltered by the
impact.
v
v
Force of impact

Momentum along the fixed surface is unaltered by the impact.
26
Elastic collision between a smooth ball
and a fixed surface

Since the collision is elastic, after the impact, it rebounds
with the same speed.
v



q
f
v
Momentum along the fixed surface is unaltered by the impact.
mv sin q = mv sin f ⇒q = f
⇒ the angle of reflection = the angle of incident
Change in momentum perpendicular to the surface
= mv cos q – (-mv cos f) = 2mv cos q.
Force of impact = 2mv cos q / t
27