1
A vector quantity
Unit: kg m s1
Momentum = mass velocity
Physicists have defined a quantity called momentum ( 動量 ) of a moving object as...
Momentum
P = mv
2
Newton’s 3rd law and Conservation of momentum
Newton’s 3rd law FBA = -FAB
FAB FBA
A B
t
umvm
t
umvm BBBBAAAA
BBBBAAAA umvmumvm
BBAABBAA vmvmumum
If there is no external force acting on a system, then the total momentum of the system is conserved
3
Example 1A 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)vv = 1 ms-1
By conservation of energy,½ mv2 = mgh½ (1)2 = 10hh = 0.05 m
100 ms-1
h
v
4
Example 1A 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
= ½ mu2 – ½ (m + M)v2
= ½ (0.01)(100)2 – ½ (0.01 + 0.99)(1)2
= 49.5 J
100 ms-1
h
v
5
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.
6
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.
v1 m2m1m1 m2
v2
Before explosion After explosion
If no external force acts on them, conservation of
momentum gives 0 = m1v1 + m2v2
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.
2
1
1
2
v
v
m
m
7
Example 2Compare 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.
0.16 ms-1
m2m1m1 m2
0.96 ms-1
Before explosion After explosion
By conservation of momentum,
0 = m1(-0.16) + m2 (0.96)
The ratio of the masses m1: m2 = 6:1
6
1
96.0
16.0
1
2 m
m
8
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.
9
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.
10
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
11
Example 3A 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.
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
gas (4 kg ms-1)
bulletriflerifle
900 ms-1
v
12
Ft = mv mu
Rearrange terms in
Impulse = change in momentum
impulse: product of force & time during which the force acts (vector)
F = t
mv mu
13
Area under F-t graph = impulse= change in momentum
a Force-time graph of impactforce / N
time / s
force / N
time / s
14
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.
15
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.
16
Different kinds of collisions Assume no external force acts on colliding bodies.
Collisions Momentum conserved?
K.E. conserved?
Elastic collisions
Inelastic collisions
Completely inelastic collision
Yes
Yes
Yes
Yes
No
No (K.E. loss is maximum)
17
Relative velocity rule (For elastic collisions only) relative speed before collision = relative speed after
collision
u1u2 v1
v2
Before collision After collision
uu11 – u – u22 = v = v22 – v – v11
18
Find the velocities of A and B after the elastic 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
5 ms-1 3 ms-1
v1v2
Before collision After collision
1 kg 2 kg
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.
By conservation of momentum, mu = mv + MV --- (1) By relative velocity rule, u = V – v --- (2) (1) + (2) x m:
2mu = (m + M)V ⇒
A BA Bu
At restv V
(1) – (2) x M:mu – Mu = (m + M)v ⇒
Mm
muV
2
uMm
Mmv
20
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.
Some interesting results
If m = M, trolley A will ___________________________
If m < M, trolley A will
________________________
If m << M, trolley B will __________________________
A BA Bu
At restMm
muV
2u
Mm
Mmv
stop after the collisionstop after the collision
move in the opposite direction move in the opposite direction after the collisionafter the collision
remain at rest after the collisionremain at rest after the collision
21
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.
Apply conservation of momentum along x-axis, mu = mv1cos + Mv2cos
Apply conservation of momentum along y-axis, 0 = Mv2sin + (-mv1sin )
mu
mv1
Mv2
x
y
m
M
22
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.
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
500 g
Before explosion
200 g200 g
100 g
0.2v1
0.2v2
45o
30o
0.1 x 40 ms-1
After explosion
23
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.
24
Mathematical proof of right-angled fork track (i.e. = 90o)
u
v2
v1
a collision between identical particles
)1(coscoscoscos 2121 vvumvmvmu
)2(sinsinsinsin 11 vumvmu
Along the line of centres,
Perpendicular to the line of centres,
For elastic collision,
)3(2
1
2
1
2
1 22
21
222
21
2 vvumvmvmu
25
)4(cos2
sincos2cos
sincossincos
2221
21
2
221
2221
221
2
221
221
2222
vvvvu
vvvvvu
vvvuu
90
0cos
0cos2
cos2
21
22
21
2221
21
vv
vvvvvv
(1)2 + (2)2:
Sub.(4) into (3):
The two particles must move at right angle to each other.
26
Elastic collision between a smooth ball and a fixed surface Momentum along the fixed surface is unaltered by the
impact.
Momentum along the fixed surface is unaltered by the impact.
v v
Force of impact
27
Elastic collision between a smooth ball and a fixed surface Since the collision is elastic, after the impact, it rebounds
with the same speed.
Momentum along the fixed surface is unaltered by the impact.
mv sin = mv sin ⇒ =
⇒ the angle of reflection = the angle of incident Change in momentum perpendicular to the surface
= mv cos – (-mv cos ) = 2mv cos . Force of impact = 2mv cos / t
v v