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.

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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

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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.

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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)

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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

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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

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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

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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

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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

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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.

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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

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)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