Momentum Student Notes

7/29/2019 Momentum Student Notes

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1. FURTHER MECHANICS

1.1 MOMENTUM

Momentum is a quantity possessed by moving objects.

Momentum is the product between mass and velocity. Being a vector quantity, it

has a direction (its direction is the same as that of velocity), and the direction isvery important when doing momentum calculations. Momentum is not a thing that wecan see, but it does explain many things that go on in physics.

Momentum (kg m/s) = mass (kg) x velocity (m/s)

p = mv

Units are kilogram metres per second (kg m/s) or Newton seconds (N s). We canshow that the units are the same.

1 N = 1 kg m s-2 (Newton II)

1 kg m s-1 = 1 kg m s-2 s = 1 N s

Example 1

What is the value of the momentum of a 10 kg ball running down a bowling alley at aspeed of 5 m/s?

Answer

Formula first:p = mv

p = 10 kg x 5 m/s = 50 kg m/s

In the example above, we only looked at the value of the momentum. This is why weused the word speed. It is very important to make sure that you pay attention to thesigns when doing momentum calculations.


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Think of a ball bouncing off a wall. It leaves the wall at the same speed as before. Letscall going from right to left negative, and going from left to right positive.

Example 2

Can you show that the change in momentum is +2 mv?

Answer

Change in momentum = momentum after - momentum before.

Change in momentum = + mv - - mv = + 2mv

Example 3

The ball has a mass of 200 g, and the value of its velocity throughout remains 6m/s. What is the change in momentum?

Answer

Change in momentum = momentum after - momentum before.Change in momentum = + mv - - mv = + 2mv

Change in momentum = + 0.2 kg x 6 m/s - - 0.2 kg x 6 m/s = +2.4 kg m/s


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ImpulseThe change in momentum is called the impulse and is given the physics code p. We

can define Newtons Second Law in terms of change of momentum:

Force = mass acceleration

Force = mass change in velocity

time interval

Change in velocity = velocity at end - velocity at start

Change in momentum = mass change in velocity

Therefore:

Force (N) = change in momentum (Ns)Time interval (s)

Impulse (Ns) = Force (N) x time interval (s)

In code:

p = Ft

If we plot a force time graph, we can see that impulse is the area under the graph.

In this graph, both impulses are the same. The forces and time intervals are different.


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Impulse and Newton's Second LawConsider an object of mass m which is subject to a force Ffor a time period oftseconds.

We can say that there is an impulse on the object:

F = p/t = mv/t = mv/t

the term v/t= change in velocity time interval = acceleration

So Force = mass acceleration which is Newton II.

Remember that wherever there is a resultant force, there is always acceleration. Achange in momentum results in acceleration, which is caused by a force.

Example 4

Can you explain how the formula F =p/

tis consistent with Newton II?

Answer

The formula says that the force = change in momentum / time interval.

Change in momentum = momentum after - momentum before.

In Physics code we write: p = mv - mu

We can write the whole thing in physics code:

F = m (v-u)

t

v-u is change in velocity, and change in velocity / time interval = acceleration

So we can write Force = mass x acceleration which is commonest expression of Newton II


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Example 5

A car is involved in a collision in which it is brought to a standstill from a speed of 24m/s. The driver of mass 85 kg is brought to rest by his seat belt in a time of 400 ms.

a) Calculate the average force exerted on the driver by his seatbelt.

b) Compare this force to his weight and hence work out the g-force

c) Comment on the likelihood of serious injury.

a) Calculate the average force exerted on the driver by his seat belt.

First we need to work out the change in momentum (impulse)p = mv - mu = 0 - 85 kg x 24 m/s = 2040 Ns

Now we can work out the forceF = p/t= 2040 Ns 0.400 s = 5100 N

b) Compare this force to his weight and hence work out the g-force

The weight of the driver = 85 kg x 9.81 N/kg = 834 NThe g-force = 5100 N 834 N = 6.1 g (i.e. 61 m/s2)

c) Comment on the likelihood of serious injury.

This will reduce the likelihood of serious injury as the body withstand accelerationsof up to 8 g. Aerobatic pilots regularly pull 5 to 6 g in their manoeuvres.


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Conservation of Momentum

We have seen how momentum is the product between mass and velocity. Wecannot see a momentum. But it is a concept that is essential in explaining what happensin collisions and explosions.

An important principle:

The t o t a l mom en tum o f a syst em rema i n s const an t p r o v i ded t ha t no ex t e r na l

fo r ces ac t on t he syst em.

Or

For a system of interacting objects, the total momentum remains constant,provided no external resultant force acts on the system.

This has important implications in the study of collisions. In simple terms, we can saythat the total momentum before = total momentum after.

The key thing is that share of the momentum may change.

The system may consist of several elements, each of which has its own momentum.The total momentum is the sum of all these different momenta. As long as the totalmomentum remains the same, momentum can be shared out differently between eachelement. In other words, each body can exchange momentum with the other bodies, andget a different share, as long as the total momentum stays the same.

We will only consider the momentum of two objects, acting in a straight line. This hasimportant implications in the study of collisions. Remember:

The t o t a l m om en tum o f a syst em rema i n s const an t p r o v i ded t h a t no ex t e r nal

fo r ces ac t on t he syst em.

When any two objects in a system interact, they exert equal and opposite forces on oneanother. (Newtons third law)

As a result, one of the objects gain momentum from the other object. The total momentum stays the same because the gain of momentum by one object is due to

an equal loss of momentum by the other.


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Consider bodies 1 and 2 below;

The force on ball 1 from ball 2 changes the velocity of 1 from u1 to v1, so

F =

Where t is the time of contact between ball 1 and 2, and m1 is the mass of 1.

An equal and opposite force F on ball 2 from 1 changes the velocity of 2 from u2 to v2,

F =

Because the two forces F and F are equal and opposite to one another, then

F = F

= - (

)

m1v1 + m2v2 = m1u1 + m2u2

total final momentum = total initial momentum

In other words, the total momentum is unchanged by the collision.


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Explosions

An explosion where objects fly apart is another example where momentum is conserved.

No external forces act during an explosion, only internal forces which push the objects apart.

Consider a bullet of mass m from a gun of mass M. After the explosion which drives the bullet out

of the gun barrel, the gun barrel recoils.

The momentum carried away by the bullet is mv where v is the speed of the bullet. The gun barrel

recoils at speed V, so the momentum carried away is MV (- since the direction is opposite to the

bullets momentum).

Therefore the total final momentum = mv MV.

Since the bullet and gun were at rest before the explosion, the total initial momentum is zero.

Hence , by the conservation of momentum principle, mv- MV = 0 which gives;

MV = mv

CollisionsMomentum is always conserved in collisions.

If objects bounce off each other, the collision is elastic. If the total kinetic energy is thesame (conserved) at the end as it is at the start, then the collision isperfectly elastic. The rebound of particles against each other tends to be perfectlyelastic. A tennis ball bouncing off the floor is not perfectly elastic as it can lose up to 25

% of its kinetic energy in doing so.

If some kinetic energy is lost, converted into heat or light, then the collision is inelastic.

Think about two objects travelling in the same direction. The table below shows theproperties of the objects:

Prope r t y Large Ob jec t Sm al l Ob jec t

Mass M m

Initial velocity u1 u2

Final velocity v1 v2


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We can show this as a diagram:

There are two important principles here:

1. Conservation of momentum:

Total momentum before = total momentum after

Mu1 + mu2 = Mv1 + mv2

2. Energy is conservedTotal energy before = total energy afterMu1

2 + mu22 = Mv1

2 + mv22 + E

The term E is the energy that is lost in the collision. In a perfectly elastic collision E= 0.

Particle collisions

When considering quantum physics, you will meet the idea that light has momentum (but it doesnt make any

sense to use the classical physics definition ofmv for light that by definition travels at the speed of light, so

Einsteins Relativity theory is needed).

When considering sub-atomic particles, you may use the equation KE =p2/2m. You will also learn how the

momentum of a particle can be read from the curvature of its track.

From;

Ek=

mv

2 (i)

And

P = mv (ii)


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And from (ii) v =

Putting v =

in (i)

Ek=

m (

)2

Ek=

For particles such as a photon (a packet/particle of light), the de Broglie wavelength;

=

From Ek=

p =

2Ekm

=

2Ekm

When doing momentum calculations, always be careful about the directions you areusing.


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Example 2

The diagram shows two cars at a fairground, before and after bumping into eachother. One car and driver has a total mass of 500 kg, while the other car and driver has atotal mass of 400 kg.

(a) What is (i) the total kinetic energy before the collision;(ii) the total kinetic energy after the collision.(iii) the total loss in kinetic energy.

(b) Is this an elastic collision? Explain your answer.

Answer

(a) What is (i) the total kinetic energy before the collision;

Kinetic Energy = 1/2 mv2 Kinetic Energy of blue car = 1/2 x 500 kg x (5 m/s)2 = 6250 J Kinetic Energy of blue car = 1/2 x 400 kg x (2 m/s)2 = 800 J Total energy = 6250 J + 800 J = 7050 J

(ii) the total kinetic energy after the collision.

Kinetic Energy = 1/2 mv2 Kinetic Energy of blue car = 1/2 x 500 kg x (3 m/s)2 = 2250 J


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Kinetic Energy of blue car = 1/2 x 400 kg x (4.5 m/s)2 = 4050 J Total energy = 2250 J + 4050 J = 6300 J

(iii) the total loss in kinetic energy.

Total loss = 7050 J - 6300 J = 750 J

(b) Is this an elastic collision? Explain your answer.

It is not elastic (or it's inelastic), because there has been energy lost.

Example 3

A second collision is shown below:

What is the speed of the 500 kg car after the collision?

Answer

Momentum before = momentum after Momentum before = m1u1 + m2u2 Momentum before = (500 kg x 5 m/s) + (400 kg x 2 m/s) Momentum before = 2500 kg m/s + 800 kg m/s = 3300 kg m/s


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Example 4

A bullet of mass 45 g is travelling horizontally at 400 m/s when it strikes a wooden blockof mass 16 kg suspended on a string so that it can swing freely. The bullet is embeddedin the block.

Calculate:

a) The velocity at which the block begins to swing;b) The height to which the block rises above its initial position;

c) How much of the bullets kinetic energy is converted to internal energy.


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Answer

Calculate:

a) The velocity at which the block begins to swing;

Momentum before = momentum after Momentum before = m1u1+ 0 Momentum before = (0.045 kg x 400 m/s) + (0) Momentum before = 18 kg m/s

Momentum after = total mass of bullet and wood x speed Momentum after = (16.000 kg + 0.045 kg) x v m/s Momentum after = 16.045 v kg m/s

Therefore:

16.045 v kg m/s = 18 kg m/s v = 18 kg m/s 16.045 = 1.12 m/s

b) The height to which the block rises above its initial position;

We use the conservation of energy principle:

Kinetic energy = potential energy 1/2 mv2 = mgh m terms cancel out.

Rearranging:

h = v2 = 1.122 (2 x 9.8 ms2) = 0.057 m2g


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c) How much of the bullets kinetic energy is converted to internal energy.

We need to know the kinetic energy of the bullet:

Ek= 1/2 mv2= 1/2 x 0.045 kg x (400 m/s)2 = 3600 J

Now we need to know the kinetic energy of the block and bullet:

Ek = 1/2 x 16.045 kg x (1.12 m/s)2 = 10.1 J

Kinetic energy lost = 3600 J - 10.1 J = 3590 J

This energy is not destroyed, but converted into internal energy.

This question is a typical question you will find in the exam. It is often called synoptic,because it tests several different concepts at the same time.

Momentum in ExplosionsAn explosion not just where something goes BANG. In physics it is any situation wherethere is zero momentum at the start. Since linear momentum is conserved, it means that

the total momentum at the end must be zero. Consider two railway wagons that arebuffered up very tightly and the springs in the buffers are ready to push them apart.

When the wagons are released, they fly apart in opposite directions as in the picturebelow:


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Since momentum is conserved and the momentum at the start was zero, we can write:

0 = m1v1 + m2v2

which we can rearrange to:

m1v1 = -m2v2

Worked example

Wagons 1 and wagon 2 are buffered up tightly together, but NOT coupled together. Thebrakes on the wagons are released at the same time. The release of the springs makeswagon 2 move to the right at a velocity of 0.10 m/s. What is the velocity of wagon 1?

Momentum at start = 0:

Momentum at end = 0

Momentum of wagon 2 = 20000 kg +0.1 m/s = +2000 kg m/s

Momentum of wagon 1 = 0 kg m/s - +2000 kg m/s = -2000 kg m/s

Velocity of wagon 1 = -2000 kg m/s 15 000 kg = -0.13 m/s from right to left.


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Example 5

A cannon has a mass of 1500 kg. It fires a shell of mass 25 kg. It recoils at a velocity of10 m/s. What is the velocity of the shell?

Answer

Momentum before = 0

Momentum after = 0 = 1500 -10 + 25 v

25 v = 15000

v = 600 m/s (from left to right)

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