5/24/2007 Collisions (© F.Robilliard) 1flai/Theory/lectures/Collisions.pdf · momentum has been transferred, during the collision, to another body of the system, so the total vector

5/24/2007 Collisions (© F.Robilliard) 1


Interactions:

Here are some examples of systems of interacting bodies -

- nuclei colliding with each other in a nuclear reaction.

- molecules colliding with the walls of a container creating gas pressure

- two cars colliding in a traffic accident

- an asteroid colliding with the earth

- the solar system, consisting of the sun and planets moving under mutual gravitational attraction.

In our earlier studies of force and work, we saw, that both these quantities arise in the context of an interaction between two bodies. We will now look more closely at such interactions.Interactions occur between two, or more, bodies over an interval of time. We call the collection of interacting bodies, the system.

-a galaxy composed of stars interacting with each other gravitationally

- a rocket engine ejecting fuel for thrust.

- a proton & an electron, held together by Coulomb attraction in the hydrogen atom.


Collisions:

We aim to predict the outcome of a collision between the bodies. That is -given the velocities of the bodies before the collision, we would like to predict their velocities after the collision.

In theory, this problem could be solved, by determining the instantaneous interaction forces between the bodies, hence finding the accelerations produced, and thence, the final velocities. However, this requires micro-detailed knowledge of the non-constant interaction forces, and is complex to undertake.

For simplicity, we will focus on systems of two bodies, that interact by collision.

We are going to look more closely at these simultaneous momentum and energy transfers.

For any such system, interaction forces act between the component bodies. Associated with these interaction forces, momentum and energy are transferred from one body in the system, to another.

An alternative is to use a systems approach. We try to relate the overall properties of the system before, to those after, the collision. The properties in question here, are momentum, and energy.


Impulse of a Force:

( )( ) ( )1................md .dt thus

mdtd

N2

vF

vF

=

=

( ) ( )2............m .dt tover

vF ∆=�∆

We will firstly look at system momentum, by revisiting Newton’s Second Law.

Consider a mass, m, on which a force, F, acts:

Say the magnitude of the force, F, varies over an interval of time, ∆t, something like -

F

t

∆t

Integrate both sides of (1) -

and ∆(mv) is the total change in momentum of the mass, that is produced by the varying force, in time interval ∆t.

.dt tover �

∆

F is called the total impulse of the force, F, on the mass, over time interval ∆t.

mF

v

Force, F, acting for tiny time interval, dt, will produce a change in momentum of d(mv)


Impulse-Momentum:

( ) ( )2............m .dt tover

vF ∆=�∆

(2) is called the Impulse-Momentum equation. It is simply Newton 2 in disguise.

Considering the total effect of the force, F, on the mass, m, over time interval ∆t:

F

t

∆t

mF

v

��

�

�

��

�

�

=��

�

�

��

�

�

produced momentumin

change Total

mmass, aon F, force, of

impulse Total

graph.t -F under the area the toequal is .dt Impulsetover �

∆

F

Equation (2) says -


During a Collision:

( ) ( ) ( )( ) ( ) 0m�M�

2 ....fromm�M� Thus

.dt.dt�t�t

=+−=

−=

−=

��

vVvV

fF

fF

m M

F

f

F t

∆t

f

force

But, by Newton 3, F and f are, at every instant, equal, and opposite.

Consider a Collision between two masses, m and M.

F = force on M, due to m, during the collision.f = force on m, due to M, during the collision.

There is no change in the total momentum of the two masses, due to the collision.

We treat the two masses as a system. We can see from the above graph, that the total impulse on the system ( = total area under the Force-time graph ), due to collisional forces, F and f, is zero, since F and f are equal, but opposite. Hence the total change in momentum of the system must also be zero.

As the bodies come into contact, each exerts a force on the other.


Comparing Before with After:

�� =

�

��

�

=

�

��

�

systemf

systemi

collision after thesystem of

momentum vector total

collision thebeforesystem of

momentum vector total

pp

Where the Σ is a sum over the system, p represents vector momentum, i = initial, and f = final.

Since the total momentum of the system is constant, we say that momentum has been conserved, for the system.

This is the principle of conservation of momentum, and is usually stated formally as follows -

In using this principle, it is critically important to treat momentum as a vector


Note:

We have assumed that the only forces acting on the bodies of the system are the interaction forces of the collision. We have assumed that there are NO net external forces, such as friction, or gravity, acting on the system, which wouldchange its overall momentum.

We have found, that, if one of the bodies loses momentum, then all of that momentum has been transferred, during the collision, to another body of the system, so the total vector momentum has not changed.

Momentum must be treated as a vector – direction is important!.

Conservation of momentum applies not only to colliding blocks, but to other interacting systems, such as the solar system, galaxies, molecules of a gas, interacting elementary particles.

We now apply conservation of momentum to the case of two blocks, firstly in a 1-dimensional, and then in a 2-dimensional, collision.


1-dimensional collision of two masses, m and M:We depict the system, before, and then after the collision.

m M

u U

Before +x

m M

v V

After +x

u = initial velocity of mU = initial velocity of M

v = final velocity of mV = final velocity of M

MVmvMUmusystem

fsystem

i

+−=−+

= �� pp

Note: the negative signs are due to those momenta being in the (-x)-direction

Note: we have made an assumption about the direction of velocities, after the collision. If our assumption is correct, we will get a positive solution, if wrong, a negative solution, for the particular velocity. We need to interpret these solutions.


2-dimensional collision of two masses, m and M:

( )1........MV.cosmv.cosmu

:direction-In x

systemf

systemi

ααααββββ +−=+

= �� pp

m Mx

y

u

Before m

M

x

y

αβ

V

v After

We need to apply conservation of momentum in both the x- and y- directions.

( )2........MV.sinmv.sin0

:direction-yIn

systemf

systemi

ααααββββ +−=

= �� pp

Note: the angles α and β will depend on the details of how the surfaces collided. Note: Equations (1) and (2) need to hold simultaneously.

m collides with a stationary M. M is knocked on at an angle α, while m rebounds at angle β.


Is Momentum Sufficient?

1kg 1kg

10m/s 10m/s

Before +x

Two 1 kg blocks approach each other at 10 m/s, along the x-axis. What will be their velocities after they collide?

Experiment shows, that there are many possible outcomes, depending on the details of the interaction at the colliding surfaces – for example, whether or not elastic limits have been exceeded, resulting in permanentdeformation of the surfaces.

For three representative outcomes, we will consider the total momentum, p, and also the total energy associated with the motion (= total KE), of the colliding blocks, both before, and after the collision.

Is momentum conservation sufficient to determine the outcome of a collision?

Let’s reconsider the collision of two blocks in 1-dimension.


3 PossibleOutcomes:

1kg 1kg

10m/s 10m/s

Before +x

1kg 1kg

10 10

After +x

1kg 1kg

0 0

After +x

1kg 1kg

5 5

After +x

Initial total p & KE:Σp = +(1x10) - (1x10)

= 0ΣKE = ½.1.102 + ½.1.102

= 100 J

Σp = +(1x10) - (1x10) = 0ΣKE = ½.1.102 + ½.1.102 = 100J

KE is conserved –Elastic collision

Σp = +(1x5) - (1x5) = 0ΣKE = ½.1.52 + ½.1.52 = 25J

75% KE lost –Inelastic collision

Σp = +(1x10) - (1x10) = 0ΣKE = 0 + 0 = 0J

All KE lost –Perfectly inelastic collision

Momentum is conserved in all three cases.

KE only conserved for “elastic collision”.


Energy:We see that momentum alone does not determine the outcome of a collision –energy must also be taken into account.

During the collision, surfaces of colliding bodies are deformed. If elastic limits are exceeded, permanent deformation of the surfaces results. The deformation forces do work, and result in a conversion of energy from the form of KE of the bodies, into heat. To predict the outcome of a collision, we need to know how much energy is converted to other forms, such as heat, and consequently lost to the system.

Elastic Collisions: no KE lost.Inelastic Collisions: some, or all, of the KE is lost

We know from our studies of energy, that the principle of conservation of energy must hold for all systems, including colliding ones.

��

�+

��

�=

��

�

lossesEnergy

collision after thesystem ofEnergy Total

collision thebeforesystem ofEnergy Total

We will assume that the only form of energy possessed by colliding bodies, before the collision, is kinetic. After the collision, this energy will either stay in the form of KE, or be converted ( lost from the system ) to some other form of energy, such as heat.


Conservation of Energy:

( ) ( ) lostfi EKEKE

lossesEnergy

collision after thesystem of KE Total

collision thebeforesystem of KE Total

+Σ=Σ

��

�+

��

�=

��

�

For all collisions -

i = initial;f = final

We need to know the energy losses, in order to use this equation

If the collision is elastic, the energy losses are zero, all energy stays in the form of kinetic, and the conservation of energy becomes -

( ) ( )fi KEKE

collision after thesystem of KE Total

collision thebeforesystem of KE Total

Σ=Σ

��

�=

��

�

(We have the “conservation of KE”.)


Summary:

is constrained by

m M

u U

Before +x

m M

v V

After +x

u = initial velocity of mU = initial velocity of Mv = final velocity of mV = final velocity of M

MVmvMUmusystem

fsystem

i

+−=−+

= �� pp

A collision between two bodies:

Conservation of Momentum:

( ) ( )

lost2222

lostfi

E21

21

21

21

EKEKE

++=+

+Σ=Σ

MVmvMUmu

Conservation of Energy:

Assuming no external forces change the overall momentum of the system

where Elost = 0 if the collision is Elastic.

Note: that momentum is a vector whose direction must be considered, whereas energy is a scalar, and KE is always positive.


To Solve Collision Problems:

1. Draw a labeled diagram of before, and after

2. Assume a positive x-direction ( and y-direction) .

3. Write conservation of momentum equation

4. Write the conservation of energy equation, if losses are known.

5. Solve the equations for unknowns.

6. Interpret your solution.


Example 1:

An arrow, of mass 0.5 kg, is fired into a stationary, freely hanging sandbag, of mass 10 kg, causing it to swing forward with a velocity of 2 m/s.Find the velocity of impact of the arrow.

Mu

rest

m

Before+x

( )

m/s 420.20.5

100.5

Vm

Mmu Thus

V Mm0mu

ppsystem

fsystem

i

=��

��

� +=

��

��

� +=

++=++

= ��

The collision is inelastic and we do not know how much energy was lost –. We can only use momentum conservation. The arrow’s velocity was 42 m/s.

M

V

m

After

+x


Example 2:A 3 kg body moving with a velocity of 2 m/s makes a 1-dimensional, elastic collision with a stationary mass of 9 kg. Find the velocities of the two masses after the collision.

In this problem, we will need to use both momentum, and energy conservation.

m=3kg M=9kg

u=2m/s

Before +x

rest v V

After +x

m=3kg M=9kg

We assumedirections of the bodies, after the collision

( )( )( )1.....3V........-v2

9V3v023system

fsystem

i

+=+−=++

= �� pp ( ) ( )

( )

( )2....................3Vv4

9V21

3v21

02321

KEKE

22

222

fi

+=

+=+

Σ=Σ

Momentum: Collision is elastic:


Example 2 :

Solve (1) and (2) simultaneously for v and V:

(1): v = 3 V -2………………….(3)(3)->(2): 4 = (3 V-2)2 + 3 V2

4 = 9V2 – 12 V + 4 + 3V2

0 = 12 V (V - 1)thus: V = 0 or +1………………….(4)

(4)->(3): v = -2 or +1 correspondingly

2m/s V=0

After1 +x

m=3kg M=9kg

1m/s 1m/s

After2 +x

m=3kg M=9kg

The two solutions are illustrated.

Clearly After1 is the same situation as Beforeand represents the situation where no collision took place, and is thus invalid.

After2 represents the valid solution.

m=3kg M=9kg

u=2m/s

Before +x

rest

2 = -v + 3V …..….(1)4 = v2 + 3V2 …..…(2)


Some Rocket Science:As a final application of the conservation of momentum, we look at the acceleration of a rocket in space – we do some rocket science!

To accelerate forward, the rocket ejects fuel particles backward. The rocket accelerates forward because of the reaction force, forward, on the rocket. However, the acceleration of the rocket depends on the rocket’s mass, which includes the mass of the on-board fuel. As fuel is ejected, however, the total mass of rocket and fuel becomes less. Therefore, even with a constant thrust forward, the acceleration of the rocket is NOT uniform.

Say a rocket, of total mass m, ejects fuel particles of mass, dm, at an ejection velocity, ve, relative to the rocket.

mdm

v

Before

mdm

v + dv

veAfter

Say the ejection of the fuel, causes the rocket’s velocity to increase from v, to (v + dv)


The Change in Velocity:

( ) ( ) ( )[ ]

( )

( )( ) ( )

��

�

=

→=+++=++

= ��

. .pos increasean is dv whereas,neg. decrease a is dm since ,1in needed is "-"

1............... m

dm -

vdv

so

0dm.dv since vdm dv m thence

dv v m v-dv v dm vdm m

e

e

e

systemf

systemi pp

Conservation of momentum:

mdm

v

Before +x

+y

mdm

v + dv

veAfter +x

+y

Velocities must be relative to the xy axes. But ve is, by definition, the velocity of the fuel, relative to the rocket. (See “After” picture, above.)

mdm

v

Before +x

+y

mdm

v + dv

veAfter +x

+y

A space journey consists of three stages - acceleration, cruise, and deceleration. We will calculate the fuel for the acceleration, then double it for the deceleration. No fuel is used to cruise.


The Rocket Equation:

m

vi

Initial velocity

m

vfa

Final velocity

Say the rocket starts with an initial velocity vi, and accelerates, over a

period of time, to a final velocity vf.

( )[ ]

( )2.... mm

ln v v- v

mln - vv

mdm

- vdv

:(1) Integrate

f

ieif

mmmm

vv

vve

m

m

v

v e

f

i

f

i

f

i

f

i

��

��

�=

=

��

�

=

==

=

=

��

Equation (2) is the “rocket equation”. We see that the total change in velocity of the rocket, for a given amount of fuel used, is proportional to the ejection velocity, ve, of the fuel particles. Therefore high fuel ejection velocities are desirable. Also, as would be expected, the more fuel used, the greater the velocity change.

Once ve is set, and the velocity, vf, in the cruise stage of a space trip (which determines the time of the trip), has been decided, the rocket equation allows us to calculate the total fuel mass needed for the trip.


Rocket Thrust:

mve

F( )

( )3.............. dtdm

v F

(1) using m

dmv

dtm

2Newton dtdv

m F Thrust

e

e

=

=

=≡The rocket thrust is easily found:

Rocket thrust is proportional to the ejection velocity, and to the rate of ejection, of fuel.

Example: Fuel particles are ejected from the thrusters of a space vehicle at 5 km/s, relative to the vehicle. Starting from rest, what velocity will the vehicle reach, after half of its total mass has been ejected at a rate of 2 kg/s, as fuel particles, by the thrusters. Find the thrust.

( )

( ) m/s 3.5x10 m

mln 5x10 v

2.... mm

ln v v- v

3

i21

i3f

f

ieif

=��

��

�=

��

��

�= ( )

( ) thrustkN 01

N 1x10 2 5x10

3 from dtdm

v F

43

e

===

=Finalvelocity: Thrust:


Summary:When bodies interact, such as in collisions, momentum and energy are transferred from one body to another.

If no momentum enters, or leaves, a system of bodies, the total vector momentum of the system remains constant -principle of conservation of momentum.

Knowing the initial state of a system, we can find the final state of that system, without having to analyse the moment-to-moment history of the interaction process itself.

If no energy enters , or leaves, a system of bodies, the total energy of the system remains constant -principle of conservation of energy.

Finally we were able to apply these ideas to some rocket science.


Documents

5/24/2007 Collisions (© F.Robilliard) 1flai/Theory/lectures/Collisions.pdf · momentum has been transferred, during the collision, to another body of the system, so the total vector