Momentum and Momentum and Collisions - Spada UNS

Momentum and Momentum and Momentum and Momentum and

Collisions

Collisions

Collisions

Collisions

Linear Momentum

„

The linear momentum of a particle or an

„

The linear momentum of a particle or an object that can be modeled as a particle of

mass m moving with a velocity v g y is defined to be the product of the mass and velocity:

‰ pp = m v

„ We will usually refer to this as “momentum”, omitting the

“linear”.

Momentum is a VECTOR

„ Momentum is a VECTOR.

„ It has components. Don’t forget that.

Linear Momentum

„ The dimensions of momentum are ML/T

„ The SI units of momentum are kg · m / s

„ Momentum can be expressed in component form:

‰ p_x = m v_x p_y = m v_y p_z = m v_z

A 3.00-kg particle has a velocity of . (a) Find its x and y components of momentum. (b) Find the magnitude and direction of its momentum.

direction of its momentum.

Newton and Momentum

„

Newton called the product Newton called the product mv mv the the quantity of quantity of motion of the particle

„

Newton’s Second Law can be used to relate the momentum of a particle to the resultant force acting on it

( )

d d m d

m m

dt dt dt

Σ = = v = v = p

F a

with constant mass.

Newton’s Second Law

„ The time rate of change of the linear momentum of g a particle is equal to the net force acting on the

particle

‰ This is the form in which Newton presented the Second

‰ This is the form in which Newton presented the Second Law

‰ It is a more general form than the one we used previously This form also allows for mass changes

‰ This form also allows for mass changes

„ Applications to systems of particles are particularly powerful

p

Two Particles (1 dimension for now)

dp dp dp

h f

l O

h

2 1

dt dp dt

dp dt

dp

_total

+

=

are system

on this acting

forces only

the

then forces

external NO

are there

IF

other on the

force

a producing particle

one are

that those

other.

on the

force

Continuing

2 1

+

total

dp dp

dp = +

dp dt dt

dt

0

0 = =

_int

=

=

^total

∑

^ernal

external

F

dt

F dp

NO EXTERNAL FORCES: p

_total

is constant

„

Force on P

₁

is from particle 2 and is F

_2Æ1

„

Force on P

₁

is from particle 2 and is F

_2Æ1

„

Force on P

₂

is from particle 1 and is F

_1Æ2

„

F

_2Æ1

=-F

_1Æ2

Summary

The momentum of a SYSTEM of particles that are

i l d f l f i f

isolated from external forces remains a constant of

the motion.

Conservation of Linear Momentum

„

Whenever two or more particles in an p isolated system interact, the total

momentum of the system remains constant

constant

‰ The momentum of the system is conserved, not necessarily the momentum of an individual not necessarily the momentum of an individual particle

‰ This also tells us that the total momentum of

i l t d t l it i iti l t

an isolated system equals its initial momentum

An Easy One…

A particle of mass m moves with momentum p Show that the kinetic momentum p. Show that the kinetic

energy of the particle is K = p2/2m. (b) Express the magnitude of the particle’s Express the magnitude of the particle s momentum in terms of its kinetic energy and mass

and mass.

Consider a particle roaming around that is

dd l bj d ki d f FORCE

suddenly subjected to some kind of FORCE that looks something like the last slide’s graph.

: NEWTON

v p

v p =

d d

m

F v a

p = = m = dt

m d dt

d

change will

then 0

If

0 if

constant p F

F p

≠

=

=

change will

then 0

If F ≠ p

Let’s drop the vector notation and stick to one dimension.

dp F

=

=

Fdt dp

dt F p

∫

^{f =}

∫

^f

i

t

t

Fdt dp

∫

=

−

f i

t i

f

i t

Fdt p

p

∫

Impulse

ti

f

Change of Momentum

NEW LAW NEW LAW

„

IMPULSE = CHANGE IN MOMENTUM

„

IMPULSE CHANGE IN MOMENTUM

A 3 00-kg steel ball strikes a wall with a speed of 10 0 m/s at A 3.00 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P9.9). If the ball is in contact

i h h ll f 0 200 h i h f d

with the wall for 0.200 s, what is the average force exerted on the ball by the wall?

LETS

TALK

ABOUT

Consider two particles:

v₁ v₂

m₁ m₂

1 22

1 2

V₁ V₂

1 2

V1 V₂

What goes on during the collision? N3

Force on m₂

=F(1( Æ2))

Force on m₁

=F(2Æ1)

=F(2Æ1)

The Forces

Equal and Opposite

More About Impulse: F-t The Graph

I l i t tit

„ Impulse is a vector quantity

„ The magnitude of the

impulse is equal to the area nder the force time c r e under the force-time curve

„ Dimensions of impulse are M L / T

„ Impulse is not a property of the particle, but a measure of the change in momentum of the particle

of the particle

Impulse

„ The impulse can alsoThe impulse can also be found by using the time averaged force

„ I = Δt

F

„ This would give the same impulse as the time-varying force does time-varying force does

Conservation of Momentum, Archer Example

„ The archer is standing on a frictionless surface (ice)

A h

„ Approaches:

‰ Newton’s Second Law – no, no information about F or a

‰ Energy approach – no, no

information about work or energy

‰ Momentum – yesy

Archer Example, 2

„ Let the system be the archer with bow (particle 1) y (p ) and the arrow (particle 2)

„ There are no external forces in the x-direction, so it is isolated in terms of momentum in the x direction is isolated in terms of momentum in the x-direction

„ Total momentum before releasing the arrow is 0

„ The total momentum after releasing the arrow is p_1f

„ The total momentum after releasing the arrow is p_1f + p_2f = 0

Archer Example, final

„

The archer will move in the opposite direction

„

The archer will move in the opposite direction of the arrow after the release

‰ Agrees with Newton’s Third Law

‰ Agrees with Newton s Third Law

„

Because the archer is much more massive

than the arrow, his acceleration and velocity

than the arrow, his acceleration and velocity

will be much smaller than those of the arrow

An estimated force-time curve for a baseball struck by a bat is shown in Figure P9.7. From this curve, determine (a) the

i l d li d h b ll (b) h f d

impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball.

Overview: Collisions – Characteristics

„ We use the term collision to represent an event p

during which two particles come close to each other and interact by means of forces

„ The time interval during which the velocity changes

„ The time interval during which the velocity changes from its initial to final values is assumed to be short

„ The interaction force is assumed to be much greater g than any external forces present

‰ This means the impulse approximation can be used

Collisions – Example 1

„ Collisions may be the y result of direct contact

„ The impulsive forces may vary in time in may vary in time in complicated ways

‰ This force is internal to h

the system

„ Momentum is conserved

co se ed

Collisions – Example 2

„ The collision need not include physical contact between the objects

„ There are still forces

„ There are still forces between the particles

„ This type of collision can be analyzed in the same way as those that include physical contact

p y

Types of Collisions

„ In an elastic collision, momentum and kinetic , energy are conserved

‰ Perfectly elastic collisions occur on a microscopic level

‰ In macroscopic collisions only approximately elastic

‰ In macroscopic collisions, only approximately elastic collisions actually occur

„ In an inelastic collision, kinetic energy is not

d lth h t i till d

conserved although momentum is still conserved

‰ If the objects stick together after the collision, it is a perfectly inelastic collision

Collisions, cont

„

In an inelastic collision some kinetic energy

„

In an inelastic collision, some kinetic energy is lost, but the objects do not stick together

„

Elastic and perfectly inelastic collisions are

„

Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types

between these two types

„

Momentum is conserved in all collisions

Perfectly Inelastic Collisions

„ Since the objects stickSince the objects stick together, they share the same velocity after the

lli i collision

„ m₁v_1i + m₂v_2i =

( + )

(m₁ + m₂) v_f

A 10.0-g bullet is fired into a stationary block of wood (m = 5.00 kg). The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination

immediately after the collision is 0.600 m/s. What was the original speed of the bullet?g p d b

Velcro couplers make the carts stick together after colliding . Velcro couplers make the carts stick together after colliding.

Find the final velocity of the train of three carts. (b) What If ? Does your answer require that all the carts collide and stick

together at the same time? What if they collide in a different order?

Now Where Were We???

Oh yeah …. Elastic Collisions

„ Both momentum andBoth momentum and kinetic energy are

conserved

1 1 2 2

1 1 2 2

i i

f f

m m

m m

+ =

+

v v

v v

2 2

1 1 2 2

1 1

2 m v ⁱ + 2 m v ⁱ =

2 2

1 1 2 2

1 1

2 m v ^f + 2 m v ^f

Let’s look at the case of equal masses.

Second Particle initially at rest

Explains the demo!

Elastic Collisions, final

„ Example of some special casesp p

‰ m₁ = m₂ – the particles exchange velocities

‰ When a very heavy particle collides head-on with a very light one initially at rest the heavy particle continues in light one initially at rest, the heavy particle continues in motion unaltered and the light particle rebounds with a speed of about twice the initial speed of the heavy particle

‰ When a very light particle collides head-on with a very

‰ When a very light particle collides head on with a very heavy particle initially at rest, the light particle has its velocity reversed and the heavy particle remains

approximately at restpp y

Collision Example – Ballistic Pendulum

„ Perfectly inelastic collision – the bullet is embedded in the block of wood

„ Momentum equation will have two unknowns

„ Use conservation of energy from the pendulum to find the velocity just after the collision

„ Then you can find the speed of the bullet

As shown in the figure, a bullet of mass m and speed v passes

l l h h d l b b f M Th b ll

completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length and negligible mass. What is the g g g

minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

Two-Dimensional Collisions

„ The momentum is conserved in all directionsThe momentum is conserved in all directions

„ Use subscripts for

‰ identifying the object

‰ indicating initial or final values

‰ the velocity components

If th lli i i l ti ti f ki ti

„ If the collision is elastic, use conservation of kinetic energy as a second equation

‰ Remember the simpler equation can only be used for one-

‰ Remember, the simpler equation can only be used for one dimensional situations

Two-Dimensional Collision, example

„ Particle 1 is moving atParticle 1 is moving at velocity v_1i and particle 2 is at rest

„ In the x-direction, the initial momentum is m v

m₁v_1i

„ In the y-direction, the initial momentum is 0 initial momentum is 0

Two-Dimensional Collision, example cont

„ After the collision, the , momentum in the x-

direction is m₁v_1f cos

θ

+ m₂v_2f cos

φ

m₂v_2f cos

φ

„ After the collision, the momentum in the y-

di ti i i

direction is m₁v_1f sin

θ

+ m₂v_2f sin

φ

Problem-Solving Strategies – Two- Dimensional Collisions

„

Set up a coordinate system and define your Set up a coordinate system and define your velocities with respect to that system

‰ It is usually convenient to have the x-axis coincide with one of the initial velocities

„

In your sketch of the coordinate system, draw and label all velocity vectors and include all

and label all velocity vectors and include all

the given information

Problem-Solving Strategies – Two- Dimensional Collisions, 2

„ Write expressions for theWrite expressions for the xx- andand yy-components ofcomponents of the momentum of each object before and after the collision

‰ Remember to include the appropriate signs for the components of the velocity vectors

„ Write expressions for the total momentum of the

„ Write expressions for the total momentum of the system in the x-direction before and after the

collision and equate the two. Repeat for the total momentum in the y-direction.

Problem-Solving Strategies – Two- Dimensional Collisions, 3

„

If the collision is inelastic kinetic energy of

„

If the collision is inelastic, kinetic energy of the system is not conserved, and additional information is probably needed p y

„

If the collision is perfectly inelastic, the final

velocities of the two objects are equal. Solve

velocities of the two objects are equal. Solve

the momentum equations for the unknowns.

Problem-Solving Strategies – Two- Dimensional Collisions, 4

„

If the collision is elastic the kinetic energy of

„

If the collision is elastic, the kinetic energy of the system is conserved

„

Equate the total kinetic energy before the

„

Equate the total kinetic energy before the

collision to the total kinetic energy after the

collision to obtain more information on the

collision to obtain more information on the

relationship between the velocities

Two-Dimensional Collision Example

„ Before the collision, the , car has the total

momentum in the x- direction and the van direction and the van has the total

momentum in the y- direction

direction

„ After the collision, both have x- and y-y

components

Th k

Thank you