Lecture 12Momentum, Energy, and

Collisions

Announcements

•EXAM: Thursday, October 17 (chapter 6-9)

•Equation sheet to be posted

•Practice exam to be posted

Linear Momentum

Impulse

With no net external force:

Center of MassThe center of mass of a system is the point where the

system can be balanced in a uniform gravitational field.

For two objects:

The center of mass is closer to the more massive object.

Think of it as the “average location of the mass”

Center of Mass

(1)

XCM

(2)

a) higher

b) lower

c) at the same place

d) there is no definable CM in this case

The disk shown below in (1) clearly has its center of mass at the center.

Suppose the disk is cut in half and the pieces arranged as shown in (2).Where is the center of mass of (2) as compared to (1) ?

(1)

XCM

(2)

CM

Center of Mass

The CM of each half is closer to the top of the semicircle than the bottom. The CM of the whole system is located at the midpoint of the two semicircle CMs, which is higher than the yellow line.

a) higher

b) lower

c) at the same place

d) there is no definable CM in this case

The disk shown below in (1) clearly has its center of mass at the center.

Suppose the disk is cut in half and the pieces arranged as shown in (2).

Where is the center of mass of (2) as compared to (1) ?

When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly (whereas it is safe to hold the gun while it is fired)?

a) it is much sharper than the gun

b) it is smaller and can penetrate your body

c) it has more kinetic energy than the gun

d) it goes a longer distance and gains speed

e) it has more momentum than the gun

Gun Control

When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly (whereas it is safe to hold the gun while it is fired)?

a) it is much sharper than the gun

b) it is smaller and can penetrate your body

c) it has more kinetic energy than the gun

d) it goes a longer distance and gains speed

e) it has more momentum than the gun

Even though it is true that the magnitudes of the momenta of the gun and the bullet are equal, the bullet is less massive and so it has a much higher velocity. Because KE is related to v2, the bullet has considerably more KE and therefore can do more damage on impact.

Gun Control

Rolling in the Rain

a) speeds up

b) maintains constant speed

c) slows down

d) stops immediately

An open cart rolls along a

frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)

Because the rain falls in vertically, it

adds no momentum to the box, thus

the box’s momentum is conserved.

However, because the mass of the box

slowly increases with the added rain,

its velocity has to decrease.

Rolling in the Rain

a) speeds up

b) maintains constant speed

c) slows down

d) stops immediately

An open cart rolls along a

frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)

Two objects collide... and stick

A completely inelastic collision: no “bounce back”

No external forces... so momentum of system is conservedinitial px = mv0

final px = (2m)vf

mv0 = (2m)vf

vf = v0 / 2

mass m mass m

Inelastic collision: What about energy?

mass m mass m vf = v0 / 2

Kinetic energy is lost! KEfinal = 1/2 KEinitial

initial

final

Collisions

This is an example of an “inelastic collision”

Collision: two objects striking one another

Elastic collision ⇔ “things bounce back”

⇔ energy is conserved

Inelastic collision: less than perfectly bouncy

⇔ Kinetic energy is lost

Time of collision is short enough that external forces may be ignored so momentum is conserved

Completely inelastic collision: objects stick together afterwards. Nothing “bounces back”. Maximal energy loss

Elastic vs. Inelastic

Inelastic collision: momentum is conserved but kinetic energy is not

Completely inelastic collision: colliding objects stick together, maximal loss of kinetic energy

Elastic collision: momentum and kinetic energy is conserved.

Completely Inelastic Collisions in One Dimension

Solving for the final momentum in terms of initial velocities and masses, for a 1-dimensional, completely inelastic collision between unequal masses:

Completely inelastic only(objects stick together, so have same final velocity)

KEfinal < KEinitial

Momentum Conservation:

Crash Cars I

a) I

b) II

c) I and II

d) II and III

e) all three

If all three collisions below

are totally inelastic, which

one(s) will bring the car on

the left to a complete halt?

Crash Cars I

In case I, the solid wall clearly stops the car.

In cases II and III, because

ptot = 0 before the collision,

then ptot must also be zero

after the collision, which means that the car comes to a halt in all three cases.

a) I

b) II

c) I and II

d) II and III

e) all three

If all three collisions below

are totally inelastic, which

one(s) will bring the car on

the left to a complete halt?

Crash Cars II

If all three collisions below are

totally inelastic, which one(s)

will cause the most damage (in

terms of lost energy)?

a) I

b) II

c) III

d) II and III

e) all three

Crash Cars II

a) I

b) II

c) III

d) II and III

e) all three

The car on the left loses the same KE in all three cases, but in case III, the car on the right loses the most KE because KE = mv2 and the car in case III has the largest velocity.

If all three collisions below are

totally inelastic, which one(s) will

cause the most damage (in

terms of lost energy)?

Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block.

vf = m v0 / (m+M)

momentum conservation in inelastic collision

KE = 1/2 (mv0)2 / (m+M)

energy conservationafterwards

PE = (m+M) g h

hmax = (mv0)2 / [2 g (m+M)2]

Velocity of the ballistic pendulum

Pellet Mass (m): 2 gPendulum Mass (M): 3.81 kgWire length (L): 4.00 m

approximation

Inelastic Collisions in 2 Dimensions

Energy is not a vector equation: there is only 1 conservation of energy equation

Momentum is a vector equation: there is 1 conservation of momentum equation per dimension

For collisions in two dimensions, conservation of momentum is applied separately along each axis:

Elastic CollisionsIn elastic collisions, both kinetic

energy and momentum are conserved.

One-dimensional elastic collision:

Elastic Collisions in 1-dimension

solving for the final speeds:

Note: relative speed is conserved for head-on (1-D) elastic collision

We have two equations: conservation of momentum conservation of energy

and two unknowns (the final speeds).

For special case of v2i = 0

Limiting cases of elastic collisions

note: relative speed conserved

Limiting cases

note: relative speed conserved

Limiting cases

note: relative speed conserved

Toy Pendulum

Could two balls recoil and conserve both momentum and energy?

Incompatible!

Elastic Collisions I

v 2v

1at rest

at rest

a) situation 1

b) situation 2

c) both the same

Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on.

In which case does the golf ball have the greater speed after the collision?

Remember that the magnitude of the relative velocity has to be equal before and after the collision!

Elastic Collisions I

v1

In case 1 the bowling ball will almost remain at rest, and the golf ball will bounce back with speed close to v.

v 22v

In case 2 the bowling ball will keep going with speed close to v, hence the golf ball will rebound with speed close to 2v.

a) situation 1

b) situation 2

c) both the same

Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on.

In which case does the golf ball have the greater speed after the collision?

A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a 0.150 kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed?

A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a 0.150 kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed?

Our simplest formulas for speed after an elastic collision relied on one body being initially at rest.

So lets try a frame where one body (the elephant) is at rest!

In the frame, what is the final speed of the ball?

Back in the frame of the ground:

What is the speed of the ball relative to the elephant?

A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a 0.150 kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed?

Our simplest formulas for speed after an elastic collision relied on one body being initially at rest.

So lets try a frame where one body (the elephant) is at rest!

In the frame, what is the final speed of the ball?

Back in the frame of the ground:

What is the speed of the ball relative to the elephant?

NOTE: Formulas for 1-D elastic scattering with non-zero initial velocities are given in end-of-chapter problem 88.

Elastic Collisions in 2-D

Two-dimensional collisions can only be solved if some of the final information is known, such as the final velocity of one object

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A proton collides elastically with another proton that is initially at rest.

The incoming proton has an initial speed of 3.5x105 m/s and makes

a glancing collision with the second proton. After the collision one

proton moves at an angle of 37o to the original direction of motion,

the other recoils at 53o to that same axis. Find the final speeds of the

two protons.

v0 = 3.5x105 m/s

initial

37o

53o

v2

v1

final

A proton collides elastically with another proton that is initially at rest.

The incoming proton has an initial speed of 3.5x105 m/s and makes

a glancing collision with the second proton. After the collision one

proton moves at an angle of 37o to the original direction of motion,

the other recoils at 53o to that same axis. Find the final speeds of the

two protons.

v0 = 3.5x105 m/s

initial

37o

53o

v2

v1

final

Momentum conservation:

if we’d been given only 1 angle, would have needed conservation of energy also!

Rotational Kinematics

Angular Position

Degrees and revolutions:

Angular Positionθ > 0

θ < 0

Arc Length

Arc length s, from angle measured in radians:

s = rθ

- Arc length for a full rotation (360o) of a radius=1m circle?

- What is the relationship between the circumference of a circle and its diameter?C / D = π

C = 2 π r

s = 2 π (1 m) = 2 π meters

1 complete revolution = 2 π radians

1 rad = 360o / (2π) = 57.3o

Angular Velocity

Instantaneous Angular Velocity

Period = How long it takes to go 1 full revolution

Period T:

SI unit: second, s

Linear and Angular Velocity

Greater translation for same rotation

Bonnie and Klyde II

ω

BonnieBonnieKlydeKlyde

a) Klyde

b) Bonnie

c) both the same

d) linear velocity is zero for both of them

Bonnie sits on the outer rim of amerry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one revolution every 2 seconds. Who has the larger linear (tangential) velocity?

a) Klyde

b) Bonnie

c) both the same

d) linear velocity is zero for both of them

Their linear speeds linear speeds vv will be

different because v = r v = r ωω

and Bonnie is located farther Bonnie is located farther

outout (larger radius r) than

Klyde.

ω

BonnieBonnie

KlydeKlyde

Bonnie and Klyde II

Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one revolution every 2 seconds. Who has the larger linear (tangential) velocity?

Angular Acceleration

Instantaneous Angular Acceleration

Rotational Kinematics, Constant Acceleration

If the acceleration is constant:

If the angular acceleration is constant:

v = v0 + at

Analogies between linear and rotational kinematics:

An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in the time t, through what angle did it rotate in the time t?

Angular Displacement IAngular Displacement I

½

a) θ

b) θ

c) θ

d) 2 θ

e) 4 θ

½

¼

¾

An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in the time t, through what angle did it rotate in the time t ?

a) θ

b) θ

c) θ

d) 2 θ

e) 4 θ

The angular displacement is θ = αt 2 (starting from rest), and

there is a quadratic dependence on time. Therefore, in half the half the

timetime, the object has rotated through one-quarter the angleone-quarter the angle.

Angular Displacement IAngular Displacement I½

¼

½

¾

Which child experiences a greater acceleration?(assume constant angular speed)

Larger r: - larger v for same ω - larger ac for same ω

ac is required for circular motion.

An object may have at as well, which implies angular acceleration

Angular acceleration and total linear acceleration