chapter 10: energy - smudclarke/phys1100/documents/chapter10.pdf · chapter 10: energy a quick...

SMU PHYS1100.1, fall 2008, Prof. Clarke 1

Chapter 10: Energy

Unlike force, displacement, or even momentum, energy is a very difficult concept to define.

Indeed, it is such an elusive concept that its use wasn’t widely recognised by physicists until the mid 19th century; 150 years after Newton’s death.

Yet its most profound property – that of conservation – is a direct consequence of Newton’s 2nd Law!

We’ll start this chapter with a very “loose” definition:

Energy is what one entity has when it has the ability to affect the motion of another. It is to dynamics what money is to economics. It is the currency of physics.

WHAT IS ENERGY?WHAT IS ENERGY?


Chapter 10: Energy

Examples:

A billiard ball in motion can affect the motion of another ball by striking it.

The gravitational attraction between water and the earth gives water above a turbine the potential of setting a hydroelectricalturbine in motion, but not water below the turbine.

Heat from a fire can cause water to boil, setting in motion molecules of water vapour.

The chemical structure of wood is such that when given a sufficient catalyst (e.g., a match), it can generate the heat (fire) necessary to boil the water and set into motion water molecules that form water vapour.


Chapter 10: Energy

Law of conservation of energy:

Energy can neither be created nor destroyed. It can, however, change its form from one type of energy to another. Indeed, physics may be described as keeping track of these changes.

The SI unit for energy is the Joule (J). 1 J = 1 Nm = 1 kgm2s–2

The two most important “types” of energy are:

Kinetic energy: the energy attributable to the motion of an individual particle.

Potential energy: the energy attributable to the position of one particle relative to another. It is always associated with a system of particles, and never to an individual particle.


Chapter 10: Energy

Other “types” of energy:

Heat is a “randomised” form of kinetic energy. Friction forces specialise in converting the “organised” kinetic energy of a macroscopic object into the “randomised” kinetic energy of microscopic objects such as individual molecules – heat.

Chemical and nuclear energy is the potential energy of the position of one microscopic particle relative to another as a result of the mutual force of attraction between them.

According to Einstein’s famous equation, E = mc2, matter itself may be thought of as “condensed” potential energy, just waiting for some brilliant physicist to learn how to unlock it.

Cosmic, life, and pyramid energy: Consult your astrologer!


Chapter 10: Energy

Fnet,x(t) = max = m ⇒⇒⇒⇒ m dvx = Fnet,x (t)dt ⇒⇒⇒⇒ ∆∆∆∆px = Jx

dvx

dt

tf

ti

vf

vi

= Jx (impulseimpulse)mvfx – mvix

= pfx – pix

The Impulse-momentum theorem in chapter 9 was obtained by:

1. imagining the net force to vary as a function of timetime; and

2. integrating Newton’s 2nd law over timetime:

Let’s see what happens instead when we:

1. allow the force to vary as a function of positionposition; and

2. integrate Newton’s 2nd Law over position.position.

Now, for a more quantitative look at energy…


Chapter 10: Energy

Based on the 2003 – 2006 grade sheets for PHYS 1210:

of the 75 students who got less than 10/30 on the midterm,

64% failed the course

Midterm statistics

Last day to withdraw from a class without

academic penalty: this Friday, Nov. 7, 2008


Chapter 10: Energy

Fnet,x(x) = max = m = m (chain rule) = m vx

dvx

dt

dvx

dx

dx

dt

dvx

dx

xf

xi

vfx

vix

vfx

vix

⇒⇒⇒⇒ Fnet,x (x)dx = mvxdvx = mvx = mvfx – mvix

12

2 12

12

22

Reminder from Chapter 2: If f(x) = axn, where a and n are constants, then the fundamental theorem of the calculus says:

f(x)dx = axndx = axn+1 = a(xfn+1 – xi

n+1), n = –1

The function g(x) = axn+1 is the anti-derivative of f(x) = axn.

g(x) = f(x)dx and = f(x)

xf

xi

xf

xi

1n +1

xf

xi

1n +1

1n +1

dg(x)dx

ended here, 30/10/08


Chapter 10: Energy

The quantity mv2 plays such an important role in physics, it is given its own symbol and name:

K = mv2 is the kinetic energy.

Kinetic Energy is the energy of motion of a particular object.

Caution: The speed of an object differs for different inertial frames of reference, and thus so does its kinetic energy.

12

12

Fnet,x (x)dx = mvf – mvi = Kf – Ki = ∆∆∆∆Kxf

xi

12

12

22

What remains is to solve the force integral , and for that we need to specify the force.

Integrating Newton’s 2nd law over position has so far yielded:


Chapter 10: Energy

e.g. 1: gravitational force near the surface of the earth:

= –mg(yf – yi) = –mgyf + mgyi = –mg∆∆∆∆y

yf

yi

yf

yi

yf

yi

Fy = –mg ⇒⇒⇒⇒ Fy dy = (–mg)dy = –mgyyf

yi

Fx = –k∆∆∆∆x where k is the “spring constant” and where ∆∆∆∆x = x – xe is the displacement from the equilibrium position, xe.

e.g. 2: “Hooke’s Law” spring (restoring force):

F > 0

y

xxe0

0

F < 0

∆∆∆∆x > 0

∆∆∆∆x < 0

Fxdx = –k(x – xe) dx = –k z dz = – kz2

= – k(zf – zi ) = – k(∆∆∆∆xf)2 + k(∆∆∆∆xi)

2

xf

xi

zf

zi

12

12

2 2

xf

xi

zf

zi

12

12


Chapter 10: Energy

Clicker question 10.1

Consider two identical cars on Highway 102 going up to Truro. Aunt Ida is in no hurry and dawdles along at 60 kph. Speedy Gonzales, on the other hand, is in a mad rush and passes her at 120 kph. Relative to the road, the kinetic energy of Speedy’s car is:

a) one half b) the same as

c) twice d) four times

the kinetic energy of Aunt Ida’s car.


Chapter 10: Energy


Consider two identical cars on Highway 102 going up to Truro. Aunt Ida is in no hurry and dawdles along at 60 kph. Speedy Gonzales, on the other hand, is in a mad rush and passes her at 120 kph. Relative to the road, the kinetic energy of Speedy’s car is:

a) one half b) the same as

c) twice d) four times

the kinetic energy of Aunt Ida’s car.

K is proportional to v2. Thus, if you

double v (and

leave the mass

the same), you quadruple K.


Chapter 10: Energy

The figure shows force vs.displacement graphs for three springs with three different spring constants (Fsp = –k∆∆∆∆x). Which spring is the “stiffest”? That is, which spring has the greatest spring constant, k?

Fsp

∆∆∆∆x

k1

k2

k3


a) k1 b) k2 c) k3

d) they’re all the same e) can’t tell from the diagram


Chapter 10: Energy

Fsp

∆∆∆∆x

k1

k2

k3


a) k1 b) k2 c) k3

d) they’re all the same e) can’t tell from the diagram

The figure shows force vs.displacement graphs for three springs with three different spring constants (Fsp = –k∆∆∆∆x). Which spring is the “stiffest”? That is, which spring has the greatest spring constant, k?


Chapter 10: Energy

For both the gravitational force and the spring force, we noticethat the force integral has the form:

Fx(x)dx = –(Uf – Ui) = –∆∆∆∆U

Like the kinetic energy, U plays such an important role in physics, it also has its own symbol and name:

U is the potential energy of the system in which Fx is acting.

For gravity near the surface of the earth: U = mgy, where y is the height above some “zero” elevation (e.g., ground level).

For springs, U = k(∆∆∆∆x)2, where ∆∆∆∆x = x – xe is the displacement from the equilibrium position, xe, where the spring is neither stretched nor compressed.

xf

xi

12


Chapter 10: Energy

Potential Energy is the energy attributable to the position of one particle relative to another, and not to an individual particle.

Thus, it is the rock-earth system that has the potential energy that can be “turned into” kinetic energy, not just the rock.

Fx(x)dx = –Uf + Ui = Kf – Ki ⇒⇒⇒⇒ –∆∆∆∆U = ∆∆∆∆Kxf

xi

⇒⇒⇒⇒ Ki + Ui = Kf + Uf ⇒⇒⇒⇒ K + U = constant

Notice the

negative sign!

Returning now to integrating Newton’s 2nd law over position:

and we have uncovered another conservation law!


Chapter 10: Energy

A quick recap:

1. kinetic energy: K = mv2

2. gravitational potential energy: Ug = mgy

where y is measured relative to some “zero” such as ground level.

3. potential energy stored in a spring: Usp = k(∆∆∆∆x)2

where ∆∆∆∆x is the displacement from the equilibrium position of the spring.

4. Subject to certain conditions (which we’ll discuss soon), the quantity K + U is conserved. That is:

∆∆∆∆K + ∆∆∆∆U = 0 ⇒⇒⇒⇒ ∆∆∆∆K = –∆∆∆∆U

12

12


Chapter 10: Energy


Two balls, one twice as heavy as the other, are dropped from the roof of a building. Just before hitting the ground, the kinetic energy of the heavier ball is

a) one half

b) the same as

c) twice

d) four times

the kinetic energy of the lighter ball.


Chapter 10: Energy


Two balls, one twice as heavy as the other, are dropped from the roof of a building. Just before hitting the ground, the kinetic energy of the heavier ball is

a) one half

b) the same as

c) twice

d) four times

the kinetic energy of the lighter ball.

Balls have the same speed just before hitting the ground. However, K is proportional to m. Thus, if you double m, you

double K.


Chapter 10: Energy


Consider two springs with identical spring constants, k. The first spring is compressed by 1 cm, the other spring is stretched by 1 cm. The potential energy stored in the compressed spring is

a) one half

b) twice

c) negative

d) the same as

the potential energy stored in the stretched spring.


Chapter 10: Energy


Consider two springs with identical spring constants, k. The first spring is compressed by 1 cm, the other spring is stretched by 1 cm. The potential energy stored in the compressed spring is

a) one half

b) twice

c) negative

d) the same as

the potential energy stored in the stretched spring.

The potential energy in a

spring is Usp = k(∆∆∆∆x)2.

Thus, only the magnitude

of ∆∆∆∆x matters, not the sign.

12


Chapter 10: Energy

“Challenge” clicker question 10.5

Consider two identical springs (same k) and two masses, m1 = m and m2 = 2m. m1 is pushed into one spring by ∆∆∆∆x, m2

is pushed into the other spring by 2∆∆∆∆x, and both springs are released. When the springs return to their equilibrium position, the velocity of m2 is

a) twice

b) 2 times

c) the same as

d) times

the velocity of m1. xe0

x

∆∆∆∆x

2∆∆∆∆x

m

2m

2

1

Usp = k(∆∆∆∆x)2

K = mv2

∆∆∆∆K = –∆∆∆∆Usp

1212


Chapter 10: Energy

“Challenge” clicker question 10.5

Consider two identical springs (same k) and two masses, m1 = m and m2 = 2m. m1 is pushed into one spring by ∆∆∆∆x, m2

is pushed into the other spring by 2∆∆∆∆x, and both springs are released. When the springs return to their equilibrium position, the velocity of m2 is

a) twice

b) 2 times

c) the same as

d) times

the velocity of m1. xe0

x

∆∆∆∆x

2∆∆∆∆x

m

2m

2

1

Usp = k(∆∆∆∆x)2

K = mv2

∆∆∆∆K = –∆∆∆∆Usp

1212

∆∆∆∆K = –∆∆∆∆Usp ⇒⇒⇒⇒ mv2 = k(∆∆∆∆x)2⇒⇒⇒⇒ v = ∆∆∆∆x

km

∆∆∆∆Usp = U

f – Ui = – k(∆∆∆∆x)2

12

0

∆∆∆∆K = Kf – K

i = mv212

0

∆∆∆∆x 2 ⇒⇒⇒⇒ v 2; m 2 ⇒⇒⇒⇒ v 2

net result: v 2.


Chapter 10: Energy

Mechanical Energy

Define the mechanical energy of a system: EM = K + U.

Then, subject to the conditions assumed while integrating Newton’s 2nd law:

∆∆∆∆EM = EMf – EMi = ∆∆∆∆K + ∆∆∆∆U = 0

That is, subject to these conditions, mechanical energy is conserved in any dynamical process.

Conditions? What conditions??

1. forces must be conservative

2. system must be isolated from external forces that canaffect (significantly) its kinetic energy.


Chapter 10: Energy

What is a conservative force?

A conservative force is one for which the conservation of mechanical energy applies.

This definition seems rather circular! A more useful definition:

A conservative force is one whose force integral can be written:

Fx(x)dx = –(Uf – Ui) = –∆∆∆∆Uxf

xi

which was central to our argument. We’ll revisit the idea of conservative and non-conservative forces in Chapter 11. In this chapter, all problems will involve conservative forces only.

examples of conservative forces: gravitation, spring force

examples of non-conservative forces: friction, drag


Chapter 10: Energy

Whether a force is external or internal to the system depends on how we define the system.

If the system is defined as the ball alone, then its mechanical energy is:

EM = K

since U = 0 (potential energy is the energy of interaction between two particles, and the ball is just one particle).

So is EM constant?

No! As the ball falls, its kinetic energy increases and so does EM. This is because there is a force external to the system (Earth’s gravity) acting on the system (the ball).


Chapter 10: Energy

Now define the system to be the earth + ball.

We can now attribute a potential energy of the earth+ball system, and

EM = K + U

Since gravity is conservative and there are no (significant) forces external to this system (we’ll ignore air drag), EM

is conserved and

∆∆∆∆EM = 0 ⇒⇒⇒⇒ ∆∆∆∆K = –∆∆∆∆U

Thus, as K increases (∆∆∆∆K > 0), it does so at the expense of U (∆∆∆∆U < 0) so that their sum (EM) is constant.

ended here, 4/11/08


Chapter 10: Energy

So which “system” is the right one?

Let the ball start at rest from yi = h, and end up at yf = 0. What is its final velocity?

EM is not conserved in the “ball only” system, so conservation of EM is not the way to go. Instead, just use kinematics:

vf = vi – 2g(yf – yi)

⇒⇒⇒⇒ vf = –2g(0 – h)

⇒⇒⇒⇒ vf = 2gh

2 2

2

They both are!!

How we define our system is simply a matter of convenience.

Conversely, EM is conserved in “earth + ball” system. So,

EM,i = Ki + Ui = 0 + mgyi = mgh

EM,f = Kf + Uf = mvf + 0

EM,f = EM,i ⇒⇒⇒⇒

mvf = mgh

⇒⇒⇒⇒ vf = 2gh

2

2

12

12


Chapter 10: Energy

External forces parallel to the velocity…

… will change the speed and thus K without changing U. Thus, ∆∆∆∆K + ∆∆∆∆U = 0 and mechanical energy is not conserved.

Does this mean we must abandon ideas in energy conservation when we have such forces? Only for now. In Chapter 11, we’ll see how the concept of work allows us to incorporate such forces into the “energy budget”.

The force of the woman pushing on the box is external to the box system and parallel to the velocity.

The tension pulling up the elevator is external to the man-elevator-earth system, and parallel to the velocity.


Chapter 10: Energy

External forces perpendicular to the velocity…

… will change only the direction of v and not its magnitude. Thus, neither K nor U are changed and ∆∆∆∆K + ∆∆∆∆U = ∆∆∆∆EM = 0.

Mechanical energy is still conservedwhen external forces perpendicular to the

velocity are applied to the system.

As m slides along the path, the normal force (external to the m + earth system) is everywhere perpendicular to the velocity.

As the pendulum swings, the tension force (external to the m + earth system) is always perpendicular to the velocity.

T

v

mv

n

n

nv

v

m


Chapter 10: Energy


A block slides down a frictionless ramp of height h. It reaches velocity v at the bottom. To reach a velocity of 2v, the block would need to slide down a ramp of height:

a) 2h

b) 2h

c) 3h

d) 4h

h


Chapter 10: Energy

mv2 = mgh ⇒⇒⇒⇒ v = 2gh

(mechanical energy is conserved even with the ramp, since the normal force is ⊥⊥⊥⊥ to v )

Thus, for 2v we need 4h.

12

A block slides down a frictionless ramp of height h. It reaches velocity v at the bottom. To reach a velocity of 2v, the block would need to slide down a ramp of height:

a) 2h

b) 2h

c) 3h

d) 4h


h


Chapter 10: Energy


A block is shot up a frictionless 40° slope with initial velocity v. It reaches height h before stopping and sliding back down. The same block is shot with the same velocity up a frictionless 20° slope. On this slope, the block reaches height:

a) 2h

b) h

c) h/2

d) can’t tell from given information


Chapter 10: Energy

A child slides down the four frictionless slides A–D shown below. Each slide has the same height. For which slide will her speed at the bottom be the greatest?

a) slide A b) slide B c) slide C d) slide D

e) all the same f) can’t tell without knowing her mass



Chapter 10: Energy

Conservation of mechanical energy, when applicable, offers an alternate way (to kinematics and/or dynamics and FBDs) to approach a problem – not a superior way, just an alternate way.

After all, conservation of mechanical energy did come directly from Newton’s 2nd Law!

So why bother? Why not just use FBDs, Newton’s laws, and kinematics all the time?

1. Sometimes energy conservation provides an easier way (and occasionally much easier) to do a dynamical problem.

2. It offers new insight into physics. Energy is a relatively new concept (not appreciated until 150 years after Newton) whose understanding has spawned our present age of technology.


Chapter 10: Energy

Example: A mass, m, is released from rest at the top of a quarter-circle frictionless slide of radius r. What is the speed of m at the bottom of the slide?

r

m

yf = 0

yi = r

vf = ?

vi = 0

vf

Define the system: m + earth.

Are there any non-conservative forces? No.

Are there any external forces? Yes, the normal force which is perpendicular to v

and has no effect on the mechanical energy.

Assess EM,i: EM,i = Ki + Ui = 0 + mgr

Assess EM,f: EM,f = Kf + Uf = mvf + 0

Equate EM,i = EM,f ⇒⇒⇒⇒ mgr = mvf

⇒⇒⇒⇒ vf = 2gr

2

2

12

12


Chapter 10: Energy

Note that 2gr is the same speed m would have if it simply dropped a distance r from rest.

Does the fact that the slide redirects the velocity from vertical to horizontal without changing its magnitude surprise you?

It shouldn’t! Remember, the normal force is perpendicularto the velocity (here acting as a centripetal force) and, assuch, can only affect its direction, not its magnitude.

By the way, should you ever doubt the usefulness of the technique of conservation of mechanical energy, try doing this seemingly simple problem directly with Newton’s 2nd

Law, FBDs, and kinematics. Good luck!


Chapter 10: Energy

Example: An effective way to measure the speed of a bullet is depicted

in the diagram. A bullet of mass mB and speed vB,0 strikes a block of wood of mass mW hanging vertically at rest on a massless rod of length L free to pivot at its top. If the maximum angle of deflection of the rod is θθθθ, what is vB,0? (This is the “theory” behind the ballistic pendulum lab.)

Divide this problem into two parts:

1. the inelastic collision between the bullet and the wood where momentum is conserved but not mechanical energy;

2. the swing of the block of wood in which mechanical energy is conserved.

vB,0

vW,0

vB,0


Chapter 10: Energy

1. Conserve momentum: The impact occurs over a very short time during which all external impulses (e.g., air drag?) can beneglected. The bullet becomes embedded into the wood, and thus the collision is inelastic (chapter 9).

vB,0 = velocity of bullet right before impact

v1 = velocity of bullet + block right after impact

pi = mBvB,0 pf = (mB + mW)v1

pi = pf ⇒⇒⇒⇒ mBvB,0 = (mB + mW)v1

⇒⇒⇒⇒ v1 = vB,0 1mB

mB + mW

vB,0

vW,0


Chapter 10: Energy

2. Conserve mechanical energy: On the swing up, consider the wood + bullet + earth system. Dissipative forces (air drag)are ignorable, and the external force (tension in the rod), is everywhere perpendicular to the velocity.

Let m = mB + mW

EMi = mv1 + mgy1 = mv1

EMf = mv2 + mgy2 = mgy2

EMi = EMf ⇒⇒⇒⇒ mv1 = mgy2

⇒⇒⇒⇒ v1 = 2gy2 = 2gL(1 – cosθθθθ) 2

1 & 2 ⇒⇒⇒⇒ vB = 2gL(1 – cosθθθθ)

= 317 ms–1

12

212

12

2

2

L

L – Lcosθ

212

mB

m


Chapter 10: Energy

Example: A massless spring hanging vertically has a length 5.0 cm. A 0.50 kg mass is attached to the end of the spring and lowered to the equilibrium position where the length of the spring is now 6.0 cm.

a) Find the spring constant.

m is in equilibrium at y = y1. Thus:

y

mg

Fsp

m

Fsp – mg = 0 ⇒⇒⇒⇒ Fsp = mg

Fsp = –k∆∆∆∆y = –k((((y1 – 0) = –ky1

⇒⇒⇒⇒ mg = –ky1

⇒⇒⇒⇒ k = – = 490 Nm–1mg

y1

y1 = –0.010

m

0y1

y


Chapter 10: Energy

b) If the mass is pulled down an additional 2.0 cm and released, what is the speed of the mass as it passes through the equilibrium position?

0y1

m v

y

y1 = –0.010y2 = –0.030

m

my2

EM,i = “initial” mechanical energy at y = y2.

EM,f = “final” mechanical energy at y = y1.

from previous slide, set mg = –ky1:

EM,i = Ui + Ki = Usp,i + Ug,i + 0 = k(y2 – 0)2 + mgy212

EM,f = Uf + Kf = k(y1 – 0)2 + mgy1 + mv212

12

ky2 – ky1y2 = ky1 – ky1 + mv222 212

12

12

⇒⇒⇒⇒ ky2 + mgy2 = ky1 + mgy1 + mv22 212

12

12

⇒⇒⇒⇒ mv2 = k(y2 – 2y1y2 + y1 ) = k(y2 – y1)21

212

12

Even though the total stretch of the spring is between y0 and y2, the spring behaves as though it were only stretched between y1 and y2.

2 2


Chapter 10: Energy

Continuing with the example…

mv2 = k(y2 – y1)2

⇒⇒⇒⇒ v = (y1 – y2) = (0.02) = 0.63 ms-1 (up)

12

12

km

4900.5

Now, the quantity has units (Nm–1 kg–1)½ = (s–2)½ = s-1 and

thus it is an angular frequency. We shall see in Chapter 14 that:

km

is the angular frequency of oscillation of the mass bouncing up and down on the spring, with the period of oscillation given by

ωωωω = (= 31.3 rad s-1 for this example)km

2ππππωωωω

T = (= 0.201 s, or about 5 oscillations per second)

We chose this to be the negative

root of that so the velocity would

be positive, i.e., up.


Chapter 10: Energy

10.6 Elastic Collisions

In most collisions, some of the initial mechanical energy is “dissipated” inside the colliding objects, and not “retained” as kinetic energy. Thus, Kf < Ki.

For some collisions, however, very little if any kinetic energy is lost (e.g., collisions of subatomic particles, even billiard balls). In the extreme case where kinetic energy is conserved, collisions are called perfectly elastic.

Of course, for both elastic and inelastic (chapter 9) collisions, momentum is conserved too.

Given the initial velocities of the particles, we can determine the final velocities with a little algebra…


Chapter 10: Energy

Consider the collision depicted below in which the “incident” mass (m1, red) moves at velocity v1i toward the “target” mass (m2, blue) initially at rest. What are the velocities of the two masses after the perfectly elastic collision? (The algebra is not as bad as the text makes it out to be!)

Conserve momentum: m1v1i = m1v1f + m2v2f 1

Conserve kinetic energy: m1v1i = m1v1f + m2v2f 2222

multplying

through by 2Solve 1 for m2v2f and square it: (m2v2f)2 = m1 (v1i – v1f )2 3

Multiply 2 through by m2: m2 v2f = m2m1(v1i – v1f ) 4

2

2222

Equate 3 and 4 , cancel an m1, and expand the “difference of squares”:

m1(v1i – v1f )2 = m2 (v1i – v1f )(v1i + v1f ) 5

⇒⇒⇒⇒ v1i (m1 – m2 ) = v1f (m1 + m2 )

m1 – m2

m1 + m2

⇒⇒⇒⇒ v1f = v1i 6

before

collision

after

v1i

v2fv1f

v2i = 0

m1 m2


Chapter 10: Energy

As for v2f, from 5 we have: m1(v1i – v1f ) = m2(v1i + v1f )

and from 1 , we have: m1v1i – m1v1f = m2v2f

The left hand sides of these equations are the same, and thus:

m2v2f = m2(v1i + v1f ) ⇒⇒⇒⇒ v2f = v1i + v1f 7

which is a curious relationship as it doesn’t involve the masses. Equation 7 must hold for any elastic collision regardless of the masses if the target mass is initially at rest, (e.g., two billiard balls, or a pebble off a truck!).

2m1

m1 + m2

⇒⇒⇒⇒ v2f = v1i 8

Substitute 6 into 7 and we get:

v2f = v1i + v1i

m1 – m2

m1 + m2

m1 – m2

m1 + m2

v1f = v1i 6 (recall)

before

collision

after

v1i

v2fv1f

v2i = 0

m1 m2


Chapter 10: Energy

v1i v2i

v1f = v1i + v2im1 – m2

m1 + m2

2m2

m1 + m2

v2f = v1i – v2i2m1

m1 + m2

m1 – m2

m1 + m2

v2i + v2f = v1i + v1f

before

collision

after

v1i

v2fv1f

v2i

m1 m2

For the case where v2i = 0, the algebra does get a little hairy (try it!). Generalisations to equations 6 , 8 , and 7 are:

perfectly elastic collision

before

collision

after

vf

m1 m2

And, from Chapter 9 we have:

m1v1i + m2v2i

m1 + m2

v1f = v2f = vf =

perfectly inelastic collision


Chapter 10: Energy


Tabulated below are the initial and final velocities of two balls (of unknown mass) that have undergone a collision. All entries are in ms–1. Which one, if any, is perfectly elastic?

a)

v1i v2fv2i v1f

b)

c)

d)

e)

2.0 0.0 –1.0 2.5

2.5 –2.0 2.0–1.5

1.5 –2.0 0.0–2.0

1.0 –0.5–3.0 –0.5

none of them are elastic


Chapter 10: Energy


Tabulated below are the initial and final velocities of two balls (of unknown mass) that have undergone a collision. All entries are in ms–1. Which one, if any, is perfectly elastic?

a)

v1i v2fv2i v1f

b)

c)

d)

e)

2.0 0.0 –1.0 2.5

2.5 –2.0 2.0–1.5

1.5 –2.0 0.0–2.0

1.0 –0.5–3.0 –0.5

none of them are elastic

v1i + v1f

= 2.5 – 2.0 = 0.5

v2i + v2f

= –1.5 + 2.0 = 0.5


Chapter 10: Energy


Tabulated below are the initial and final velocities of two balls (of unknown mass) that have undergone a collision. All entries are in ms–1. Which one, if any, is perfectly inelastic?

a)

v1i v2fv2i v1f

b)

c)

d)

e)

2.0 0.0 –1.0 2.5

2.5 –2.0 2.0–1.5

1.5 –2.0 0.0–2.0

1.0 –0.5–3.0 –0.5

none of them are inelastic


Chapter 10: Energy


Tabulated below are the initial and final velocities of two balls (of unknown mass) that have undergone a collision. All entries are in ms–1. Which one, if any, is perfectly inelastic?

a)

v1i v2fv2i v1f

b)

c)

d)

e)

2.0 0.0 –1.0 2.5

2.5 –2.0 2.0–1.5

1.5 –2.0 0.0–2.0

1.0 –0.5–3.0 –0.5

none of them are inelastic

v1f = v2f = –0.5


Chapter 10: Energy

Example 10.9 from text: A steel ball of mass mA = 0.200 kg on a string

of length L = 1.00 m is released from rest at a 45o

angle. At the vertical position, the ball strikes a steel block of mass mB = 0.500 kg resting on a frictionless surface. To what angle does the steel ball rebound?


Chapter 10: Energy

Part 1. Conserve EM on the way down (no friction, Fext ⊥⊥⊥⊥ v )

= hh0

Part 2. Treat the collision as perfectly elastic.

mA – mB

mA + mB

v2A = v1A

= –1.027 ms-1

mv1A = mgh0

⇒⇒⇒⇒ v1A = 2gh0 = 2gL(1 – cosθθθθ0)

= 2.396 ms-1

12

2


Chapter 10: Energy

Part 3. Finally, conserve EM on the way up (again, no friction, Fext ⊥⊥⊥⊥ v )

y

L co

s θθ θθ3

h3

0

mv2A = mgh3

⇒⇒⇒⇒ h3 = = L(1 – cosθθθθ3)

⇒⇒⇒⇒ cosθθθθ3 = 1 – = 0.9462

⇒⇒⇒⇒ θθθθ3 = 18.9o

12

2

v2A2

2gv2A

2

2gL

The solution was as easy as it was because of how the problem was set up: A good pictorial diagram, sensible variables defined, and the recognition that the problem should be examined in three parts.

Does this problem seem easier than you thought it might be at first?


Chapter 10: Energy

10.7 Energy Diagrams

Since potential energy is the “energy of position”, a graph of U(s) vs. s can be very useful in visualising a problem.

For gravitational potential energy, Ug(y) = mgy, and a plot of Ug(y) vs. y is a straight line with slope mg, and an intercept through the origin (Ug(0) = 0).

If EM is conserved, the EM graph is a horizontal line. (Inexplicably, the texthas started to refer to mechanical energy

as total energy. This is in error.)

The kinetic energy is the difference between the EM andUg graphs.

Once Ug = EM, K = 0 and the particle is (perhaps momentarily) at rest.


Chapter 10: Energy

Consider a mass launched vertically. It’s history in both pictorial diagrams and E vs. y graphs is shown below. As mrises, its potential energy increases until Ug = EM. Here, K and thus v are 0, the mass is momentarily at rest, and it can go no higher. In Chapter 2, we referred to this as a “turning point”.


Chapter 10: Energy

For a spring, Usp = k(x – xe)2, and the U vs. x graph is a

parabola opening upward with its apex located at (x,U) = (xe,0)

The level of the horizontal line representing EM (“TE” on the graph) is determined by the initial compression of the spring.

There are two turning points corresponding to the maximum stretch and maximum compression of the spring.

12


Chapter 10: Energy

Generalisations for more complicated systems:

When Ug = EM, K = 0 and the object is at rest (turning points).

When Ug is a minimum, K is a maximum (maximum speed).

If EM = E2, an object at x2 must be at rest. Small perturbations will cause it to oscillate gently back and forth ⇒⇒⇒⇒ stable equilibrium.

If EM = E3, an object at x3 is also at rest but this time perturbations will cause it to accelerate away from x3

(like balancing a ball on top of a hill) ⇒⇒⇒⇒ unstable equilibrium.

chapter 10: energy - smudclarke/phys1100/documents/chapter10.pdf · chapter 10: energy a quick...

Documents