Physics 8 — Friday, September 20, 2013

I Hand in HW3 at start of class. I’ll put HW4 online tomorrow.HW3 took many people 4+ hours. I’ll make HW4 shorter.

I Today: finish ch5 (energy); start ch6 (relative motion)

I For Monday: read ch7 (interactions). Very few equations!

I After ch9, the pace of reading will slow down to ≈ 1ch/wk!

I HW help sessions: Wed 7pm DRL 2N36, Thu 7pm DRL 3W2.

I My summary of key equations (“equation sheet”):positron.hep.upenn.edu/wja/phys008_2013/equations.pdf

I Expect a quiz on HW2 next Weds. I’ll try to make it shorter.It’s just an incentive to make sure you understand your HW.Quiz total counts only 5% of term grade, and semester quiztotal ≥ 80% gets full credit.

I I try hard to make it possible for you to do well in this coursewhether or not you’re someone who enjoys math/physicsexams. Many other courses make exams 100% of the grade.That’s why I put so much emphasis on homework.

Where are we (going)?

To analyze structures, you need a thorough understanding offorces, torques (a.k.a. “moments” of forces), and vectors.

Where are we (going)?(Richard Wesley: “A course should tell a story.”)

I To analyze structures, you need a thorough understanding offorces, torques (a.k.a. “moments” of forces), and vectors.

I To understand forces well, you need a solid grasp ofI How forces affect motion.I How different forces relate to one another.I How objects interact with one another via forces.

I In ch2–3, we studied the key concepts of motion: position,velocity, acceleration, etc.

I In ch4–5, we studied two key conservation laws (momentum,energy) and some of the restrictions they place on howcolliding objects can interact with one another.

I Finally in ch8 we’ll discuss forces! We’re preparing your mindfor forces in ch6–7 by learning a few more of the restrictionsimposed, as a consequence of momentum & energyconservation, on how objects can interact with one another.

You walk due east 3000 meters from home to a restaurant in 2500seconds, stay there for 6000 seconds, and then walk exactlyhalfway home, taking another 1500 seconds, where you set up atent for the night (halfway between your home and the restaurant).

Working with one or two of your neighbors, draw a sketch of x asa function of t.

You walk due east 3000 meters from home to a restaurant in 2500 seconds, stay there for 6000 seconds, and then

walk exactly halfway home, taking another 1500 seconds, where you set up a tent for the night.

What feature of this graph represents the instantaneous velocityat t = 9990 s?

Work together with your neighbor(s) on this!

You walk due east 3000 meters from home to a restaurant in 2500seconds, stay there for 6000 seconds, and then walk exactlyhalfway home, taking another 1500 seconds, where you set up atent for the night (halfway between your home and the restaurant).

The instantaneous velocity at t = 9990 s is

(A) 1.50 m/s, moving east

(B) 1.50 m/s, moving west

(C) 1.00 m/s, moving east

(D) 1.00 m/s, moving west

(E) 1.0067 m/s, moving east

(F) 1.0067 m/s, moving west

(G) 1.51 m/s, moving east

(H) 1.51 m/s, moving west

A compact car and a large truck collide head on and sticktogether. Which undergoes the larger momentum change?

(a) car

(b) truck

(c) The momentum change is the same (in magnitude) for bothvehicles.

(d) Can’t tell without knowing the final velocity of combined mass.

A compact car and a large truck collide head on and sticktogether. Which undergoes the larger velocity change?

(a) car

(b) truck

(c) The velocity change is the same (in magnitude) for bothvehicles.

(d) Can’t tell without knowing the final velocity of combined mass.

If all three collisions in the figure shown here are totally inelastic,which bring(s) the car on the left to a halt?

(a) I

(b) II

(c) III

(d) I, II

(e) I, III

(f) II, III

(g) all three

You’ve now had lots of practice using the fact that the totalmomentum of an isolated system (e.g. two objects that interactonly with one another, not with outside world) does not change:

m1v1x ,i + m2v2x ,i = m1v1x ,f + m2v2x ,f

You’ve also had lots of practice using the following energy-relatedstatements (usually involving relative speed |v1x − v2x |) todescribe various types of collisions:

I elastic: (v1x ,f − v2x ,f ) = −(v1x ,i − v2x ,i )

I totally inelastic: (v1x ,f − v2x ,f ) = 0

I if e is given: (v1x ,f − v2x ,f ) = −e(v1x ,i − v2x ,i )

I if change in internal energy is given:

1

2m1v

21i +

1

2m2v

22i + Ei ,internal =

1

2m1v

21f +

1

2m2v

22f + Ef ,internal

What if you try to solve the general case of a two-body elasticcollision? Here’s the input:

because an isolated system’s momentum is constant:

m1v1x ,i + m2v2x ,i = m1v1x ,f + m2v2x ,f

because the collision is elastic:

(v1x ,f − v2x ,f ) = −(v1x ,i − v2x ,i )

Here’s the output. It’s messy!

v1x ,f =(m1 −m2)v1x ,i + 2m2v2x ,i

m1 + m2

v2x ,f =(m2 −m1)v2x ,i + 2m1v1x ,i

m1 + m2

http://www.wolframalpha.com/input/?i=m1*v01+%2B+m2*v02+%3D+m1*v1+%2B+m2*v2%2C+v1-v2+%3D+v02-v01%2C+m1%3E0%2C+m2%3E0%2C+solve+for+v1%2Cv2+real

Now suppose instead I look at this collision from the “zeromomentum” frame, which is the frame in which the center-of-massis at rest. Here’s the input:

m1v1x ,i + m2v2x ,i = 0 → m1v1x ,f + m2v2x ,f = 0

(v1x ,f − v2x ,f ) = −(v1x ,i − v2x ,i )

Here’s the output. It’s clean!

v1x ,f = −v1x ,i v2x ,f = −v2x ,i

If you observe the collision from the ZM frame, the elastic collisionjust flips the sign of each velocity. That’s it!

Even the general inelastic case is neat and tidy in the ZM frame!

v1x ,f = −ev1x ,i v2x ,f = −ev2x ,i

animation of Earth frame vs. ZM frame

Computer programming for visual artists & graphic designers:processing.org

Chapter 6 included a few key ideas, some of which were obscuredby the notation and equations.

Law of inertia — this is big deal! (a.k.a. Newton’s law #1.)

I In an inertial reference frame, an isolated object at restremains at rest, and an isolated object in motion keepsmoving at a constant velocity.

I You can’t “feel” the difference between being at rest in Earth’sframe vs. being at rest in some other inertial reference frame.

I Imagine that you are in a jet airplane that has just landed andis in the midst of screeching to a stop on the runway. Is theframe of reference in which your body is at rest an inertialframe? Why or why not?

I (Illustrate with “ball popper.”)

More chapter 6 key ideas

I Center of mass.

~rcm =m1~r1 + m2~r2 + · · ·

m1 + m2 + · · ·

xcm =m1x1 + m2x2 + · · ·

m1 + m2 + · · ·I CoM of an object lies along axis of symmetry (if there is one).

I When analyzing the motion of a complicated object(composed of many pieces), it is often useful to considerseparately the motion of its CoM and the motion of thevarious internal parts w.r.t. the CoM.

I Illustrate by tossing complicated object across room.

More chapter 6 key ideas

I “Center-of-mass velocity” is the velocity of the CoM of asystem of objects:

~vcm =m1~v1 + m2~v2 + · · ·

m1 + m2 + · · ·

vx ,cm =m1vx1 + m2vx2 + · · ·

m1 + m2 + · · ·I An isolated system’s CoM velocity cannot change!

I You can see this by noticing that the numerator in ~vcm is thesystem’s total momentum, which you know is constant for anisolated system.

More chapter 6 key ideas

I An isolated system’s CoM velocity cannot change!

I A somewhat obscure consequence of this fact is that even in atotally inelastic collision, it is not necessarily possible toconvert 100% of the initial kinetic energy into heating up,mangling, etc. the colliding objects.

I Momentum conservation requires that the CoM velocitycannot change, so if the CoM is moving initially, it has tokeep moving after the collision.

I Textbook: “convertible” vs. “translational” parts of asystem’s kinetic energy.

I That idea is worth remembering, but the math is not.

I (Can illustrate using colliding carts.)

Zero-Momentum (ZM) frame for two-object collisions

I Very useful (but difficult to visualize) tool: ZM frame.

I Elastic collision analyzed in ZM (“∗”) frame:

v∗1i ,x = v1i ,x − vZM,x , v∗

2i ,x = v2i ,x − vZM,x

v∗1f ,x = −v∗

1i ,x , v∗2f ,x = −v∗

2i ,x

v1f ,x = v∗1f ,x + vZM,x , v2f ,x = v∗

2f ,x + vZM,x

I Inelastic collision analyzed in ZM frame (restitution coeff. e):

v∗1f ,x = −ev∗

1i ,x , v∗2f ,x = −ev∗

2i ,x

I Step 1: shift velocities into ZM frame, by subtracting vZM,x

I Step 2: write down (very simple!!) answer in ZM frame

I Step 3: shift velocities back into Earth frame, by adding vZM,x

Which of these systems are isolated?

(a) While slipping on ice, a car collides totally inelastically withanother car. System: both cars (ignore friction)

(b) Same situation as in (a). System: slipping car

(c) A single car slips on a patch of ice. System: car

(d) A car brakes to a stop on a road. System: car

(e) A ball drops to Earth. System: ball

(f) A billiard ball collides elastically with another billiard ball on apool table. System: both balls (ignore friction)

(g) (a) and (f)

(h) (a), (c), and (f)

(i) (a), (b), (c), and (f)

(j) all

Read ch7 (interactions) for Monday

I When two objects interact with one another, the resultingchanges in the two objects’ momenta are equal in magnitudeand opposite in direction: ∆p1x = −∆p2x

I Therefore ∆v1x/∆v2x = −m2/m1

I Therefore (this is hugely important) when 2 objects interact,

a1x

a2x= −m2

m1

I When the medicine ball and I push apart from one another,we both accelerate: in opposite directions, and in inverseproportion to our masses.

Read ch7 (interactions) for Monday

I Lifting an object up a height ∆x in Earth’s gravity changes itsgravitational potential energy by

∆UG = mg∆x

I I usually remember U = mgh where h is height

I Basketball: back & forth between 12mv2 and mgh

I Kinetic energy, potential energy: “coherent”

I Thermal energy, source (i.e. fuel, food) energy: “incoherent”

I Source energy (e.g. from my muscles) can be transformedinto a system’s kinetic or potential energy (basketball)

I Inevitably, “dissipative” interactions such as friction (thingsthat don’t make sense when the movie runs backward) convertcoherent mechanical energy into incoherent thermal energy.

Physics 8 — Friday, September 20, 2013

I Handed in HW3 today. I’ll put HW4 online tomorrow. HW3took many people 4+ hours. I’ll make HW4 shorter.

I For Monday: read ch7 (interactions). Very few equations!

I After ch9, the pace of reading will slow down to ≈ 1ch/wk!

I HW help sessions: Wed 7pm DRL 2N36, Thu 7pm DRL 3W2.

I My summary of key equations (“equation sheet”):positron.hep.upenn.edu/wja/phys008_2013/equations.pdf

I Expect a quiz on HW2 next Weds. I’ll try to make it shorter.It’s just an incentive to make sure you understand your HW.Quiz total counts only 5% of term grade, and semester quiztotal ≥ 80% gets full credit.

I I try hard to make it possible for you to do well in this coursewhether or not you’re someone who enjoys math/physicsexams. Many other courses make exams 100% of the grade.That’s why I put so much emphasis on homework.

I I’ll put today’s slides up on Canvas this afternoon.