Physics 8 — Monday, September 19, 2011

I will hand back homework #1 during class, while people arediscussing some of the Peer Instruction questions.

If you scored 52/60 or better (as 85% of you did), you’re doinggood work. If you scored below 52/60, try working with a friend ortwo to make it easier to catch silly mistakes.

Continuing with the inclined air track . . .

I Reminder: the x axis points downhill, and the point on thetop of the ramp is called x = 0.

Which of the following shows the expected shapes of x(t) [blue]and vx(t) [red] if I shove the cart upward from x = 2 m? (Wedecided on Friday that answer D was correct.)

I shove the cart uphill and watch it travel up and back. At the topof its trajectory (where it turns around), vx is

(a) positive

(b) negative

(c) zero

(d) infinite

(e) undefined

Which of the following shows the expected shapes of x(t) [blue]and vx(t) [red] if I shove the cart gently upward from x = 1 m?

A

C

B

D

Which of the following shows the expected shapes of x(t) [blue]and vx(t) [red] if I shove the cart gently downward from x = 0 m?

A

C

B

D

What are the expected shapes of x(t) [blue] and vx(t) [red] for atennis ball tossed upward, when the x axis points upward?

A

C

B

D

Tennis ball

What are the values of vx and ax at the top of the tennis ball’strajectory (assuming that the x axis points upward)?

(a) vx = 0, ax = −9.8 m/s2

(b) vx = 0, ax = 0

(c) vx = 0, ax = +9.8 m/s2

(d) vx = 0, ax is undefined

(e) vx < 0, ax = −9.8 m/s2

(f) vx < 0, ax = 0

(g) vx < 0, ax = +9.8 m/s2

(h) vx < 0, ax is undefined

Ball thrown downward

If you throw an object downward in the absence of air resistance,its initial vx and ax are (remember that we’ve chosen x axis topoint upward)

(a) vx = 0, ax = −9.8 m/s2

(b) vx = 0, ax = 0

(c) vx = 0, ax = +9.8 m/s2

(d) vx = 0, ax is undefined

(e) vx < 0, ax = −9.8 m/s2

(f) vx < 0, ax = 0

(g) vx < 0, ax = +9.8 m/s2

(h) vx < 0, ax is undefined

Today’s reading: Momentum!

“I think momentum in general is a pretty interesting concept. It’sa physics concept/quantity that is much less intuitive than say,velocity, inertia, acceleration, etc., and yet it can be so welldemonstrated in such basic experiments and phenomena ineveryday life.”

“Momentum is the product of the inertia and the velocity of anobject. This concept is useful because momentum is conserved sothat the momentum of a system can only change due to a transferof momentum between the environment and the system.”

“Momentum is an object’s capacity to affect the motion of theother objects in a collision. This takes into account (a) the amountof matter that is in motion and (b) how fast that matter ismoving.”

“The change of x component of the velocity of the cart m isnegative three times the change of x component of the velocity ofthe cart 3m.”

Inertia

An object tends to resist changes in its velocity. This tendency ismeasured by an object’s inertia, m.

I Inertia is just another word for mass. The author is preparingyour brain for the fact that mass plays two key roles in physics:(1) a tendency to resist acceleration (i.e. inertia), and (2) agravitational role related to weight, which we’ll study later.

When two objects of inertias m1 and m2 collide, their changes invelocity are inversely related to their inertias:

∆v1x

∆v2x= −m2

m1

Inertia

An object tends to resist changes in its velocity. This tendency ismeasured by an object’s inertia, m.

I Inertia is just another word for mass. The author is preparingyour brain for the fact that mass plays two key roles in physics:(1) a tendency to resist acceleration (i.e. inertia), and (2) agravitational role related to weight, which we’ll study later.

When two objects of inertias m1 and m2 collide, their changes invelocity are inversely related to their inertias:

∆v1x

∆v2x= −m2

m1

Inertia

An object tends to resist changes in its velocity. This tendency ismeasured by an object’s inertia, m.

I Inertia is just another word for mass. The author is preparingyour brain for the fact that mass plays two key roles in physics:(1) a tendency to resist acceleration (i.e. inertia), and (2) agravitational role related to weight, which we’ll study later.

When two objects of inertias m1 and m2 collide, their changes invelocity are inversely related to their inertias:

∆v1x

∆v2x= −m2

m1

Momentum

The product ~p = m~v (called momentum) is extremely useful inphysics, because it is always conserved in an isolated system.

I A system is isolated if nothing in the external world is pushingor pulling on it.

I Imagine how it feels to throw a very heavy ball.

I Now imagine that you are standing on a sheet of ice!

I The difference is the impulse you get from the interactionbetween your shoes and the non-slippery floor.

It is often (but not always!) easy to choose “the system” such thatthe external influences are small enough to be safely ignored.

Then~pi = ~pf

m1~v1i + m2~v2i = m1~v1f + m2~v2f

Momentum

The product ~p = m~v (called momentum) is extremely useful inphysics, because it is always conserved in an isolated system.

I A system is isolated if nothing in the external world is pushingor pulling on it.

I Imagine how it feels to throw a very heavy ball.

I Now imagine that you are standing on a sheet of ice!

I The difference is the impulse you get from the interactionbetween your shoes and the non-slippery floor.

It is often (but not always!) easy to choose “the system” such thatthe external influences are small enough to be safely ignored.

Then~pi = ~pf

m1~v1i + m2~v2i = m1~v1f + m2~v2f

Momentum

The product ~p = m~v (called momentum) is extremely useful inphysics, because it is always conserved in an isolated system.

I A system is isolated if nothing in the external world is pushingor pulling on it.

I Imagine how it feels to throw a very heavy ball.

I Now imagine that you are standing on a sheet of ice!

I The difference is the impulse you get from the interactionbetween your shoes and the non-slippery floor.

It is often (but not always!) easy to choose “the system” such thatthe external influences are small enough to be safely ignored.

Then~pi = ~pf

m1~v1i + m2~v2i = m1~v1f + m2~v2f

Momentum

The product ~p = m~v (called momentum) is extremely useful inphysics, because it is always conserved in an isolated system.

I A system is isolated if nothing in the external world is pushingor pulling on it.

I Imagine how it feels to throw a very heavy ball.

I Now imagine that you are standing on a sheet of ice!

I The difference is the impulse you get from the interactionbetween your shoes and the non-slippery floor.

It is often (but not always!) easy to choose “the system” such thatthe external influences are small enough to be safely ignored.

Then~pi = ~pf

m1~v1i + m2~v2i = m1~v1f + m2~v2f

Momentum

The product ~p = m~v (called momentum) is extremely useful inphysics, because it is always conserved in an isolated system.

I A system is isolated if nothing in the external world is pushingor pulling on it.

I Imagine how it feels to throw a very heavy ball.

I Now imagine that you are standing on a sheet of ice!

I The difference is the impulse you get from the interactionbetween your shoes and the non-slippery floor.

It is often (but not always!) easy to choose “the system” such thatthe external influences are small enough to be safely ignored.

Then~pi = ~pf

m1~v1i + m2~v2i = m1~v1f + m2~v2f

Momentum

The product ~p = m~v (called momentum) is extremely useful inphysics, because it is always conserved in an isolated system.

I A system is isolated if nothing in the external world is pushingor pulling on it.

I Imagine how it feels to throw a very heavy ball.

I Now imagine that you are standing on a sheet of ice!

I The difference is the impulse you get from the interactionbetween your shoes and the non-slippery floor.

It is often (but not always!) easy to choose “the system” such thatthe external influences are small enough to be safely ignored.

Then~pi = ~pf

m1~v1i + m2~v2i = m1~v1f + m2~v2f

Momentum

The product ~p = m~v (called momentum) is extremely useful inphysics, because it is always conserved in an isolated system.

I A system is isolated if nothing in the external world is pushingor pulling on it.

I Imagine how it feels to throw a very heavy ball.

I Now imagine that you are standing on a sheet of ice!

I The difference is the impulse you get from the interactionbetween your shoes and the non-slippery floor.

It is often (but not always!) easy to choose “the system” such thatthe external influences are small enough to be safely ignored.

Then~pi = ~pf

m1~v1i + m2~v2i = m1~v1f + m2~v2f

Momentum

If you write down

m1~v1i + m2~v2i = m1~v1f + m2~v2f

and work only with the x component for now:

m1v1i ,x + m2v2i ,x = m1v1f ,x + m2v2f ,x

you can just rearrange the terms to get

∆v1x

∆v2x= −m2

m1

So you really only have to remember one thing from this chapter:momentum ~p = m~v is conserved for an isolated system (i.e. for asystem with no external pushing/pulling).

Momentum

If you write down

m1~v1i + m2~v2i = m1~v1f + m2~v2f

and work only with the x component for now:

m1v1i ,x + m2v2i ,x = m1v1f ,x + m2v2f ,x

you can just rearrange the terms to get

∆v1x

∆v2x= −m2

m1

So you really only have to remember one thing from this chapter:momentum ~p = m~v is conserved for an isolated system (i.e. for asystem with no external pushing/pulling).

Momentum

If you write down

m1~v1i + m2~v2i = m1~v1f + m2~v2f

and work only with the x component for now:

m1v1i ,x + m2v2i ,x = m1v1f ,x + m2v2f ,x

you can just rearrange the terms to get

∆v1x

∆v2x= −m2

m1

So you really only have to remember one thing from this chapter:momentum ~p = m~v is conserved for an isolated system (i.e. for asystem with no external pushing/pulling).

Momentum

If you write down

m1~v1i + m2~v2i = m1~v1f + m2~v2f

and work only with the x component for now:

m1v1i ,x + m2v2i ,x = m1v1f ,x + m2v2f ,x

you can just rearrange the terms to get

∆v1x

∆v2x= −m2

m1

So you really only have to remember one thing from this chapter:momentum ~p = m~v is conserved for an isolated system (i.e. for asystem with no external pushing/pulling).

Momentum: collisions on air track

Let’s make∆v1x

∆v2x= −m2

m1

or equivalently

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

come to life more visually on the air track!

What are the expected shapes of v1,x(t) [blue] and v2,x(t) [red]when m2 = m1, and initially cart 1 is moving to the right andcart 2 is stationary?

A

C

B

D

What are the expected shapes of v1,x(t) [blue] and v2,x(t) [red]when m2 = m1, and initially cart 1 is moving to the right andcart 2 is moving to the left?

A

C

B

D

What are the expected shapes of v1,x(t) [blue] and v2,x(t) [red]when m2 = 2m1, and initially cart 1 is moving to the right andcart 2 is stationary?

A

C

B

D

What are the expected shapes of v1,x(t) [blue] and v2,x(t) [red]when m2 = 2m1, and initially cart 1 is stationary and cart 2 ismoving to the left?

A

C

B

D

What are the expected shapes of v1,x(t) [blue] and v2,x(t) [red]when m2 = 2m1, and initially cart 1 is moving to the right andcart 2 is moving to the left?

A

C

B

D