Physics 1: exam 3

CQ: 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)

*Slows down* -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 -Δp = mfvf - mivi -can't come to a complete stop (zero) unless it has an infinite mass

examples of conservative forces

-Gravity -Springs -Electrical forces

Example: slide 4 Work done by friction on a closed path is not zero:

-closed path in a flat plane = friction (blue arrows = direction of motion and friction is always opposite the direction of motion) -looking at friction force for this example -fk = μkN (N is = mg) -W = Fd -W = μkmgd -work with friction will always be negative -in all segments: force and displacement the same (W = -μkmgd) -total work = -4(μkmgd) -non-conservative (no net work done = negative = not zero, so non-conservative

slide 6: before and after a teddy bear falling versus a bouncy ball

-free fall = no external force -ball = collision and bounce back -bear = collision and stops -before/falling = both have velocity down (in y direction) -after/hitting = ball has velocity up and teddy has zero velocity (in the y direction)

Example: slide 3 Work done by gravity on a closed path is zero:

-mg is weight -total work = sum of work from each segment -no horizontal force, only looking at gravitational force for this example -A --> B: work is zero (W = 0) because force is perpendicular to the direction of motion (gravity) = no work -only forces along the direction of motion do work -B --> C: work is negative (W = -mgh) because it is going against the direction of force (up) -C --> D: work is zero (W = 0) (same explanation as A --> B) -D --> A: work is positive (W = mgh) because it is going with the direction of force -total work = zero (because positive + negative + 0 + 0 = 0) <-- positive and negative cancel out -total work = 0 = conservative

A giant "egg" explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with piece B moving in the positive x direction. The masses of both pieces are indicated in (Figure 1), shown traveling in opposite directions (half A is 10 kg going left and half B is 30 kg going right) 1.) What is the magnitude of the momentum |p⃗ Ai| of piece A before the explosion? 2.) During the explosion, is the magnitude of the force of piece A on piece B greater than, less than, or equal to the magnitude of the force of piece B on piece A? 3.) The component of the momentum of piece B, pBx,f, is measured to be +500 kg⋅m/s after the explosion. Find the component of the momentum pAx,f of piece A after the explosion

1.) 0 kg•m/s (because, p = m x 0) (Similarly, piece B has zero momentum before the collision. The total momentum of the "egg," the sum of the two individual momenta, is also zero.) 2.) equal to 3.) -500 kg⋅m/s

An iron block with mass mB slides down a frictionless hill of height H. At the base of the hill, it collides with and sticks to a magnet with mass mM 1.) What is the speed v of the block and magnet immediately after the collision? 2.) Now assume that the two masses continue to move at the speed v from Part A until they encounter a rough surface. The coefficient of friction between the masses and the surface is μ. If the blocks come to rest after a distance s, which of the following equations would you use to find s?

1.) 1/2mBv^2 = mBgH v^2 = mBgH / 1/2mB vB = √2gH mB(√2gH) = (mB + mM)vM mB(√2gH) / (mB + mM) = vM 2.) ((mB^2)/mB+mM)gH= u(mB+mM)gs

A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 100-g pebble. 1.) Is the change of the boulder's momentum in one second greater than, less than, or equal to the change of the pebble's momentum in the same time period? 2.) Choose the best explanation for the question above

1.) F = ma --> 200 = 100 a --> a = 2 --> a = v/t --> 2 = v/1 --> v = 2 p = mv --> p = 100 x 2 --> p = 200 F = ma --> 200 = 0.1 a --> a = 2000 --> a = v/t --> 2000 = v/1 --> v = 2000 p = mv --> p = 0.1 x 2000 --> p = 200 *equal to the change of the pebble's momentum* 2.) Equal force means equal change in momentum for a given time

Suppose you throw a rubber ball at a charging elephant (not a good idea). 1.) When the ball bounces back toward you, is its speed greater than, less than, or equal to the speed with which you threw it? 2.) Explain.

1.) Greater than initial speed 2.) This is an elastic collision where the momentum is conserved. This means that the final momentum is equal to the initial momentum. But, since momentum is mass multiplied by the velocity, and final momentum is the momentums of the 2 objects added together. In order for the momentum to be the same before and after the collision, the velocity of the ball and the elephant together will have to be greater than the ball alone, because the balls mass does not change

A force is conservative if one of the two following statements is true:

1.) If the work it does on a moving object is independent of the path between the object's initial and final position. 2.) If no net work is done on an object moving around a closed path (starting and ending at the same point). -conserving mechanical/total energy = conservative -these 2 rules distinguish between conservative and non-conservative: 1 is conservative (zero force) and 2 is non-conservative (not a zero force)

A 0.22 kg apple falls from a tree to the ground, 4.0 m below. Ignore air resistance. Take ground level to be y = 0. 1.) Determine the apple's kinetic energy, K, the gravitational potential energy of the system, U, and the total mechanical energy of the system, E, when the apple's height above the ground is 4.0 m. 2.) Determine the apple's kinetic energy, K, the gravitational potential energy of the system, U, and the total mechanical energy of the system, E, when the apple's height above the ground is 3.0 m. 3.) Determine the apple's kinetic energy, K, the gravitational potential energy of the system, U, and the total mechanical energy of the system, E, when the apple's height above the ground is 2.0 m. 4.) Determine the apple's kinetic energy, K, the gravitational potential energy of the system, U, and the total mechanical energy of the system, E, when the apple's height above the ground is 1.0 m. 5.) Determine the apple's kinetic energy, K, the gravitational potential energy of the system, U, and the total mechanical energy of the system, E, when the apple's height above the ground is 0 m.

1.) PE = mgh --> 0.22 x 9.81 x 4 = PE --> *PE = 8.6328 J* *KE = 0 J* (because velocity is zero at the top) E = PE + KE --> *E = 8.6 J* 2.) PE = mgh --> PE = 0.22 x 9.81 x 3 --> *PE = 6.4 J* *E = 8.6 J* KE = E - U --> KE = 8.6 - 6.4 --> *KE = 2.2 J* 3.) PE = mgh --> PE = 0.22 x 9.81 x 2 --> *PE = 4.3 J* *E = 8.6 J* KE = E - U --> KE = 8.6 - 4.3 --> *KE = 4.3 J* 4.) PE = mgh --> PE = 0.22 x 9.81 x 1 --> *PE = 2.2 J* *E = 8.6 J* KE = E - U --> KE = 8.6 - 2.2 --> *KE = 6.4 J* 5.) PE = mgh --> 0.22 x 9.81 x 0 --> *PE = 0 J* *E = 8.6 J* KE = E - U --> KE = 8.6 - 0 --> *KE = 8.6 J*

We have made 2 explicit assumptions about our study of motion:

1.) The motion of the object is translational -Vertical, horizontal, along a parabola, or any other shape. There just aren't any twists or rotations. 2.) The object is a point particle -Real objects are big. Mass is spread out over the volume; there is some surface area that can interact with things. Until now we have assumed that everything is a point and that all forces act exactly on that point. -translate = moves -did not care about shape, just the center of mass (point where the object was balanced) -we will look at rotation instead of translation in this chapter (rotation vs linear)

In inelastic collisions, we know the "lost" kinetic energy has to go somewhere. Where does it go?

1.) Work done by friction (Heat) 2.) Deforming objects 3.) Sound energy -These are 3 major sources of energy "loss" during inelastic collisions. -momentum us more useful than KE

A piece of sheet metal of mass M is cut into the shape of a right triangle, as shown in the figure(Figure 1). A vertical dashed line is drawn on the sheet at the point where the mass to the left of the line (M/2) is equal to the mass to the right of the line (also M/2). The sheet is now placed on a fulcrum just under the dashed line and released from rest. 1.) Does the metal sheet remain level, tip to the left, or tip to the right? 2.) choose the best explanation from the following -Equal mass on either side will keep the metal sheet level. -The metal sheet extends for a greater distance to the left, which shifts the center of mass to the left of the dashed line. -The center of mass is to the right of the dashed line because the metal sheet is thicker there

1.) You can think of the center of mass as the "average" location of the system's mass. The mass M/2 on the right, on average, is located closer to the fulcrum than the mass M/2 on the left. The center of mass must therefore be to the left of the dashed line, and we conclude that the metal sheet will *tip to the left* 2.) The metal sheet extends for a greater distance to the left, which shifts the center of mass to the left of the dashed line.

We conserve both momentum and kinetic energy during an elastic collision:

1.) conservation of momentum: pf = p0 pAf + pBf = pA0 + pB0 mAvAf + mBvBf = mAvA0 + mBvB0 2.) conservation of kinetic energy: KEf = KE0 1/2mAvAf^2 + mBvBf^2 = 1/2mAvA0^2 + 1/2mBvB0^2

Ball 1 is thrown to the ground with an initial downward speed; ball 2 is dropped to the ground from rest. 1.) Assuming the balls have the same mass and are released from the same height, is the change in gravitational potential energy of ball 1 greater than, less than, or equal to the change in gravitational potential energy of ball 2? 2.) Choose the best explanation from among the following: -The gravitational potential energy depends only on the mass of the ball and the drop height. -Ball 1 has the greater total energy, and therefore more energy can go into gravitational potential energy. -All of the initial energy of ball 2 is gravitational potential energy.

1.) equal to the change in gravitational potential energy of ball 2 2.) The gravitational potential energy depends only on the mass of the ball and the drop height.

You throw a ball upward and let it fall to the ground. Your friend drops an identical ball straight down to the ground from the same height. 1.) Is the change in kinetic energy of your ball greater than, less than, or equal to the change in kinetic energy of your friend's ball? 2.) Choose the best explanation from among the following: -The change in gravitational potential energy is the same for each ball, which means the change in kinetic energy must be the same also. -Your ball is in the air longer, which results in a greater change in kinetic energy. -Your friend's ball converts all its initial energy into kinetic energy

1.) equal to the change in kinetic energy of your friend's ball 2.) The change in gravitational potential energy is the same for each ball, which means the change in kinetic energy must be the same also.

A 730-kg car stopped at an intersection is rear-ended by a 1780-kg truck moving with a speed of 15.0 m/s 1.) If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of the truck. 2.) find the final speed of the car

1.) m1v1i + m2v2i = m1v1f + m2v2f --> (1780)(15) + (730)(0) = (1780)(v1f) + (730)(v2f) --> 26700 = 1780(v1f) + 730(v2f) 26700 = (1780)v1f + (730)(15 + v1f) --> v1f = *6.27 m/s* (Here we learn how to determine the final speed of an object after an elastic collision with another object.) 2.) (1780)(15) = (1780)(6.27) + (730)(v2f) --> *v2f = 21.3 m/s* (Here we learn how to apply energy and momentum conservation to determine the speed of an object initially at rest after elastic collision with another object.)

A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 33 m/s . 1.) Assuming that air resistance can be ignored, find the initial speed of the rock. 2.) Find the greatest height of the rock as measured from the base of the cliff.

1.) mghi + 1/2mvi^2 = mghf + 1/2mvf^2 --> 81.6 + 1/2(0.26)vi^2 = 0 + 141.57 --> *vi = 21.478 m/s* 2.) mghi + 1/2mvi^2 = mghf + 1/2mvf^2 --> (0.26)(9.81)(32) + 1/2(0.26)(441) = (0.26)(9.81)h + 0 --> *h = 57 m*

A bullet of mass mb is fired horizontally with speed vi at a wooden block of mass mw resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed vf. (Figure 1) 1.) What best describes this collision? 2.) Which of the following quantities, if any, are conserved during this collision? 3.) What is the speed of the block/bullet system after the collision?

1.) perfectly inelastic 2.) momentum only 3.) mb • vi / (mb +mw)

Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of the ramp to the moment it stops?

1/2mv^2i + Wnc = 1/2mv^2f

Which form of the law of conservation of energy describes the motion of the block when it slides from the top of the table to the bottom of the ramp?

1/2mv^2i + mghi = 1/2mv^2f + mghf

An example: The ballistic pendulum (slide 13) A bullet is fired into a block that is suspended from the ceiling by a massless string. The block has a mass M and is initially at rest. The bullet has a mass m and is fired into the block with an initial speed v0 and embeds itself. What is the maximum height of the pendulum and bullet?

The height h can be found using conservation of mechanical energy after the object is embedded in the block. mv0 = (m+M)vf; ½ (m+M)vf2 = (m+M)gh

and the work is the change in potential energy:

W = PE0 - PEf --> W = mgh0 - mghf -ex: tossing an apple upwards = work is negative

Catching a wave, a 77-kg surfer starts with a speed of 1.1 m/s , drops through a height of 1.60 m , and ends with a speed of 8.3 m/s . How much nonconservative work was done on the surfer?

W = mgh - ½m(vf² - vi²) W = (77)(9.81)(1.6) - 1/2(77)(68.89 - 1.21) *W = 1397.088 J

Forces and time

We have talked about forces acting over distances, say a force acts for a certain time, Δt. We can define a quantity called the impulse W = Fd, can have forces over time (impulse)

Bouncy collisions

When the two objects do not stick together, we still conserve momentum, but we need more information

If our system includes non-conservative forces, (like the brakes on a car or the click-track that starts a roller coaster) the work energy theorem tells us that the work done by these forces is given as:

Wnc = Ef - E0 in this case, Ef ≠ E0 --> non-conservative force (because Wnc isn't zero)

We can manipulate our work-energy equation just a little bit:

Wnc = KEf - KE0 + PE0 - PEf Wnc = (KEf + PEf) - (KE0 + PE0) Wnc = Ef - E0 -E is mechanical energy/total energy

Is momentum conserved in each type of collision?

YES! Collision = Conservation of momentum

Impulse is the change in momentum resulting from

a collision. -This is the impulse-momentum theorem: ∑F Δt = maΔt --> m (Δv/Δt) Δt --> mΔv = Δp = mvf - mvi so... J = mvf - mvi -These are all vector quantities! The impulse in the x-direction comes from forces in the x-direction, the y-direction from force in the y-direction, ...

A collision or explosion is

an isolated event in which two or more bodies exert relatively strong forces on each other over a short time compared to the period over which their motions take place. They only experience forces subject to Newton's 3rd Law. All of the internal forces sum to zero. -collision = crashing -explosion = flying away -isolated = with more than one object, the 2 objects effect each other, but nothing else effects them (no external forces)

Momentum is constant in

an isolated system! This is the conservation of momentum!

If we examine the arc produced by a point on a rotating object, we can relate the distance the point travels to the distance from the center of rotation. This is the

angle of rotation

We need three reference points for a collision:

before, during, and after.

Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy.

change / potential

2 types of collisions

collisions where the objects bounce off each other and collisions where the objects stick together

Newton was initially interested in the study of

collisions. -He defined a quantity known as *momentum*.

Potential energy comes from only

conservative forces. -We can think of the work on an object as the sum of the work from conservative forces plus the work from non-conservative forces (example: gravitational force + friction force) W = Wc + Wnc W = KEf - KE0 --> W = ΔKE Wc = PE0 - PEf --> Wc = -ΔPE *Wnc = ΔKE + ΔPE* -non-conservative work = KEf - KE0 + PE0 - PEf

law of conservation of momentum

constant/unchanging momentum -momentum is never different after from before

The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion.

distance / kinetic -It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball.

Example: A particle starting from point A (h = 3 m) is projected down a curved runway. Upon leaving the runway, the particle is traveling straight upwards and reaches a height of 4.00 m above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A

draw a picture (in notes) -h1 = 3, h2 = 4, vf = 0, v0 = ?, Ui = mgh1 Ki = 1/2mv^2 Uf = mgh2 Kf = 0 mgh1 + 1/2mv^2 = mgh2 + 0 h1 + 1/2v^2 = h2 3 + 1/2(v0)^2 = 4 *v0 = 4.43m/s*

To illustrate the work-energy concept, consider the case of a stone falling from xi to xf under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone.

force / kinetic

What force is responsible for the decrease in the mechanical energy of the block?

friction

angular displacement

have a reference line (start of movement) Δθ = θfinal - θinitial

Momentum has units of

kg•m/s

example: slide 10 You raise your keys to a certain height then drop them, which one is initial? which one is final?

left image: -Ui = mgh and KEi is zero (when holding the keys = initial) -final position: Uf = 0 and KEf = 1/2mv^2 -Wnc = 0 (because gravitational force is the only force acting = no non-conservative energy) -Ef = Ei --> constant energy *conservation of mechanical energy equation*: U0 + K0 = Uf + Kf (U0 + K0 is E0 and Uf + Kf is Ef) -mgh + 0 = 0 + 1/2mv^2 --> gh = 1/2v^2 (because m cancels out) --> v = square root (2gh)

A 95-kg astronaut and a 1000-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving it a speed of 0.16 m/s directly away from the shuttle. Seven-and-a-half seconds later the astronaut comes into contact with the shuttle. What was the initial distance from the shuttle to the astronaut?

m1v1 = m2v2 (95)(v1) = (1000)(0.16) v1 = 1.68 m/s d = vt d = 1.68 x 7.5 d = 12.6 m

What happens to the cart when it stops raining?

maintains a constant speed/uniform motion/momentum

CQ: A system of particles is known to have a total kinetic energy of zero. What can you say about the total momentum of the system?

momentum of the system is zero -because the total kinetic energy is zero, this means that all of the particles are at rest (v = 0). Therefore, because nothing is moving, the total momentum of the system must also be zero -no KE = no motion = no momentum -KE = 1/2mv^2 -p = mv

A 1210 kg car drives along a city street at 22.0 miles per hour (9.83 m/s ) What speed must the 0.142-kg baseball have if its momentum is to be equal in magnitude to that of the car?

p = mv p = 1210 x 9.83 p = 11894.3 kg•m/s 1894.3 = 0.142 x v v = 83762.7 m/s

Two ice skaters stand at rest in the center of an ice rink. When they push off against one another the 70-kg skater acquires a speed of 0.64 m/s . If the speed of the other skater is 0.87 m/s , what is this skater's mass?

p = mv --> p = 70 x 0.64 --> 44.8 p = mv --> 44.8 = m x 0.87 --> *m = 51.49 kg* Here we learn how to apply the conservation of momentum to determine the mass of an object.

Consider a collision of two objects A and B:

p0 = pf PA0 + PB0 = PA+Bf (mA)(VA0) + (mB)(VB0) = (mA + mB)(Vf) -Kinetic energy is not conserved! Collision is completely inelastic! -2 things come together and stop = no KE after (not conserved) = momentum is always the same before and after (conserved) (p0 = pf) -momentum of thing A before + momentum of thing B before = momentum of thing A + B after -added them together because now 1 object (stuck together = mass of one is the same as the mass of the 2 together)

Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v1 = v and has mass m1 = 2m. Object 2 moves with speed v2 = 2√v and has mass m2 = m 1.) Which object has the larger magnitude of its momentum? 2.) Which object has the larger kinetic energy?

p1 = 2mv p2 = m2√v KE1 = 1/2(2m)(v^2) KE2 = 1/2(m)(2√v)^2 --> 1/2(m)(2v) 1.) Object 1 has the greater magnitude of its momentum. 2.) The objects have the same kinetic energy. -Many students confuse the quantities momentum and kinetic energy and think that they are the same thing. However, as this problem has demonstrated, they are two very different physical quantities and even if two objects have different momenta, they can still have the same amount of kinetic energy.

IP Calculate the work done by gravity as a 5.8 kg object is moved from A to B in the figure(Figure 1) along paths 1 and 2. path 1 = over 5 m and down 1 m path 2 = down 1 m and over 5 m How do your results depend on the mass of the block? Specifically, if you increase the mass, does the work done by gravity increase, decrease, or stay the same?

path 1: W1 = mgh --> W1 = 5.8(9.81)(1.0) --> W1 = 56.898 J path 2: W2 = mgh --> W2 = 5.8(9.81)(1.0) --> W2 = 56.898 J increase

elastic collision

pf = p0 pAf + pBf = pA0 + pB0 mAvAf + mBvBf = mAvA0 + mBvB0 -bounce back = *conservation of momentum and kinetic energy (because energy isn't lost)*

To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes the kinetic energy change.

position

Example of A completely inelastic collision with Assembling Freight Train

slide 10: -before: initial velocity going to the right for car one and the velocity of car 2 (to the right of cart one) is zero (at rest) -after: they move together, velocity final moves right -p0 = m1v1i + m2v2f -pf = (m1 + m2)vf these two equations are equal to each other - m1v1i + m2v2f = (m1 + m2)vf -going to be slower than the first car was initially after they collide -won't completely stop unless an external force is acted, same speed with opposite direction

Using conservation of energy, find the speed vb of the block at the bottom of the ramp

square root (V^2 + 2gh)

This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______

sum / conserved

The net work done by external non-conservative forces is equal to

the change in total energy! -If this work is 0, then there is no change in energy (this means the energy is conserved!) -if no non-conservative forces (like friction) --> Wnc = 0 --> Ef = E0 (means energy is conserved)

momentum

the product of the mass and velocity of an object p ≡ mv ≡ means it is identical to/defined by momentum is a vector (has direction and magnitude)

We can think, instead, that an object at a certain height has some possible energy that could be released if it is lowered. Similarly, if the object is raised,

there would be more available energy if dropped. We call this available energy *the gravitational potential energy*

If we lose (or gain) no energy from external forces,

we are freely exchanging kinetic and potential energies. -ex: Amusement parks are energy conservation demos on an epic scale. Roller coasters are sent up large hills, raising the potential energy. This energy is then traded for kinetic energy as the coaster goes down hill. If we ignore friction during the ride, the total energy of the roller coaster would never change! -sum of KE and PE have to be equal

The concept of the center mass is

what allows us to make these assumptions

From the work-energy theorem,

when we raise or lower an object causing work, we get a change in energy. W = KEf - KE0 -if difference/Work is positive = final > initial

Potential energy is a concept that builds on the

work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. -The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. -In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy: Kf+Uf=Ef=Wnc+Ei=Wnc+Ki+Ui,where Uf and Ui are the final and initial potential energies, and Wnc is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy

example: The carbon monoxide molecule (CO) consists of a carbon atom and an oxygen atom separated by a distance of 1.13x10^-10 m. The mass mc of the carbon atom is 0.750 times the mass of the oxygen atom. Determine the location of the center of mass of this molecule relative to the carbon atom.

xcm = (mc xc + mo xo)/(mc + mo) We let the position of the carbon atom equal 0. mc = 0.75mo x = 1.13x10^-10 m xcm = ? (0.75mo)(0) + (mo)(1.13x10^-10) / (0.75mo + mo) = xcm xcm = (mo)(1.13x10^-10) / mo + 1.75 xcm = 1.13x10^-10 / 1.75 xcm = 6.46x10^-11m

If there are no external forces, Newton's 2nd law tells us that:

∑FΔt = pf - pi = 0 p = constant so pf = pi

Collisions

-2 or more things bumping into each other -isolated group of objects (no net external force) means ∑F = 0 (no force = no rate in change of momentum = constant/stays the same (conserved) -law of conservation of momentum

Example: Two vehicles A and B are traveling west and south, respectively, toward the same intersection, where they collide and lock together. Before the collision, A (weight 12.0 kN) has a speed of 64.4 km/h = 17.9 m/s and B (weight 16.0 kN) has a speed of 96.6 km/h = 26.8 m/s. Find the magnitude and direction of the (interlocked) vehicles velocity immediately after the collision.

-A: 12 kN --> 12000 N --> m = W/g --> m = 1224.49 kg -B: 16 kN --> 16000 N --.> m = W/g --> m = 1632.65 kg pA0 = mAvA --> pA0 = (1224.49)(64.4) --> pA0 = 21,918 kg•m/s west pB0 = mBvB --> pB0 = (1632.65)(96.6) --> pB0 = 43,755 kg•m/s south pA+Bf = pA0 + pB0 --> 48,937 kg•m/s --> *17.1 m/s* tanθ = pB/pA --> *63.4º south of west*

examples of non-conservative forces

-Friction -Air resistance -Tension -Normal forces -Rocket forces

Changes in momentum

As is always the case in physics, change is the final condition minus the initial condition. The change in momentum is given as: Δp = pf - pi = mfvf - mivi

Three balls are thrown upward with the same initial speed v0, but at different angles relative to the horizontal, as shown in the figure. Ignoring air resistance, indicate which of the following statements is correct: -At the dashed level all three balls have the same speed. -At the dashed level ball 3 has the lowest speed. -At the dashed level ball 1 has the lowest speed. -At the dashed level the speed of the balls depends on their mass

At the dashed level all three balls have the same speed

When the objects stick together it is easy to see what happens.

Consider a collision of two objects A and B

Definition of mechanical energy:

E = U + K -U is the same as potential energy (PE)

Or equivalently:

E = constant if Ef = E0 --> constant energy

Using this definition and considering only conservative forces, we find:

Ef = E0

An example: Free-fall revisited A penny starts at rest at the top of the Empire State building, 381 m above the ground. What is the speed of the penny just before it hits the ground?

Force = mg, distance = height, PE = mgh -h = 381, vf = ? mgh + 0 = 0 + 1/2mv^2 gh = 1/2v^2 (9.8)(381) = 1/2v^2 *v = 86.4 m/s*

Chapter 8

Gravitational Potential Energy and Conservative forces

Example: A 5.0 g high bounce ball undergoes an elastic collision with the floor. The ball starts at rest at a height 1.0 m above the floor. a) What is the change in momentum of the ball? b) To what height does the ball return?

If the collision is elastic, energy is conserved in the collision. Earth is really big compared to the ball, so Earth's velocity does not noticeably change during the collision. The speed of the ball does not change, but the direction changes. a.) PE = mgh = 1/2mvi^2 --> vi = √2gh --> change in momentum = m1vf - m1vi --> p = (0.005)(√2(9.8)(1)) - (0.005)(-√2(9.8)(1)) --> *p = 0.0443 kg•m/s* (vi = -vf because its the same but in opposite directions) b.) The ball has the same speed before and after the collision, so it will return to the same height, 1.0 m

Finding the center of mass

In problems in which objects interact with each other, the mass of the system is located in several places. One can speed of a kind of average location, i.e., center of mass for the total mass. In 1 D: Xcm = (mAxA + mBxB) / (mA + mB) Vcm = (mAvA + mBvB) / (mA + mB) If we were to place an object's center of mass over a pivot, it would balance.

impulse

J = FΔt impulse = force times change in time -force is a vector, so impulse is a vector

As the block slides across the floor, what happens to its kinetic energy K, potential energy U, and total mechanical energy E?

K decreases; U stays the same; E decreases.

As the block slides down the ramp, what happens to its kinetic energy K, potential energy U, and total mechanical energy E?

K increases; U decreases; E stays the same

Work-Energy Theorem The work-energy theorem states:

Kf = Ki + Wall where Wall is the work done by all forces that act on the object, and Ki and Kf are the initial and final kinetic energies, respectively.

chapter 9

Linear Momentum & Collisions

Momentum and collisions

Momentum is conserved in all collisions!

CQ: Two objects are known to have the same momentum. Do these objects necessarily have the same kinetic energy?

NO -if object 1 has mass m and speed v and object 2 has mass 1/2m and speed 2v, they will both have the same momentum. However, because KE = 1/2mv^2, we see that object 2 has twice the KE of object 1, due to the fact that the velocity is squared -KE = p^2/2m

CQ: A system of particles is known to have a total momentum of zero. Does it necessarily follow that the total kinetic energy of the system is also zero?

NO -momentum is a vector, so the fact that p total = 0 does not mean that the particles are at rest! They could be moving such that their momenta cancel out when you add up all of he vectors. In that case, because they are moving, the particles would have non-zero KE -total vectors

Is kinetic energy conserved in a collision?

NO! Only in certain collisions, elastic. -sart to finish -can be different or the same

Impulse has units of

N•s = kg•m/s

center of mass: In problems in which objects interact with each other, the mass of the system is located in several places.

One can speed of a kind of average location, i.e., center of mass for the total mass. The center of mass of a system is the point where the system can be balanced in a uniform gravitational field: if m1 = 4 g and m2 = 2 g m1 --|CM|---------m2 m1----|CM|----m1 m2 ----------|CM|--m1

gravitational potential energy

PE = mgh -h = height at which you raise the object (surface/floor is zero height)(changing position vertically)

Find the gravitational potential energy of an 80 kg person standing atop Mt. Everest at an altitude of 8848 m. Use sea level as the location for y = 0.

PE = mgh PE = 80 x 9.8 x 8848 PE = 6936832 J

At an amusement park, a swimmer uses a water slide to enter the main pool If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.66 m , what is her speed at the bottom of the slide?

Potential energy lost = Kinetic energy gained so mgh = 1/2mv^2 m(9.81)(2.66) = 1/2mv^2 mass cancels out v^2 = 2(9.81)(2.66) *v = 7.22 m/s* Here we learn how to apply the conservation of mechanical energy to determine the speed of an object sliding from a given height.

chapter 10

Rotational Kinematics & Energy

Find the amount of energy E dissipated by friction by the time the block stops.

Since the block comes to a stop at the level where U=0, the energy lost to friction is E = 1/2 mv^2 + mgh

If the mass remains constant:

∑F = ma = m(Δv/Δt) Δp = mΔv

Newton's 2nd Law with momentum

∑F = Δp/Δt