HW06 - conservation of energy and momentum: 8-33, 8-37, 8-41, 8-64,8-x12, 9-CQ6, 9-4, 9-9, 9-14, 9-42.

Chapter 8, problem 33: A spring with is at the top of a frictionless incline of angle. The lower end of the incline is distance from the end of the spring, which is at its relaxed length. A canister is pushed against the spring until the spring is compressed and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

The kinetic energy of the canister is given either by the work-energy theorem, or by conservation of energy,

What results is , where and ; using any one you want, you get a speed of,

When the canister reaches the lower end, gravity does an additional amount of work ; we just use the work-energy theorem, and redo the steps in ,

chapter 8, problem 37: A uniform cord of length and mass is initially stuck to a ceiling. Later, it hangs vertically from the ceiling with only one end still stuck. What is the change in the gravitational potential energy of the cord with this change in orientation? (Hint: Consider a differential slice of the cord and then use integral calculus.)

The textbook hint kind of sucks. Use my hint, and start from , and use , and integrate over , in which, due to the cord being uniform, we have ,

We can let the cord hanging from the ceiling have a potential energy of zero, and let the cord hanging have a nonzero potential energy. CAUTION: now it remains to get the sign of potential energy right. I have not mentioned a coordinate systemI just asserted that . Using the equation , we see that , so we made down the negative direction. This means that , and ! Plugging these into , we finally have,

That is, in order to check our final result, we massaged it into an expression that makes it clear that is given by times the displacement of the center-of-mass coordinate of the cord: very obviously a displacement of for a length-L cord initially with its entire length stuck at , and having a final state of hanging straight-downwards (in the negative direction).

chapter 8, problem 41 (|| 8.40): A single conservative force F(x) acts on a 1.0 kg particle that moves along an x axis. The potential energy U(x) associated with F(x) is given by where x is in meters, , and . At x = 5.0 m the particle has a kinetic energy of 2.0 J. (a) What is the mechanical energy of the system? (b) Make a plot of U(x) as a function of x for , and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of x the particle can reach and (d) the greatest value of x the particle can reach. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f ) the value of x at which it occurs. (g) Determine an expression in newtons and meters for F(x) as a function of x. (h) For what (finite) value of x does F(x) = 0? Solution: (a) The (mechanical) energy at x = 5.0 m is,

(b) A plot of the potential energy curve (SI units understood) and the energy E (the horizontal line) is shown for 0 x 10 m. Because is constant, we find that , and we can plot that too,

(c) and (d) The lowest and highest reachable by the particle are obviously where its kinetic energy is zero (good thing we plotted it, even so we didnt have to! Its always useful to solve extra of a problem, because sometimes its easier to answer a question of your own designing rather than someone elses). This is obviously at and (or, where the red-dashed-curve passes through zero).

(e) One can either use calculus, or read off the maximum value of the kinetic energy curve shown in (man, its so awesome that we graphed itmade our lives easier!). We use the former to check the latter. Using the product rule (), and the chain rule (), we have,

Solving for , we immediately get ; indeed, this is corroborated by the plot ; kinetic energy evaluated at this yields,

(f) As mentioned in the previous part, the minimum of the U curve occurs at .(g) The force implied by a potential energy is given by following the exact same machinations as in in the formula I gave in class, and jumped up and down about how great it was,

Notice that both sides of are vectors: I illustrated it that way (in contrast to the solution manual) to avoid confusing you. (h) We have already done this in : .

Chapter 8, Problem 64: In the figure, a block is released from rest at height d = 40 cm and slides down a frictionless ramp and onto a first plateau, which has length d and where the coefficient of kinetic friction is 0.50. If the block is still moving, it then slides down a second frictionless ramp through height d/2 and onto a lower plateau, which has length d/2 and where the coefficient of kinetic friction is again 0.50. If the block is still moving, it then slides up a frictionless ramp until it (momentarily) stops. Where does the block stop? If its final stop is on a plateau, state which one and give the distance L from the left edge of that plateau. If the block reaches the ramp, give the height H above the lower plateau where it momentarily stops. (a)The block stops on the: 1 - First plateau; 2 - Second plateau; 3 - Ramp. Give the number of the correct answer. (b) Give L or H, whichever is appropriate.

Notice I have labeled the points 1, 2, 3, 4, 5. We are going to calculate the kinetic energy at each of those points, due to (1) work done by friction, and (2) work done by gravity,

We pause for a moment, and notice that because for any conceivable coefficient of kinetic friction. Therefore, we calculate the kinetic energy at points 3 and 4,

We conclude that for exactly the same reasoning that we did . Thus, there indeed is a distance that the block slides up the ramp, which we calculate forthwith using ,

Chapter 8, Problem X12: Figure (a) shows a molecule consisting of two atoms of masses m and M (with M >> m) and separation r. Figure (b) shows the potential energy U(r) of the molecule as a function of r. (a) Describe the motion of the atoms if the total mechanical energy E of the two-atom system is greater than zero (as is E1). (b) Describe the motion of the atoms if the total mechanical energy E of the two-atom system is less than zero (as is E2). (c) For E1 = 1 x 10-19 J and r = 0.3 nm, find the potential energy of the system. (d) For E1 = = 1 x 10-19 J and r = 0.3 nm, find the total kinetic energy of the system. (e)ForE1= = 1 x 10-19J andr= 0.3 nm, find the force (magnitude and direction) acting on each atom. (f)For what values ofris the force repulsive? (g)For what values ofris the force attractive? (h)For what values ofris the force zero?

(a) The atoms will move apart and permanently separate. (b) The distance between the atoms will oscillate, but the atoms will not permanently separate

(c) (d) (e) (f) (g) (h)

Chapter 9, Concept Question 6: Figure 9-28 shows four groups of three or four identical particles that move parallel to either the x axis or the y axis, at identical speeds. Rank the groups according to center-of-mass speed, greatest first.

For identical masses (i.e., ), the center of mass speed in 2D is given by,

The velocities in the Figure can be approximated as,

Hence, we have the vector-resultants using the formula ,

Hence, we obviously have,

Chapter 9, Problem 4 SN: In the figure, three uniform thin rods, each of length L = 22 cm, form an inverted U. The vertical rods each have a mass m = 14 g; the horizontal rod has a mass 3m = 3*(14g) = 42 g. What are (a) the x coordinate and (b) the y coordinate of the system's center of mass? Give your answer in terms of the variables given.

We know that ; in this formula, ; e.g., it is as if all the individual rods are replaced by point-masses whose center of mass is (obviously) located at the rods midpoint,

Chapter 9, Problem 9: A stone is dropped at t = 0. A second stone, with 2 times the mass of the first, is dropped from the same point at t = 160 ms. (a) How far below the release point is the center of mass of the two stones at t = 300 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time?

If we set , then . Then,

Plugging in specific numbers for this problem,

The velocity of the center of mass at all times follows from taking the time-derivative of the position at all times; indeed, this part of the problem follows quite readily from the same machinations,

Chapter 9, Problem 14 SN: In the figure, two particles are launched from the origin of the coordinate system at time t = 0. Particle 1 of mass m = 5.00 g is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed v = 10.0 m/s. Particle 2 of mass m2 = 3/5m = 3.00 g is shot with a velocity of magnitude v2 = 2v = 20.0 m/s at an upward angle such that it always stays directly above particle 1 during its flight. (a) What is the maximum height Hmax reached by the center of mass of the two-particle system? What are the (b) x and (c) y components of the velocity as well as the (d) x and (e) y components of the acceleration of the com when the com reaches Hmax? Express your answer in terms of the variables given and g.

(a) The y-coordinate of the center of mass of the system at any time is given quite readily by the following, in which we plug in straight away, in which is plugged in at the end of the problem,

However, we do not know what the maximum height is; we thus use , in which we have and , in which , in which and ,

Placing in , and noting , subsequently implying ,

(b), (c), (d), and (e) The velocity and acceleration of the center of mass is,

Or, numerically evaluating , we have,

Chapter 9, problem 42 (|| 44): An object, with mass M and speed v relative to an observer, explodes into two pieces, one times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observers reference frame?

In the observers frame, the momentum is conserved, and the kinetic energy increases. The object of mass (in which , implying ) has a velocity given by momentum-conservation, as,

The change in kinetic energy, then, is,