As discussed the first day, the purpose of these quizzes is to help you understand that day’s material by encouraging you to read the book before class. The answer to the questions should be found in the before-class assigned reading for that day.

An answer without an explanation is not an answer! Besides, it is likely to be incorrect.

Also, equations by themselves do not constitute an explanation. If you use an equation not discussed in class, you need to explain how it arises so an intelligent but ignorant person would understand and then determine it. Each quiz is worth 8 points. Use g = 10 m/s2 in answering any questions for the quizzes (as you know, it is actually 9.82 m/s2 in our neck of the woods).

Solutions are posted on this page after class.

Please put 1250 in the subject line when you email me. I divert the quizzes to a file in my emailer. Thanks. Email addresses,, or work equally well.

Week 16

3 December

A neutron can decay into a proton, an electron, and an electron antineutrino. The antineutrino has a very, very small mass (we will take it to be zero).

a. If the neutron is at rest originally, how much energy is it possible for the three decay particles to have? The mass of the proton is 1.6726 x 10-27 kg, the mass of the neutron is 1.6749 x 10-27 kg, and the mass of the electron is 9.1094 x 10-31 kg. Explain

b. The panti-νe = 3.1 x 10-22 kg m/s i-hat, the electron momentum is pe = 2.4 x 10-22 kg m/s j-hat, and the antineutrino kinetic energy is 9.3 x 10-14 J. What is the kinetic energy of the proton and the electron? Explain.

c. Would any of the decay particles (proton, electron, and electron antineutrino) be relativistic? Explain how you would find out.


a. The neutron has a mass-energy in its rest frame of mneutronc2. All this energy is in the mass-energies of the proton and electron and the kinetic energies of the three decay products.

b, c. The proton momentum is the negative of the sum of the other momenta because the momentum on the system (the original neutron) is conserved at 0. Thus, pproton = - 3.1 x 10-22 kg m/s i-hat - 2.4 x 10-22 kg m/s j-hat. The proton kinetic energy is (p2c2 + m2c4)1/2. The first term is 1.38 x 10-26 J2, and the second term is 2.20 x 10-20 J2, much bigger. Hence the kinetic energy is very small for the proton and it is non-relativistic. For the electron, the first term is 5.18 x 10-27 J2, and the second term is 7.47 x 10-44 J2, much smaller. Hence, the electron is relativistic.

The kinetic energy is E - mc2, so we obtain the proton kinetic energy as (1.38 x 10-26 J2 + 2.20 x 10-20 J2)1/2 - 1.51 x 10-10 J = 4.59 x 10-17 J (287 eV << mprotonc2) and the electron kinetic energy as (5.18 x 10-27 J2 + 7.47 x 10-44 J2)1/2 - 2.73 x 10-22 J = 7.2 x 10-14 J (450 keV = 0.45 MeV ~ melectronc2).

Week 15

29 November

Relativity lab

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 2.50 x 10-28 kg, and that of the other is 1.67 x 10-27 kg. If the lower-mass fragment has a speed of 0.893c after the breakup, what is the speed of the higher-mass fragment? Explain how to find the answer and determine it.


Momentum is conserved, so the momenta of the two decay particles must have the same magnitude. The momentum magnitude of the lower-mass fragment is given by p = γmv. Because v = 0.893 c, γ = 2.21 and plower mass = (2.21)(2.5 x 10-28 kg)(0.893 c) = γ'(1.67 x 10-27)v'. Hence, γ'v' = (2.21)(0.893 c)(2.5 x 10-28 kg/1.67 x 10-27 kg) = 0.297 c. Because γ' = (1 - v'2/c2)1/2, we may write v'2/(1 - v'2/c2) = (0.297 c)2, so v'2 = (0.297 c)2 - (0.297 v')2, or v'2 = (0.297 c)2/(1 + 0.2972), so v' = 0.284 c.

28 November

Figure P39.28 (problem 28, page 1180) shows a jet of material (at the upper right) being ejected by galaxy M87 (at the lower left). Such jets are believed to be evidence of supermassive black holes at the center of a galaxy. Suppose two jets of material from the center of a galaxy are ejected in opposite directions. Both jets move at 0.750c relative to the galaxy center. Determine the speed of one jet relative to the other and do so. Explain your reasoning.


As explained for the last quiz, the jets move in opposite directions, so we need to write one as +0.750 c and the other as -0.750 c, giving from the Lorentz transformation equation for velocity vrelative = (0.750 c + 0.750 c)/(1 + 0.750*0.750) = 1.500 c/1.5625 = 0.96 c.

26 November

Two spacecraft A and B are moving in opposite directions. An observer on the Earth measures the speed of spacecraft A to be 0.550c and the speed of spacecraft B to be 0.950c. Find the velocity of spacecraft B as observed by the crew on spacecraft A. Explain your reasoning.


This is a question about relative velocity. Galilean relativity says that v1 with respect to 3 = v1 with respect to 2 + v2 with respect to 3. Einsteinean relativity says v1 with respect to 3 = (v1 with respect to 2 + v2 with respect to 3)/(1 - v1 with respect to 2 v2 with respect to 3/c2).

So, if A moves in what we can call the positive direction, then B moves in the negative direction because they are moving in opposite directions. This means that the desired velocity of B as measured by A is given by (0.550 c + 0.950 c)/(1 + 0.550*0.950) = 0.985 c.

Week 14

Happy Thanksgiving!

19 November

The relativistic momentum (the true definition for momentum) is p = γmv. The factor γ is equal to (1 - v2/c2)-1/2, where c is the speed of light in vacuum. Determine the relative difference between the non-relativistic (NR) and the relativistic (R) momentum (pR - pNR)/pNR for
a. v = 300 m/s (a bullet of mass 250 g)
b. v = 11 km/s (escape speed from Earth of a rocket of mass 120 tonnes)
c. v = 250 km/s (approximate speed of the sun around galactic center; msun = 2 x 1030 kg)
d. v = 299,900,000 m/s (speed of a proton in the LHC; mproton = 1.67 x 10-27 kg)


Because pR = γ pNR, the ratio r = (pR - pNR)/pNR = γ - 1.

a. r = (1 - [300 m/s]2/[300 million m/s]2)-1/2 - 1 = (1 - 10-12)-1/2 -1 = 1 - 1 = 0.

b. r = (1 - [11,200 m/s]2/[300 million m/s]2)-1/2 - 1 = (1 - 1.39 x 10-9)-1/2 - 1 = 1 + 6.97 x 10-10 - 1 = 6.97 x 10-10, essentially 0.

c. r = (1 - [250,000 m/s]2/[300 million m/s]2)-1/2 - 1 = (1 - 6.94 x 10-7)-1/2 - 1 = 1 + 3.47 x 10-7 - 1 = 3.47 x 10-7, essentially 0.

d. r = (1 - [299.9 million m/s]2/[300 million m/s]2)-1/2 - 1 = (1 - 0.999333)-1/2 - 1 = 38.73 - 1 = 37.73.

Week 13

15 November

A bar of gold is in thermal contact with a bar of silver of the same length and area. One end of the compound bar is maintained at 80.0 °C, and the opposite end is at 30.0 °C. When the energy transfer reaches steady state, what is the temperature halfway across at the junction between the gold and silver?


Suppose each bar is of length L. The thermal power is given as P = kA dT/dx, with k thermal conductivity, A the cross sectional area, and a temperature gradient of dT/dx.

So, for the gold bar, we have Pgold = kgoldAgold dT/dx|gold and Psilver = ksilverAsilver dT/dx|silver. But Pgold = Psilver = P, Agold = Asilver = A. So, P/A = ksilver (80 °C - Tmiddle)/l and P/A = kgold (Tmiddle - 30 °C).

We did not say which metal was touching the 80 °C temperature reservoir and which was touching the 30 °C temperature reservoir. That matters. Gold has a lower thermal conductivity than silver--314 vs. 427 W/m °C. If it is at the high temperature reservoir, the middle temperature will be lower than if silver is at the high temperature reservoir.

In the first case, we have (314 W/m °C + 427 W/m °C) Tmiddle = (314)(80) W/m + (427)(30) W/m = 43580 W/m, so Tmiddle = (43580 W/m)/(741 W/m °C) = 58.8 °C. In the second case, (314 W/m °C + 427 W/m °C) Tmiddle = (314)(30) W/m + (427)(80) W/m = 37930 W/m, so Tmiddle = (37930 W/m)//(741 W/m °C) = 51.2 °C.

14 November

A Carnot engine goes through the following cycle: A gas undergoes an isothermal expansion from point A to point B at the high temperature reservoir. This expansion happens while heat is absorbed and work is done--for example, a piston is raised. Then it undergoes an adiabatic expansion from B to C, which is at the low temperature reservoir. As this happens, no heat is transferred (it is adiabatic), and more work is done (as in the example, by continuing to raise the piston) The gas expands from C to D at the low temperature, while heat is expelled and, in our example, the piston does work on the gas. Finally, the gas expande adiabatically from D to A, the starting point. No heat is transferred, and more work is done on the gas.

To summarize, heat Qin is absorbed at the high temperature reservoir and heat Qout is expelled at the low temperature reservoir while net work is done on the gas, equal to the difference between heat Qin and heat Qout according to energy conservation (the first law of thermodynamics).

a. Determine the efficiency of this engine.

b. Given that Qin/Qout = Thigh/Tlow, show that efficiencyCarnot = (Thigh - Tlow)/Thigh.


a. By definition, efficiency is ΔW/Qin.

b. Because ΔW = Qin - Qout, the above equation may be written as efficiency = 1 - Qout/Qin = 1 - Tlow/Thigh = (Thigh - Tlow)/Thigh.

12 November--Veteran’s Day. Enjoy the vacation.

Week 12

8 November

A heat engine takes in 440 J of energy from a reservoir at high temperature and does 40.0 J of work during one cycle. What is the efficiency of this engine? What happens to the low temperature reservoir during one cycle? Explain.


Our definition of efficiency is a measure of the work done divided by the energy input. Here, that is 40.0 J/440 J = 0.091 = 9.1%. The low temperature reservoir gets 400 J from the engine. We know this because we know that energy is conserved.

7 November

Two moles of a diatomic ideal gas originally at 300 K expand slowly as 222 J of heat is added at a constant pressure.
a. How much work is done on the gas? Explain.
b. What happens to the gas internal energy? Explain.
c. Can the final temperature be determined? If so, what is it? If not, why not?


It is convenient to do the answers in reverse order.

c. Because it is an ideal gas, pV = nRT. We know ΔQ = ncpΔT. Because the gas is diatomic, cp = 7/2 R. So, ΔT = 222 J/(2 mol x 7/2 x 8.314 J/mol K) = 3.81 °C. The final temperature is 303.81 K.

b. The internal energy increased because the temperature increased. It is ΔEinternal = 5/2 nRΔT = 158.4 J.

a. As 222 J of heat was added and the internal energy increased by 158.4 J, there was work done BY the gas (the negative of work done ON the gas) of 222 J - 158.4 J = 63.6 J. Thus, the work done on the gas is -63.6 J.

5 November

A system that is isolated has 220 J of work done on it, while during the same time period its internal energy decreases by 450 J.

a. What other process must have taken place to have had this result?

b. What amount of energy was transferred during the process of part a?


a. The change in internal energy is the heat added plus the work done on the system. Therefore, heat must have been added or removed.

b. ΔEinternal = - 450 J = ΔQ + ΔW = = ΔQ + 220 J. This implies that ΔQ = - 450 J - 220 J =- 670 J.

Week 11

1 November

Exam to follow the first part of class at 11.

A 100.0 gram ice cube at 0 °C is placed in water at a temperature of 30.0 °C.
a. Explain how to find the final temperature of the system.
b. Determine that final temperature.


In calorimetry, energy is conserved. When applied to systems having differing temperatures, we call conservation of energy the first law of thermodynamics. The water is warmer than the ice, so energy is transferred as heat to other parts of the system that are at lower temperatures.

a. Energy gained by ice equals the energy lost by the warm water. The energy lost by the water is its mass times its specific heat (4.186 J/kg °C) times the temperature difference. The energy gained by the ice is its latent heat of fusion (3.33 x 105 J/kg) times the mass of the ice. Thus, to melt all the ice requires an input heat transfer of 0.1000 kg x (3.33 x 105 J/kg) = 3.33 x 104 J. The maximum energy available from the warm water is M (its mass) x (4.186 J/kg °C) x 30 °C = M (125.6 J/kg). M would have to be a bit over 265 kg to melt all the ice. Probably some ice would melt.

b. The final temperature would be that of an ice-water mixture, or 0 °C, unless there is more than 265 kg of water--about 265 liters--a lot of water.

31 October

In a pipe, the fluid in a region of cross-sectional area 5.0 cm2 and y = 0 m is moving at a speed of 4.5 m/s. What is the speed of the fluid in a region of cross-sectional area 3.5 cm2 located at a height y = 20 cm?


We might be tempted to use Bernoulli’s equation, P + 1/2 ρv2 + ρgz = constant, to answer this question.The pressure at both points is atmospheric and the potential energies are the same.

The equation of continuity, which can be expressed either as the conservation of volume or mass (for an incompressible fluid) says that vA = constant, is a far better choice.

For the first position, we have v1A1 = (5.0 cm2)(4.5 m/s) = v2A2 = (3.5 cm2)v2. This gives v2 = (22.5 cm2 m/s)/(3.5 cm2) = 6.43 m/s.

29 October

Andy, a physics student, has heard that Archimedes’ principle says that objects weigh less in water. He wonders what would happen if he puts a beaker with water on a scale and attached a kilogram mass to a spring scale (of course that reading is about 10 N) and then allows the mass to be immersed in the water in the beaker. The beaker and water weigh 20 N. Explain to Andy what each scale will read when Andy puts the kilogram completely into the water.


Assume that Andy does not allow the kilogram to touch the bottom of the beaker, in which case the scale under the beaker would read about 30 N. Then, the scale Andy is holding will read less than 10 N because Andy is displacing a volume of water, which is equal to the volume of the kilogram. Whatever that volume is, the weight of the water that would have had that volume is added to the scale reading. This lessens the scale reading of the scale he holds in his hand by that same amount.

Week 10

25 October

What is the difference between a homogeneous solution to the differential equation we find from looking for the forces that act a mass bob on a spring as the oscillator (ma = F(t) - kx - Rv, where F(t) = F0 cos ωt), A0e-(R/2m)t cos (ω-prime t + δ-prime), where ω-prime differs from ω0, the natural angular frequency, to driven SHM and the particular solution as the driving angular frequency ω is varied? How would this oscillator respond as the driving frequency approaches the natural frequency ω0 of the mass on the spring?


The homogeneous solution is the product of a time-dependent amplitude and a sinusoidal function of a modified angular frequency. The driven solution is the product of an amplitude that depends on the driving frequency and a sinusoidal function of a modified frequency and a phase angle that depends on the driving angular frequency.

As the driving frequency approaches the natural frequency, the amplitude increases. The maximum amplitude of response is greatest at the driving frequency.

24 October

Describe the acceleration at the different positions listed below and the corresponding energy transformations at each of the positions in the SHM of a pendulum of mass m and length L during one-half of a cycle. The pendulum is puled back 20° and released.
a. At the point of release
b. Halfway between the point of release and the lowest point of the motion
c. At the lowest point
d. At the subsequent highest point for the motion


a. The potential energy is greatest here, and the initial kinetic energy is zero. The acceleration can be found from the free body diagram and we find Fperpendicular = T sin θ0 = W tan θ0, so maperpendicular = mLα = mg tan θ0 ~ mg θ0 = 3.35 m/s2.

Hence, we can say in general for any θ smaller than θ0 that aperpendicular ~ g θ = 1.70 m/s2.

b. From the diagram and the above considerations, aperpendicular ~ g θ0/2. As for energy here, some potential energy has been lost and that has shown up as kinetic energy, which is no longer zero.

c. Clearly from the diagram (and part a above), aperpendicular = 0. Now all the energy is kinetic energy. While the acceleration is zero at the bottom of the swing, the speed is greatest there.

d. The answer is the same as for part a, except that the direction is reversed. The energy is completely potential again, and the kinetic energy is zero.

22 October

no quiz because of postponed lab from Thursday.

Week 9

18 October

A disk having a moment of inertia of 20.0 kg m2 rotates with an angular velocity ω = 10.0 rad/s about a frictionless, vertical axle. A second disk of moment of inertia 30.0 kg m2 has a central hole that allows it to be gently dropped onto the first disk; originally the second disk is not rotating. Friction between the disks allows the two to reach a final angular velocity.

a. What is that final angular velocity?

b. What is the ratio of final kinetic energy to initial kinetic energy?


a. In this case, the system consists of the two disks. Therefore, angular momentum is conserved. L0 = LF. Now, L0 = (20.0 kg m2) (10.0 rad/s) = 200 kg m2/s. Afterward, the two disks rotate together with the same angular velocity ωF, so LF = (20.0 kg m2 + 30.0 kg m2) ωF = (50.0 kg m2) ωF, Because L0 = LF, 200 kg m2/s = (50.0 kg m2) ωF and so ωF = 4.00 rad/s. b. Initial KE is 1/2 Iω2 = 1/2 (20.0 kg m2) (10.0 rad/s)2 = 1000 J. The final KE is 1/2 IFωF2 = 1/2 (50.0 kg m22 = 400 J. The ratio is 0.4. 60% of the original kinetic energy went into increased internal energy (thermal energy).

17 October

How does a uniform solid disk and a hoop of the same radius and mass compare in terms of their acceleration down an inclined plane oriented at an angle of 30° to the horizontal? Explain any differences in their accelerations--and the reason(s) for that difference--clearly.


We need to realize that the disk and the hoop rotate, and the force has to speed up the angular velocity as well as the translational velocity, so less of the acceleration is available to increase the velocity compared to a box sliding on a frictionless surface.

Drawing a free bosy diagram, we have Fplane on disk = mg cos 30° in the perpendicular-to-the-plane direction and mg sin 30° - fk = ma. Also, τnet = 0 + 0 + fkR = Iα, where R is the disk radius (the first two torques are zero because there is no displacement from the center of rotation). Hence, using a = Rα, we have fk = Ia/R2. Thus (m + I/R2)a = mg sin 30°. For the disk, I = 1/2 mR2, so we have
(m + 1/2 m)a = 2/3 a = mg sin 30°, or a = 2/3 g sin 30°.

For the hoop, the reasoning is exactly the same up to finding the moment of inertia, I = mR2, giving (m + m)a = mg sin 30° or a = 1/2 g sin 30°.

15 October

Find the tension in the strings in the diagram below. The hanging lantern has a weight of 23.50 N.


We will set the sum of forces equal to zero. If necessary, we will set the sum of the torques to be equal to zero.

Call the string tensions TL and TR, respectively. Then setting the sum of forces in the horizontal direction to zero, we find TL cos 15° = TR cos 22°. Hence, 0.966 TL = 0.927 TR. Now, we set the sum of forces in the vertical direction to zero. TL sin 15° + TR sin 22° = W = 23.50 N. Then TR (0.927*0.259/0.966 + 0.375) = 23.50 N. This gives TR = 37.72 N and thus TL = 36.21 N.

We do not need to worry about rotation here.

Week 8

11 October

A rope hangs from a cylinder over which it is wound, of radius 0.20 m and mass 40.0 kg, and is attached to a hanging mass of weight 210.0 N. What torque does the hanging mass exert on the cylinder? NOTE: You may wish to know the moment of inertia of a cylinder.


This problem involves both force and torque. The hanging mass is accelerated. Therefore, we must write Newton’s Second Law as mg - T = ma, where T is the tension in the rope and m is the mass having weight mg = 210.0 N and we have taken a to be positive downward. The torque on the cylinder is thus TR, where the force T makes a right angle to the edge of the cylinder of radius R. Clearly, T < mg = 210.0 N. The applied torque is consequently less than 42.0 N m.

Referring to he table in the book, we see that a cylinder of radius R and mass M has a moment of inertia I = 1/2 MR2. You might recall that I already introduced this quantity I for the rotational analog of inertia. In our case, I = 0.5(40.0 kg)(0.20 m)2 = 0.80 kg m2. Therefore, torque is TR = I(angular acceleration) = Iatangential/R, or atangential = TR2/I. Now the tangential acceleration of the edge of the cylinder is exactly equal to the acceleration of the hanging mass, so
mg - T = mTR2/I, or
T(1 + mR2/I) = mg, or
T = mg/[1 + mR2/I] = mg/[1 + mgR2/Ig] = 210.0 N/[1 + (210.0 N)(0.20 m)2/0.80 kg m2 x g]
= 210.0 N/[1 + (8.40 kg m3/s2)/0.80 kg m2 x g] = 210.0 N/[1 + (10.5 m/s2)/g].
Taking g = 10 m/s2, we find
T = 210.0 N/2.05 = 102 N.

The torque on the cylinder is thus 102 N x 0.20 m = 20.5 N m.

10 October

A grindstone rotates at a constant angular acceleration of 0.35 rad/s2. At time t = 0, its angular velocity is ω0 = -4.6 rad/s and it is at θ = 0.
a. At what time after t = 0 is the angle θ relative to the horizon (which is an angle θ of zero) equal to 10 π?
b. What is the grindstone’s rotation angle between t = 0 and t = 32 s?
c. At what time would the grindstone (momentarily) be stopped?


a. So 10 π rad = -4.6 rad/s x t + (1/2)(0.35 rad/s2)t2, so the solution to the quadratic equation is 32 s.

b. It starts at θ0 = 0, goes negative because ω0 is negative (-4.6 rad/s), but the positive angular acceleration means it stops and reverses itself to become positive. Δθ = (-4.6 rad/s)(32 s) + 1/2 (0.35 rad/s2) (32 s) = 32 rad (about 10π or 1800°).

c. We can use α = Δω/Δt. When ω = 0, we have Δt = t (using our usual convention that t starts when the clock is set to zero), so
t = (0 - ω0)/α = 4.6 rad/s / 0.35 rad/s2 = 13 s (rounded off from 13.14 s).

8 October

An old phonograph record accelerates from rest to a final angular velocity of 78.0 rpm (revolutions per minute) in 0.350 s. The record has a radius of 15.0 cm.
a. Express the final angular velocity in units of rad/s.
b. What is the angular acceleration of the record during this time?
c. A dust speck lies 8.36 cm from the center of the record. What is the final speed of the speck?
d. Through what total angle does the record rotate as it accelerates?
e. What is the tangential acceleration of the speck during this time?


a. 78.0 rpm is 78.0 rev/min x 2π rad/rev x 1 min/60 s = 156.0pi/60 rad/s = 8.16 rad/s.
b. The angular speed begins at 0 and goes to 8.16 rad/s in 0.350 s. The change in angular velocity is thus 8.16 rad/s, and angular acceleration is change in angular velocity divided by the time this change requires: (8.16 rad/s)/(0.350 s) = 23.3 rad/s2.
c. The velocity is all tangential in this case. The speed is the angular velocity times R, the distance from the center of rotation, or (8.16 rad/s)(8.36 cm) = 68.3 cm/s = 0.683 m/s.
d. The angular displacement is the average angular velocity times the time required to make the change (because we define the average angular velocity by angular displacement divided by the time required to make the change), or (8.16 rad/s + 0)/2 x 0.350 s = 1.42 rad.
e. The tangential acceleration is R x angular acceleration, or 8.36 cm x 23.3 rad/s2 = 195 cm/s2 = 1.95 m/s2.

Week 7

4 October

Earth and the sun are separated by 150 million kilometers. Earth’s mass is 6 x 1024 kg and the sun’s mass is 2 x 1030 kg. Locate the center of mass of the Earth-sun system.


The centers of Earth and the sun are separated by 150 million kilometers, so (as usual) we choose the coordinate system to be a single axis originating at the center of the sun and stretching to Earth. Our relation for xCM is (msunxsun + mEarthxEarth)/(msun + mEarth). Because the origin is at the sun’s center, xsun = 0. Hence, xCM = mEarthxEarth/(msun + mEarth) = (6 x 1024 kg)(1.5 x 1011 m)/(2 x 1030 kg + 6 x 1024 kg) = (6 x 1024 kg)(1.5 x 1011 m)/(2 x 1030) = 4.5 x 105 m = 450 km. This is deep inside the sun, but NOT at its center.

3 October

Three carts of masses m1 = 4.00 kg, m2 = 10.0 kg, and m3 = 3.00 kg move on a frictionless, horizontal track with speeds of v1 = 3.00 m/s to the right, v2 = 4.00 m/s to the right, and v3 = 5.00 m/s to the left as shown in the diagram. Velcro couplers make the carts stick together after colliding. Find the final velocity of the train of three carts.


Because of the velcro couplers, all the masses will stick together eventually. Therefore, the system I choose includes all the masses. Then the total system momentum is the sum of the individual momenta. Thus, p0 = (4.00 kg)(3.00 m/s, R) + (10.0 kg)(4.00 m/s, R) + (3.00 kg)(5.00 m/s, L) = 12.00 kg m/s, R + 40.00 kg m/s, R + 15.00 kg m/s, L = 42.00 kg m/s, R - 15.00 kg m/s, R = 27.00 kg m/s, R.

Because the masses stick together at the end, all the masses move with the same velocity, call it v. We can say then that the momentum is constant because there are no external forces. Therefore, p0 = 27.00 kg m/s, R = pF = (m1 + m2 + m3)v. This means that v = (27.00 kg m/s, R)/(m1 + m2 + m3) = (27.00 kg m/s, R)/(17.00 kg) = 2.18 m/s, R.

1 October

A 77.0 kg diver stands on the edge of a pool. She pushes off horizontally with an average force on the pool’s edge of 900 N for 0.25 s. What impulse has been delivered to her? What horizontal speed would you expect her to have? Explain your reasoning.


The impulse is (900 N)(0.25 s) = 225 kg m/s.

The impulse is equal to the change in momentum. Her final momentum component in the horizontal direction is (77.0 kg)vhorizontal, which is equal to 225 kg m/s. Therefore,
vhorizontal = (225 kg m/s) /(77.0 kg) = 2.92 m/s.

Week 6

27 September

An older-model car accelerates from 0 to speed v in a time interval of Δt. A newer, more powerful sports car accelerates from 0 to 2v in the same time period. Assuming the energy coming from the engine appears only as kinetic energy of the cars, compare the power of the two cars.


Power is dW/dt, so we need to compare how much work was being done in the two cases (the time interval is the same), force dot displacement.

For the older-model car and from kinematics, we know that the force (assumed constant here) is mcara, where a = (v - 0)/Δt. The distance through which the force acts is d = vaverageΔt by definition, so that d = 1/2(v + 0) Δt. The displacement is in the same direction as the force in this case, so the dot becomes simple multiplication. We then find ΔW = mcar (v - 0)/Δt x 1/2(v + 0) Δt = 1/2 mcar v2 (surprise!). That is why the hint is written in the last sentence of the question (I just went the long way around in answering).

It will come as no further surprise that ΔW = 1/2 mcar (2v)2 = 2 mcar v2 for the newer car of the same mass. This is four times as great as in the case of the older car, so the newer car is four times more powerful than the older car.

26 September

A crate of mass 12.5 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.350, and the crate is pulled 6.00 m.
a) How much work is done by the gravitational force on the crate?
b) How much work is done by the 100-N force on the crate?
c) What is the change in kinetic energy of the crate?
d) What is the speed of the crate after being pulled 6.00 m?


If a free body diagram is drawn, there is a force of 100 N parallel to the incline (given), a frictional force in a direction opposite the the dragging, a perpendicular force from the surface on the crate (the normal force), and the weight, W. Orienting the free body diagram so the axis parallel to the plane is horizontal, the weight makes an angle of 20° left of vertical. Hence, Newton’s first law says that the normal force is equal to W cos 20°. The pulling force is to the right, W sin 20° and fk are to the left. According to our model for friction, fk = μk (normal force) = μk W cos 20°. Then the forces are ma = Fpull - μk W cos 20° - W sin 20°. We would have to work the kinematics, etc., etc. This is quite messy.

But, this chapter is about how to apply ENERGY considerations to solve problems. We need to see how to apply these considerations to solve this problem. Overall, if we can use energy ideas to solve, the solution will prove to be easier. This is why we are doing this.

a. The work done by the gravitational force on the crate is -mg (rise in height) = -mg (6.00 m) sin 20°. It is negative because the gravitational potential energy is increasing, which takes work input. Numerically, this is -256.5 J.

b. This is simply (100 N)(6.00 m) = 600 J.

c. To find the change in kinetic energy of the crate, we will have to find the work done by the frictional force. From the introduction above, we know the frictional force is μk W cos 20°, which is 41.1 N. The work done by fiction is negative because the angle between the pull direction and the frictional force is π: -246.7 J. The sum of all the works is 600 J - 256.6 J - 246.7 J = 96.8 J. This, according to the work-energy theorem, is the change in kinetic energy.

d. So, 1/2 m vfinal2 = 1/2 m v02 + 96.8 J, giving vfinal2 = (1.50 m/s)2 + 2(96.8 J)/12.5 kg = 2.25 m2/s2 + 15.5 m2/s2 = 17.7 m2/s2. Therefore, vfinal = 4.21 m/s.

24 September

A 0.35-kg stone is held 1.53 m above the top edge of a water well and then dropped into it. The well has a depth of 5.10 m. Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone-Earth system
a. before the stone is released and
b. when it reaches the bottom of the well?
c. What is the change in gravitational potential energy of the system from release to reaching the bottom of the well?


a. The stone is configured so that the zero of gravitational potential energy (GPE) is at the top edge of the well, and the stone is 1.53 m above the top edge, so its GPE is (0.35 kg)(10 m/s2) (1.53 m) = 5.36 kg m/s2 = 5.36 J.

b. It is 5.10 m below the edge at the bottom of the well, so its GPE is (0.35 kg)(10 m/s2)(-5.10 m) = -17.85 J.

c. The change in GPE is final minus initial GPE, or -17.85 J - 5.36 J = - 23.21 J.

Week 5

20 September

a. How much work is done against gravity by a 90.0-kg person climbing up a 2.00-m high staircase at constant velocity of 1.50 m/s? The person moves 4.00 m to the right as well as 2.00 m up while climbing the stairs. Explain.
b. What is the gravitational potential energy of a 32.4-kg object at rest on a plane inclined at an angle of 30° to the horizontal a distance 10.00 m up the plane? How do you know this?
c. Assuming the plane of (b) is frictionless, if the object is released, what kinetic energy will it have when it reaches the bottom of the plane? How do you know this?


a. The person is lifting herself up a distance of 2.00 m in the direction parallel to the force (upward, in the opposite direction to the weight) she is exerting. The horizontal distance is perpendicular to the force and so does no work. Only the vertical distance is connected to the work. The work done is her weight (900 N) times the distance upward she has moved (2.00 m), so work done = (900 N) x (2.00 m) = 1800 N m = 1800 J.
b. The gravitational potential energy is determined by the weight and the vertical distance above the reference level (here the ground). This is the vertical component of the 10 m plank, or 10.00 m x sin 30° = 5.00 m. The gravitational potential energy is then (32.4 kg) x g x (5.00 m) = 1620 N m = 1620 J (using g = 10 m/s2).
c. It must have all 1620 J of energy it began with; the energy was transformed from potential to kinetic energy. The total energy, as long as no friction acts, is fixed.

19 September

Gordon has an antique turntable that plays vinyl disks at 33 1/3 revolutions per minute.
a. What is the speed (in cm/s) at the outer edge of a record if that edge is 15 cm from the center of rotation?
b. While the turntable runs without the stylus or record (it is the same size as a record), a penny is carefully placed on the turntable and travels with it remaining in place. Explain whether or not there is any centripetal force on the penny.
c. Where on the turntable is the frictional force on the penny the greatest? Explain.


a. The speed is distance traveled divided by the time required to travel that distance, so the distance traveled is the circumference, 2 x 3.14159265 x 15.0 cm = 94.2 cm and the time is found from time for one revolution = 1 minute/(33 1/3) = 0.03 minute = 1.8 s. The speed is 94.2 cm / 1.8 s = 52.4 cm/s.
b. Because the penny does not move with respect to the turntable, it is being carried around by the turntable through static friction. This force is centripetal, that is, it is the force causing the penny to move in the circle.
c. Because v gets bigger as we move from the center of the turntable (it is proportional to R, the distance from the center) and because the centripetal acceleration is v2/R, the centripetal acceleration is greatest at the periphery.

17 September

A racecar goes completely around a circular track of radius 4.0 km in 600 s.
a. What is its average speed?
b. What is its average velocity?
c. Is the racecar accelerated? If so, how big is the acceleration? If not, explain why not.


a. Average speed is distance traveled divided by the time required to travel that distance, so the distance traveled is the circumference, 2 x 3.14159265 x 4.0 km = 25.1 km and the time is given. The average speed is given either in km/h or m/s. The time 600 s is one-sixth of an hour, so vav = 25.1 km/(1/6 h) = 150.1 km/h. Alternatively, the distance is 25,133 m and so vav = 25,133 m / 600 s = 41.9 m/s.
b. Because the racecar returned to where it started, the displacement is zero, and thus the average velocity (displacement divided by time required to make that displacement) is zero.
c. The velocity is changing because its direction is changing, This means that there must be an acceleration. The centripetal acceleration is v2/R, so (41.9 m/s)2/ 4000 m = 0.44 m/s2.

Week 4

13 September

One side of the roof of a house slopes up at 37.0°. A roofer kicks a round, flat rock that has been thrown onto the roof by a neighborhood child. The rock slides straight up the incline with an initial speed of 15.0 m/s. The coefficient of kinetic friction between the rock and the roof is 0.400. The rock slides 10.0 m up the roof to its peak. It crosses the ridge and goes into free fall, following a parabolic trajectory above the far side of the roof, with negligible air resistance. Determine the maximum height the rock reaches above the point where it was kicked.


There are two parts to the problem. First, we need to deal with the rock sliding on the rough roof. Then the rock is launched into the air and travels upward to its maximal height, which is simple projectile motion. (I did not realize it would be so long when I chose this. Sorry.)

The rock’s free body diagram will show three forces: the weight of the rock, the perpendicular force of the roof on the rock, and the kinetic friction force. Rotating the coordinate system so that the y-axis is in the perpendicular direction and the x-axis is along the roof, we will find
Froof on rock = Wrock cos 37.0° because of Newton’s first law acting in the vertical direction, and
Fnet,x = -Wrock sin 37.0° - fk.

Our phenomenological model for friction gives fk = μk Froof on rock = μkWrock cos 37.0°. Therefore, Fnet,x = -Wrock sin 37.0° - μkWrock cos 37.0°,so that
ax = -g(sin 37.0° + μk cos 37.0°).

Numerically, ax = -0.92 g = - 9.2 m/s2 for g = 10 m/s2 or - 9.02 m/s2 for g = 9.8 m/s2. We know that the rock travels 10.0 m along the roof up to the point it launches into the air, which will take a time (say) t*. Then
10.0 m = v0xt* - 1/2 axt*2, giving solutions 0.923 s and 2.40 s for g = 9.8 m/s2 or 0.905 s and 2.36 s for g = 10 m/s2. The smaller time is clearly the correct solution--the rock would have to have traveled much farther than 10.0 m and come back for the other solution, which makes no sense. The final x-velocity is, respectively, 11.16 m/s or 11.23 m/s, both of which are 11.2 m/s when rounded off.

With that smaller time, we can find the velocity vector as the rock flies off the roof. At this point, it is necessary to change coordinate systems because we are going to ask how high above the start point the rock rises. While on the roof, it travels upwards 6 m as it reaches the peak. Then it is subject to only the gravitational acceleration, g. Let the new coordinates be horizontal and vertical. The initial horizontal velocity is 8.9 m/s and the new vertical velocity is 6.7 m/s. Setting the vertical velocity equal to zero at maximum height, it will take, respectively, 0.683 s or 0.674 s to reach that height. During that time, the rock will rise, respectively, 2.29 or 2.27 m. The rock will have traveled just about 8.3 m upward from its starting point.

12 September

A 25.0-kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion, after which a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find a. the coefficient of static friction, and b. the coefficient of kinetic friction between the block and the surface.


After constructing the free body diagram, we have in he vertical direction the weight downward and the force of the table on the block (the normal force) upward. Because there is no CHANGE in motion in the vertical direction, Newton’s first law says that these are equal-- that the normal force is 250 N in magnitude (the weight is 250 N).

a. In the horizontal direction, there are two forces, the applied force and the static frictional force. The sum is zero as long as the applied force does not exceed the maximum force of static friction. The force just barely greater than the maximum force of static friction is 75.0 N, so we may say that fs,MAX = 75.0 N. The coefficient of static friction is defined as μs x normal force = fs,MAX, so μs = 75.0 N/250 N = 0.300.

b. In this case, to move the block at constant speed Fnet = 0, so Fapplied = 60.0 N = fk = μk x normal force = μk x 250 N. Hence, μk = 60.0 N/250 N = 0.240.

10 September--we are doing the lab today, no quiz

Week 3

6 September

Block A, which weighs 880 N, rests on a rough table (that designation “rough” means that there is friction between the table and the block A). Block A is connected by an ideal rope and pulley to block B, which hangs over the table’s edge. A student sees that the rope connected between the blocks and running over the pulley supports block B, of weight 55.0 N, which hangs unmoving in midair as a result.
a. Find all the forces acting on each of the two blocks, A and B.
b. Explain what the agents are for each of these forces.
c. List the Newton’s Third Law forces corresponding to each of the forces in (b).


a. On A: weight of A, force of table (perpendicular) on A--the normal force, frictional force on A, force of rope on A.
On B: weight of B, force of rope on B.

b. These are the only forces; weight is noncontact in both cases, on B only the rope is in contact, while on A, both the table and the rope are in contact. The agent for weight is Earth; for the rope on A block B and for the rope on B block A, for the normal force, the table is the agent, and for the frictional force, the table is agent.

c. Newton’s Third Law forces: Weight of A, force of A on Earth; weight of B, force of B on Earth; rope on A, force of A on the rope; rope on B, force of B on the rope; force of table on the block A (normal force), force of block A on the table.

Note that the rope merely transmits the force between bodies. It acts to the right (say) on A and up on B. The rope puts its forces on the two blocks in different directions because of the pulley

5 September

A book rests in your hand, which is not moving. Your hand exerts a force of 50 N on the book. What is the size of any other force(s) exerted on the book? Explain.


The book is not moving, which means that it is not CHANGING ITS MOTION. According to Newton’s First Law, we know that if motion is not changing, all forces balance, i.e., they add to zero. The hand is in contact with the book; nothing else is. Therefore, the other force must be a noncontact force, the weight (caused by Earth). This must be 30 N also, and in the direction opposite that of the hand (which hand force, presumably, is upward--we know the weight is downward).

3 September--Labor Day, no classes

Week 2

30 August

Silwan stands at the edge of a steep cliff, 50.0 m above the surface of a lake. He throws a stone with an initial horizontal velocity of 40 m/s (He is a pitcher for his college baseball team. The fastest pitch ever recorded for a baseball pitcher was 46.0 m/s, set by Mark Wohlers in 1995 during a Spring Training game, according to
a. Determine the velocity with which Silwan’s stone strikes the water.
b. The speed of sound in air is 345 m/s. How much time elapses between the stone leaving Silwan’s hand and the instant the sound of the stone hitting the water reaches Silwan’s ears?


a. The kinematic equations for motion with constant acceleration are known:
v = v0 + a t
r = r0 + v0 t + 1/2 a t2

Let us say up is the z-direction and horizontal is the x-direction. Then the above vector equations have the following components:
vz = vz0 - gt = - gt because vz0 = 0
vx = vx0 = 40 m/s because ax = 0
z(t) = z0 - 1/2 gt2 = 50.0 m - 1/2 gt2
x(t) = vx0 t = 40 m/s t

The stone hits when z(t*) = 0, or 50.0 m - 1/2 (10 m/s2)t*2, or t*2 = 100.0 m/(10 m/s2) = 10 s2, or t* = 3.16 s. Therefore, the x-velocity is 40 m/s and the z-velocity is 31.6 m/s at t*. So, v = 40 m/s i-hat + 31.6 m/s k-hat.

We could also give the result for v by specifying the length v and the angle, θ. In this case, v = 51 m/s and θ = tan-1 31.6 m/s/(40 m/s) = tan-1 0.669 = 38.3°.

b. The displacement of the stone at t* is - 50.0 m k-hat + (40 m/s)(3.16 s) i-hat or - 50 m k-hat + 126.5 m i-hat. Using the Theorem of Pythagoras, the distance from the start point is 132.7 m. At 345 m/s, sound will travel back to the start in 0.38 s. The total time for the stone to hit and send back the sound is, therefore, 3.16 s + 0.38 s = 3.55 s.

29 August

Janine plans to ride her bike up Groangroan Hill, a distance of 15.00 km and have a picnic lunch with a friend who lives over the other side of the hill. She rides up the hill at a constant speed of 10 km/h. Immediately when she gets to a scenic overlook 5.00 km up, she realizes that she has forgotten her food, and immediately turns around and pedals home (at the very bottom of the hill) at a speed of 30 km/h, grabs her bag as she executes a spectacularly fast turn, and rides back up the hill.

a. What was Janine’s average speed for the abortive trip (home to home)? Explain your reasoning.

b. What was Janine’s average velocity for that same trip? Explain your reasoning.


a. Janine traveled 5 km at 10 km/h, which took her 5 km/(10 km/h) = 1/2 h. She then traveled another 5 kn at 30 km/h, which took her 5 km/(30 km/h) = 1/6 h. For the entire trip, the average speed is (5 km + 5 km)/ (1/2 h + 1/6 h) = 10 km/(2/3 h) = 15 km/h.

b. Janine arrived back at the same point she began, so Janine’s displacement is 0. Because average velocity is displacement divided by the time required to make the displacement, her average velocity is zero.

27 August

A car slows down smoothly toward a stop sign. While you were watching, it covered a distance you know to be 45.0 m long during an interval you timed to be 9.0 s long. Determine the car’s acceleration and the car’s original speed.


The average acceleration is defined as change in velocity divided by the time required to make that change. In this case, it is one-dimensional and so the direction is indicated by a plus or minus sign. The change in velocity is vfinal - v0 = 0 - v0 = -v0. Meanwhile, the average velocity times the time is 45.0 m. Because it begins at v0 and ends at 0, the average velocity is 1/2 v0, so 45.0 m = 1/2 v0 times 9.0 s, so v0 = 2 x 45.0 m/9.0 s = 10 m/s.

The time interval is 9.0 s, so the average acceleration is -10.0 m/s / 9.0 s = -1.1 m/s2.

Week 1

24 August

Explain the difference between average speed and average velocity.


Average speed involves distance and the time involved in attaining that distance. Average velocity involves displacement, a vector quantity, and the time required as well. So, average velocity is a vector while speed is a scalar. [latest revision, 19 November 2012]