Homework problems are taken from the 7th edition of Serway and Jewett text. 

The due dates/day for the homework, along with the problem numbers, are listed below.  Problems in bold red are graded in detail out of 5 points each.  Problems in green italics are not due. You are encouraged to attempt these problems before the solutions are put up. The rest of the problems are graded out of 2 points.  An almost correct solution or a valiant attempt gets 2 points while a decent attempt with the right method and equations and some progress, gets 1 point. The starred problems have extra questions associated with them. Look below the table for the extra questions associated with the starred problems.

Please make sure that all answers have the proper units. And refer to the homework instructions for guidelines to be used while solving these problems


Hw #





Week 1


June 2






June 3





June 4


HW#1 1.5, 1.14, 1.21, 1.28, 1.31, 1.33, 1.9, 1.12



June 5



2.1,2.10, 2.11, 2.21, 2.33*, 2.59



June 6



2.3, 2.23, 2.39, 2.41, 2.57, 3.1, 3.6, 3.21, 3.15,






Week 2


June 9


HW#4 2.24, 2.28, 2.46, 2.40, 3.25, 3.29, 3.41, 3.57, 3.62, 3.43, 3.34, 4.1, 4.2, 4.5



June 10



4.11, 4.17, 4.18, 4.19, 4.59, 4.54



June 11



5.1, 5.2, 5.3,5.5, 5.9, 5.14, 5.16, 5.19



June 12






June 13



5.20,5.19, 5.24, 5.25, 5.26, 5.39, 5.43, 5.44, 5.54 (No class. Turn this in on Monday along with the Monday HW).


Week 3


June 16



4.29, 4.32, 4.31,4.28, 6.3, 6.5, 6.11, 6.14, 6.15, 6.55, 6.53, 6.52, 6.61,6.50

June 17 Tuesday
June 18 Wednesday HW#9
7.1, 7.6, 7.9, 7.13, 7.17, 7.29, 7.30, 7.31
June 19 Thursday HW#10
5.54(ignore d),5.67, 5.61, 8.3, 8.4,8.7, 8.11, 8.13,


June 20 Friday HW#11
8.19, 8.21,8.25, 8.15, 8.29,8.32, 8.33, 8.43,
Week 4
June 23
8.55, 8.57, 8.62, 8.64, 9.5, 9.7, 9.9, 9.11, 9.17, 9.18, 9.19, 9.28,9.29, 9.34
June 24
9.67, 10.1, 10.5, 10.9, 10.12, 10.17, 10.21, 10.22
June 25
10.33, 10.32,10.38, 10.43, 10.71, 10.37, 10.58
June 26
June 27
11.3, 11.4,, 11.12, 11.27, 11.32, 11.35, 11.50
Week 5
June 30
HW#16 11.11, 11.29, 11.31, 12.2, 12.16, 12.19, 12.21, 12.37, 12.47, 12.39, 12.42
July 1
HW#17 13.5, 13.13, 13.17, 13.31, 13.33, 13.44, 13.30, 13.32, 13.35,
  18 July 2
14.3, 14.8, 14.12, 14.16, 14.22
July 3
No HW.
Week 6!
July 7

HW19 part 1,

HW19 part 2

14.27, 14.28, 14.33, 14.37,14.42,14.45,14.46, 14.51, 14.66,14.69, 15.2, 15.3, 15.5, 15.6, 15.4, 15.10, 15.16, 15.17, 15.49, 15.67,                      REVIEW PROBLEMS:2.24, 2.57, 3.57, Ex 4.8, 4.32, 4.11, 4.29, 4.59

July 8
HW#20 19.2, 19.9, 19.10,19.17, 19.21, 19.34,19.39, 19.43                    REVIEW PROBLEMS:5.28,5.31, 5.44, 5.45,5.62, 6.53, 6.61, 6.14                               
July 9
HW#21 20.9, 20.10, 20.15,20.16, 20.21, 20.22, 20.26,20.28 20.34,20.39            REVIEW PROBLEMS: 7.57,8.19, 8.25, 8.57,9.67, 9.19, 9.18, 9.57,
HW#22 20.38, 20.61, 20.46, 21.4, 21.9, 21.13, 21.17, 21.23                           REVIEW PROBLEMS: 8.58, Ex10.12,Ex10.13, 10.71, 10.58, 11.35, 11.27, 11.32

Exam 5.30 pm - 7.30 pm (22.1, 22.3,22.12 22.13, 22.16, 22.20, 22.23)

   REVIEW PROBLEMS:12.39, 12.47, ex12.3, 13.17, 13.32, 11.31, 7.15, 9.11

    July 12   HWrev SOLUTIONS TO new review problems



2.33 Also draw the distance vs. time, velocity vs. time, acceleration vs. time graphs for the car and the bike. Convert all units to m/s^2 for acceleration.


These problems are to help you prepare for the final. You DON't HAVE to turn them in.Most of these problems are from the homework. Some of them are new and their solutions will be posted soon in the "July 12"row. The final exam material dhapters 1-13 will be BASED on these problems, or extensions to them or variations of them. Students who want to improve their standing can write DETAILED SOLUTIONS to these problems WITH explanations and turn it in for extra credit of 1.5 %. if you do not turn these is in, your grade will not be affected. So you will not fall behind if you do not turn these problems in. "Ex" is an example from the chapter.

Answers to Even numbered problems


dot product of vectors A and B is 28.9 units


a) 33.8 J

b) 135 J

P8.4            (a) 5.94m/s; 7.67 m/s      (b) 147

P8.32         (a) 5.91 kW     (b) 11.1 kW

9.18  v=(4M/m)*sqrt(gL)  What is conserved here? Consider the collision and the circular motion separately

9.28  He is a liar!

9.34  velocity of blue puck = 5.89m/s       velocity of green puck = 7.07m/s

10.12 a) v= .605m/s

10.32 tau= 168N*m (clockwise or into the page)

10.38  mu= .312

10.58 Use energy conservation!

a) v= 2.38m/s
b) v= 4.31m/s

11.12  L= -22kg*m^2/s k-hat (into the page)

11.32  You are told the moment of inertia of the student+stool only, not with the weights.
a) 1.91rad/s
b)K,init= 2.53J
K,fin = 6.44J

11.50 What is conserved here?

a)  omega=6mv/(Md+3md)
b) fractional loss = M/(M+3m)

P12.2      Fy+Ry-Fg=0, Fx-Rx=0;    Fy*l*cos(theta)- Fg*l/2*cos(theta) – Fx*l*sin(theta)

P12.16    (a) 35.5 kN    (b) 11.5 kN to the right     (c) 4.19 kN down

P12.42    (a) 2.71 kN     (b) 2.65 kN     (c) You should lift “with your knees” rather than “with your back.” In this situation, you can make the compressional force in your spine about ten times smaller by bending your knees and lifting with your back as straight as possible

P13.30    (a) 42.1 km/s      (b) 2.20 * 10^(11) m

P13.32    469 MJ.  Both in the original orbit and in the final orbit, the total energy is negative, with an absolute value equal to the positive kinetic energy.  The potential energy is negative and twice as large as the total energy.   As the satellite is lifted from the lower to the higher orbit, the gravitational energy increases, the kinetic energy decreases, and the total energy increases.  The value of each becomes closer to zero.  Numerically, the gravitational energy increases by 938 MJ, the kinetic energy decreases by 469 MJ, and the total energy increases by 469 MJ.

P13.44    2/3 Sqrt(GM/R); 1/3 Sqrt(GM/R)

P14.8      255 N


P14.12    (a) 29.4 kN to the right      (b) 16.3 kN.m counterclockwise


P14.16    (a) 20.0 cm     (b) 0.490 cm


P14.22    3.33 * 10^3 kg/m^3

P14.28    (a) 6.70 cm    (b) 5.74 cm  


P14.42    (a) 3.93*10^(-6) Sqrt(Delata_P)where Delta_P is in pascals.   (b) 0.305 L/s     (c) 0.431 L/s     (d) The flow rate is proportional to the square root of the pressure difference.


P14.46 6.80*10^4 Pa   

20.16 All of the ice melts, final temperature is 40.4C


20.28 Energy difference = 42.9kJ


P = 1.34kW

Assume the asphalt only loses heat as radiation upward.  Again, energy in = energy out.
T = 364K

Pressure = Force/Area
Force = ma = change in momentum/change in time


2.11. a) You have been given the acc-time graph. You can find the velocity at any time by calculating the area under the graph till that time. Remember, the area is negative if it is below the x axis. ANOTHER method. There are three intervals : AB + acc, BC no acc, CD - acc. write the v vs. t and x vs t equations for AB, BC and CD. remember that the final velocity and final distance for AB is the initial valocity and initial distance for BC. similarly for BC and CD. This will allow you to calculate the diatance travelled.

IMPORTANT. When you write the equations for say the third phase CD, the "t" that goes in to the equations (for instance v=v0+at and so on) is the time from the start of the interval. so if you are writing x=x0+v0t +1/2*a*t^2 for the third interval, then if the real time is 20 seconds, the time t in the equation would be 20-15= 5 seconds, since that is the time for which the car has travelled for "With and acceleration of -3.00 m/s^2

2.33.We know that the bycycle is going to be ahead for sure during the phase when the byccicle is accelerating. The car passes the bike some time when the bike has reached a constant speed.The bike is ahead of the car when V_bike >v_CAR. Find the time when V_bike (which is 20 mi/hr) is equal to V_car = 0+a_car*t. Find where there bike and the car are at this time.


There is also another method

step 1. calculate the time it takes for the bike to reach its max speed from any of the 4 equations. It should come to about 1.54 seconds. calculate X.

step 2. second phase of motion of bike.

Final X from first step becomes inital x (X0)for second step. Equation for X is X_bike=X0+v0(t-1.54). where v0=20 mi/h

step 3. THe cars motion is a simple equation with initial velocity of 0 and acceleration of 9 mi/h.s.

So X_car= 0+1/2*9 miles/hour.sec*t^2

take the difference between X_bike-X_car.

We want this to me maximum. remember if a function has to be a maximum, then derivative with respect to the variable (here t) is zero. That gives you t as well as X_bike -X_car at that value of t.

OR. the time should be the same as in previous method.

59. Try your best. it is not due. let us know how you got started and we wil let you know if the equations seem right, or what mistake you might be committing. It is lik ethe cop, motorist problem we discussed in class.

Chapter 4.

18. If X is the horizontal distance covered and (Y-Yo) (call it Y') is the vertical distance fallen, you have been given X^2+Y'^2=3.25^2. calculate X. Use the X equation and Y equation to eliminate t to get an equation in theta. (like we did in class). Use sin/cos = tan, and 1/cos^2(theta)= sec^2(theta)= 1+tan^2(theta). You get a quadratic euqation in tan(theta).

59. This is almost similar to example 4.5. You are trying to find the distance d from the ramp to the point where the skier lands. dcos(50)=X and dsin(50)= Y. Write the X and Y equations and you have two unknowns d and t. elimitate t and solve for d.

#8.3  Hint : Use conservation of energy between the initial point and the final point A. Once that is found, use the Newton’s equation for the looping bead at the point A to find the normal force.

# 8.4 Hint : Use conservation of Energy between any two points. Work done by gravity is the negative of change in PE (or change in the Kinetic Energy of the mass) between any two points.

# 8.11 Hint : let k_short be the spring constant for the shorter (5m) cord and k_long be the spring constant for the longer cord (Lm).  As spring constant is inversely proportional to the length of the spring, Shorter the spring, stiffer is the spring constant.

k_short . (5) = k_long . (L)

Use Newton’s law for the shorter spring to find k_short. From the above equation, compute k_long. (you will have the mass as an unknown)

Use conservation of energy for the bungee jumper between the initial and final states of the jump with this new spring constant k_long.

Problems 8. 15, 19, 21  are based on the work done by non-conservative forces. The sign due to the work done is decided based on the relative direction between the force and the displacement.

8.55  This problem is very simple to analyze without the “rough” portion. How does this rough portion affect the energy of the system?

8.62  Not all systems conserve energy! Does this one? For part b you'll need to use energy considerations and your knowledge of circular motion. “To swing a complete circle” means that it won't fall when it's at the top. Consider a free body diagram at the top of the circle (the place where it is most likely to fall)

9.9  Use the Impulse-Momentum theorem. Make sure you realize that momentum is a vector.

9.11  Use the plot for part a and that answer should make the rest pretty simple. For part d consider the simplest representation of the Impulse-Momentum theorem.

9.17  Why doesn't the actor's force violate conservation of momentum?


HW 13

9.67  Make the assumption that the bullet is all the way through the block before the block starts moving. You can then come up with a momentum equation for the whole system and an energy equation just for the block.

10.9  You need to add the angular displacements for the time where it is speeding up AND slowing down. Remember that both parts of the motion are under constant acceleration. Think about the simplest way to find angular displacement in that case.

10.12  The equations are very simple, but you must take care to use the correct information. Convince yourself of what determines the speed of the chain with respect to the bike frame, etc.

10.21 This problem is simple when you know what the moment of intertia equation is (relationship 10.15

HW 14

10.33 Adopt some system where, for example, a force tending to spin the wheel clockwise applies a positive torque, etc. Torque is a cross-product, so it too has direction. All forces are applied tangentially.

10.32 How is torque related to angular acceleration? Remember, this is a spinning disk. Don't forget what you know about friction. What type of force is applied?

10.37 This is not an equilibrium situation, so the tensions are not the same. Why else would the pulley turn?

10.58 Use energy conservation!

Problem 11.11: The magnitude of angular momentum depends on the choice of the origin of coordinates. This is true, in general, for all quantities (like angular momentum, torque, moment of inertia) which have an explicit position vector (r) in their definition.

Problems 11.29:  The required condition for the conservation of Angular momentum is that the ‘net external torque has to be zero’. The friction between the two discs is an internal force of the system. Here , the system is the two discs. Also, the Kinetic Energy is not conserved because the two discs start rotating together. Please note the analogy with the case of inelastic collision where two bodies collide and stick and move together.  There, Linear momentum is conserved by Kinetic Energy is not.  Whether Kinetic Energy is conserved or not decides whether the collision is ELASTIC or INELASTIC. 

Problem 11.31: The total moment of inertia of a system is the sum of the moments of inertia of its individual components. Therefore, the total moment of inertia of the (disc + clay- lump) will be the sum of moments of inertia of each object about the axis of rotation. The lump of clay is taken to be a point mass.

Problem 12.2 through 12.47 :  The condition for equilibrium for a POINT object is that the  ‘vector sum of external forces acting on it is zero’. But, for an extended body, the conditions for equilibrium are ‘the vector sum of both external forces and external torques acting on the body is zero’. By choosing the origin of coordinates and the axes cleverly, computation of torques can be simplified to a great extent. Also, one should not forget that a torque is a vector and hence there is a sign associated with it . A counter-clockwise torque can be assigned a positive sign and a clock-wise torque can be assigned a negative sign. Though the absolute sign is not important in problems of equilibrium, it is good to have a fixed convention because the sign does matter in problems where there is a net-torque and hence a net angular acceleration, the direction of which is the same as the direction of the torque.

Note that gravity always acts through the centre–of-gravity. The centre-of-gravity is that point where the entire mass of the system SEEMS to be concentrated. So, an extended object, for all mathematical purposes, can be taken as a point object with all of its mass located at the centre-of-gravity.

Problem 13.5 :  Note that the distance r in the Newton’s law of gravitation ( F = (G M m)/ (r^2) ), the distance ‘r’ stands for the distance between the CENTRES of the spheres.  In general, for arbitrarily shaped objects, it is the distance between the centres-of-gravity for the two objects. Also, the direction of the force is always along the line of separation of the centres-of-gravity and is always attractive.
Problem 13.13: Each star exerts its gravitational pull on the other star. The stars are moving in a circle of radius r about a fixed centre. The only force that exists between them is the gravitational pull on each star due to the other star. hence, the necessary centripetal acceleration for their circular motion is provided by this force of gravity.

Problem 13.17: Compute the orbital velocity of a satellite at a general height d above the planet’s surface.  Be careful to add the radius of the planet to this altitude while applying Newton’s law of gravitation. And, the height d should be such that the linear velocity (orbital speed) of the satellite at this height will give it an angular speed  exactly the same as the angular speed of the planet.


Problem 14.3 through 22: Pressure is a scalar whereas force is a vector. So pressure is the same in all directions at a given point in the fluid.

Problem 14.12: The force due to water varies with the depth because the hydrostatic pressure changes with the depth. Hence, you need to integrate from the top of the hatch to the bottom of the hatch to compute the net force.  Similarly, with respect to the lever, the force due to the water at particular depth on the hatch has a torque associated with it. This torque changes with depth and hence needs to be integrated from the lever to the bottom of the hatch.

Don't neglect kinetic energy.

20.16 All of the ice melts, final temperature is 40.4C
Consider the phase transitions (melting/condensing) separately from changing temperature.


For each step, you need to consider the total change in internal energy. Make sure you consider both work done on/by the system AND energy given off/absorbed as heat.
Use what you know about isobaric/isothermal/adiabatic processes. For example, internal energy is constant for an isothermal process.

Energy difference = 42.9kJ


Do you know anything about the product, PV for an ideal gas?

In steady state, the net energy flowing across the boundary is zero. Energy in = Energy out
(look at equation 20.15 in the chapter summary).

P = 1.34kW

Assume the asphalt only loses heat as radiation upward.  Again, energy in = energy out.
T = 364K

Pressure = Force/Area
Force = ma = change in momentum/change in time

Use the listed values for Cp and Cv.

Write Q=Q1+Q2 where the two components come from the two stages of the warming. Consider how pressure and volume changing affects temperature.

Use equation 21.18. for diatomic gas, gamma = 7/5