Ch6_3 #2

See Ch 4 Sheet 24, Ch 6 Sheet 19' and 28.

Ch6_3 #4

Se Ch 6 Sheet 15. Here in addition to gravity the frictional force in the atmosphere does work Wfr and equation (6.4) on Sheet 15 holds. See on Ch 6 Sheet 18 how you can write the same equation in terms of the total mechanical energy E = KE + PE with the conservation of E being E2 - E1 = 0. With friction the energy loss is Wfr = E2 - E1.

Ch6_3 #3

See Ch6 Sheet 24 for the exact expression of the Gravitational Potential Energy. Use it to calculate the change in potential energy, the difference in potential energy for the 2 positions. Compare this with mgh, which is an approximate change of the potential energy neglecting the variation of g along the path of the object (See Ch 6 Sheet 24').

Ch6_3 #1

See Ch6 Sheet 8 and 28.

Ch6_2 #4

See Ch6 Sheet 5, 28 and 31. Here the motion is vertical.

Ch6_2 #3

See Ch6 Sheet 18'. There is no friction. What matters, the distance along the inclined plane or the vertical distance ?

Ch6_2 #2

See Ch 6 Sheet 13 and 15.

Ch6_2 #1

See Ch 6 Sheet 3. What is the angle between the head-force and the displacement ?

Ch6_1 #4

See Ch 6 Sheet 8.

Ch6_1 #3

See Ch6 Sheet 8.

Ch6_1 #2

See Ch 6 Sheet 12. Watch out, Death Valley is below sea level.

Ch6_1 #1

See Ch6 Sheet 3.
Part A: Is there a displacement x here ?
Part B,C: Make a sketch. Watch the angle ! The work is on the boy.

Ch5_3 #4

See Ch5 Sheet 30 and 31.

Ch5_3 #3

See Ch 5 Sheet 12. Replace the circle shown by the equator and visualize the man standing on the equator. The pole is the center of the circle. The man is held in orbit by the Centripetal Force which is the vector sum of the Normal Force by the earth on the man and the gravitational force . Write down this vector equation paying attention to the relative signs of the vectors. The apparent weight is the Normal Force.

Ch5_3 #2

See Ch 5 Sheet 26 and the algebraic equation under "Note" in Ch5_3 #1. What do you set g' equal to ? Solve for h.

CH5_3 #1

See Ch 5 Sheet 26. Write down the difference deltag = g_e - g' with g_e and g' replaced by their corresponding expressions on Sheet 26. (deltag is the given numerical value). Note that you can write for the expression for g': GM/(Re+h)^2 = (GM/Re^2)/(1+h/Re)^2 = g_e/(1+h/Re)^2. Use this and solve for h.

Ch5_2 #4

Part A: See Ch 5 Sheet 19. The problem here is, as in the notes, for zero banking angle and only friction supplying the centripetal force.
Part B: See Ch 5 Sheet 17. The problem here is, as in the notes, with no friction and only the horizontal component of the Normal Force, exerted by the road on the car, supplying the centripetal force.

Ch5_2 #3

See Ch 5 Sheet 14'. Assume that the space station is far away from any gavitational pull as in the notes.

Ch5_2 #2

See Sheet 12 and replace the box at the bottom with the ball, held in orbit by the Centripetal Force which is the vector sum of the tension in the string and the gravitational force at the bottom of the circle. Write down this vector equation paying attention to the relative signs of the vectors. The apparent weight is the force of the support on the object, the tension in the string. Take the direction of the centripetal force as positive. Solve for the speed.

Ch5_2 #1

See Sheet 12 and replace the box at the top with the car inside the circle, held in orbit by the Centripetal Force which is the vector sum of the Normal Force by the track and the gravitational force at the top of the circle. Write down this vector equation paying attention to the relative signs of the vectors. The apparent weight is the Normal Force. Set it equal to the regular weight ("true" weight). Take the direction of the centripetal force as positive. Solve for the speed. Watch out , the diameter of the loop-the-loop is given, not the radius.

Ch 5_1 #4

See Sheet 4.

Ch5_1 #3

See Ch5 Sheet 4 for the centripetal acceleration for part A. The velocity of the hand and thus the ball at the time of release is given.

For part B see Sheet 10.

Ch5_1 #2

Convert the rpm (= revolutions per minute) into radians/sec. Multiplying by t gives you the angle swept through in radians. What do you have to do with the initial angle ?
Don't forget that the problem asks for an answer between 0 and 2pi.

Ch5_1 #1

See Ch 5 Sheet 4 for the definitions of Period, Frequency, linear velocity at a given radial distance fron the center, angle swept through for a full circle (which is the same as a "rev" (revolution)) in radians (see Sheet 3 for the definition of a radian).
Part A: You can leave the "rev" and simply divide the two given quantities, the # of rev and the rev/sec.
Part B: Use Sheet 4.

Lab 4 Prep #4

Lab 4 Prep #3

Lab 4 Prep #2

Lab 4 Prep #1

Ch6: Comments on Lecture Notes only

Ch5: Comments on Lecture Notes only

Ch4: Comments on Lecture Notes only

Ch3: Comments on Lecture Notes only

Ch5_3 #3

See Ch 5 Sheet 12: Replace the circle made by the airplane by the equator
which spins about an axis through the center of the circle and perpendicular to the paper plane. Visualize the black rectangle being the person standing on the equatuator. On Sheeet 12 the expression Wapparent = m(g-a_c) with the vectors g and a_c is a generally valid expression for any relative orientation of the vectors. In the case of the person on the equator the vectors g and a_c are parallel to each other, i.e. the expression without vector signs is Wapparent = m(g-a_c) where g and a_c are now magnitudes of the accelerations. At the pole there is no rotation and thus no centripetal acceleration a_c and Wapparent = mg there. For Part B you take the difference between mg and Wapparent at the equator.
You have to calculate a_c as given on Sheet 4 using the circumference of the equator and the daily period of rotation.

Ch4_3 #4

This is a carbon copy of Example 4.5 in Ch 4 Sheet 31.

Ch4_3 #3

This question is a simpler version of what is done in Ch4 Sheet 32, but the basic principle is the same: for the red dot on Sheet 32 not to move the 3 forces have to add up to zero net force. Ditto for the 3 forces in the question here.
The tensions are active as shown below:

Add the force vector components with their appropriate sign (!) to yield a net x-component = 0 and net y-component = 0 (see Ch 3 Sheet 6 extended to 3 vectors).

Ch4_3 #2

See Ch 4 Sheet 19. Example 4.3 is very similar to the question here,
with the exception that here friction is included. ("Kinetic" friction refers to a moving object sliding on a surface, "static" friction refers to an object at rest on a surface. We do not distinguish between the two in this course and simply refer to friction.) Watch out how you use the labels m1 and m2 relative to the question here when you inspect equations in Example 4.3 ! The easiest way to proceed is to add the frictional force to the parallel weight component in the equation at the bottom of Sheet 23 with the correct sign (!) and solve for the mass asked for.
The expression obtained there is the application of Newton's Law II for the combined system (m1 + m2). I recommend strongly to study the procedure on Sheet 21 which yields the tension T in addition to the acceleration a (e.g. in case the tension T is asked for.)

Ch4_3 #1

See Ch 4 Sheet 20. There you have the expression for the weight component of the block parallel to the inclined plane given. Thus you know it in the question here for both blocks. Those 2 weight components make up the net force on the system of 2 blocks taken together. What do you know about the net force since the system is at rest ?

Ch4_2 #4

See Ch 4 Sheet 27. Replace the block by the land rover and see Sheet 28.
Warning! To simply use the solution on Sheet 28 to get the coefficient of friction will work for you but would be a silly short cut. You will be required in the exam to understand the equations on Sheet 26, so study Example 4.4!

Ch4_2 #3

See Ch 4 Sheet 19: turn the inclined plane such that it is horizontal (the 30 degrees go to 0 degrees.) and introduce friction (see Sheet 24) between the bottom of block A and the supporting surface. See Sheet 21 equation (2). How do you have to modify that equation to be valid for your block B (Hint: B moves with constant velocity. Is there an acceleration ?) ? What does your modified equation (2) say about the tension T in the string, which is also the tension pulling on block A ? The tension T acting on block A and the frictional force acting on it together make up the net force acting on block A which moves with constant velocity too. What is then the net force on A ? Write down your condition for this net force. You need to express the frictional force as is done on Sheet 24 using for the Normal Force Sheet 11. You obtain an equation where all variables are known except the coefficient of friction asked for. Solve for it.

Ch4_2 #2

See Ch 4 Sheet 18" for the case when an object is suspended (instead of sitting on a support (see Sheet 18')).
Part B: does the given velocity matter ?

Ch4_2 #1

See Ch 4 Sheets 16,17,18 for the Apparent Weight of on object. You don't have to convert the "lb" into SI units. When you set up the ratio the apparent weight when the elevator accelerates over the apparent weight when the elevator is at rest the units of the weight drop out. How about the mass m ? Solve the resulting equation for the acceleration.

Ch4_1 #4

See Ch 4 Sheet 8,10,11 for Part A.
See Ch 4 Sheet 13 for Part B.

Ch4_1 #3

See Ch 4 Sheet 6 for Newston's Law I.

Ch4_1 #2

See Ch 3 Sheet 6 and extend the formulae from 2 to 3 vectors. See Ch 4 Sheet 5 for the component treatment of Newton's Law II.

Ch4_1 #1

See Ch 3 Sheet 7,8 for the "Head-To-Tail" Rule for vector addition. See Ch 4 Sheet 5 the sketch for a vector sum of forces. You add the 2 forces together and then balance the resulting sum vector with a 3rd vector.
The vectors in Mastering Physics are treated as done in Ch 3_1 #4 (bottom):
In order to manipulate the vector F
1. Click on vector F
2. Click “add vector”
3. Select “unlabelled vector”
4. With the cross hair draw a vector on top of F
6. Manipulate the vector you drew

Lab 3 Prep #5

.

Lab 3 Prep #4

This is equal to what you did already in Lab 1 #5.

Lab 3 Prep #3

Lab 3 Prep #2

Lab 3 Prep #1

The missing figure is

It is essentially the sketch for horizontal launch in Ch 3 Sheet 25'. See Ch 3 Sheet 19 (the equations for Projectile Motion). The first 3 entries are all given. Customize them for the situation here (horizontal launch.)
For the 4th entry consult the Introduction in the Lab 3 Manual. You should, however, be able to derive this relation from your previous correct entries.

Lab 2 Prep #7

There is a missing figure which you don't really need. The hints in the text of the question tells you what to do. What is the constant factor "a" in (1.3) in your case where v=(1/t)d ? How do you convert a relative error into an absolute error (see (1.4))?

Lab 2 Prep #6

The error writeup equation (1.3) will get you to the result. Easier is equation (1.1): what is the constant factor "a" in (1.1) in your case, where d=D/N ?

Lab 2 Prep #4,5

See Ch 2 Sheet 5' (after the sheet with 5 and 5" on it):
Notice that the blue curve is the position x as a function of time:x = x(t), and that x(t) given on Sheet 7 is an example similar to the blue curve. Consult Sheet 5 for the slopes. Consult Sheet 13 and 14 for the expressions giving x(t) and v(t). Notice further that the acceleration is positive and constant.

Lab 2 Prep #1 to #3

See Ch 2 Sheets 13,14,15.
Mislabelled as Sheet 14 for question 1 and 3.

Ch3_3 #4

You should never attempt a projectile motion problem without having the 5 equations on the bottom half of Sheet 27 in front of you. I strongly recommend to make a sketch as on Sheet 21 in Ch3 and enter all given quantities from the problem here into it. This should include your knowledge of the y-component of the velocity at the top of the trajectory. It will be helpful to replace the green vector close to the end of the trajectory by a line representing the net.
Part A:
How do you judge whether the ball clears the net ? Think in terms of the y-component of the ball position at the location of the net. Can you calculate its value ?
(Hint: get the flight time first from the x-equations after you have identified all the given quantities for the horizontal motion. Then use it to calculate the vertical position of the ball at the net.)

Ch3_3 #3

You should never attempt a projectile motion problem without having the 5 equations on the bottom half of Sheet 27 in front of you.
Part A:
Visualize in the example on Sheet 21 y0 = 0 for the case here (the golf ball is hit from the ground.) Can you calculate the y-component of the initial velocity from the data given ? What do you know about the y-component of the velocity at the top of the trajectory ? Given those two can you get the rise-time by going "shopping" among the y-equations ? What do you think is the relation of the rise-and fall-time and thus the total flight time ?
Part B:
Can you get the horizontal distance from Part A and going "shopping" among the x-equations ?
Part C:
If you followed the hints in Part A you found the equation which gave you the flight time. Did it contain the gravitational acceleration ? What is then the flight time on the moon ? What would the horizontal distance be for that flight time ?

Ch3_3 #2

You should never attempt a projectile motion problem without having the 5 equations on the bottom half of Sheet 27 in front of you.
Part A:
What is the y-component of the initial velocity when the rifle is aimed horizontally (See Ch 3 Sheet 25') ? Go "shopping" among the y-equations to get the "sinking-time" the bullet takes to traverse the change in the y-component of the position.
Part B:
Go "shopping" among the x-equations to get the x-component of the initial velocity.
Is the speed, i.e. the magnitude of the full initial velocity vector, different from only the x-component of the initial velocity ?

Ch3_3 #1

You should never attempt a projectile motion problem without having the 5 equations on the bottom half of Sheet 27 in front of you.
See Ch 3 Sheet 25': Visualize the plane attached to the horizontal blue launch vector. When you drop the package what is the initial velocity of the package (magnitude and direction: notice that the package flies with the same velocity vector as the plane). Get the fall-time from the "appropriate" y-equations for the "sinking" motion in the vertical direction (go "shopping" among the equations).
Then get the horizontal distance between the point on the ground where the package was launched and the point on the ground where the package landed from the "appropriate" x-equation (go "shopping" among the equations).

Ch3_2 #4

You should never attempt a projectile motion problem without having the 5 equations on the bottom half of Sheet 27 in front of you.
Part A and B:
See the example Quiz 3.3 on Sheet 25'. Does the fall - time depend on the horizontal component of the initial velocity ?
Part C and D:
Once you have the fall-time how do you get the horizontal distance ?

Ch3_2 #3

See Ch 3 Sheet 16''': the sketch on the left and the example on Sheet 16 is the case
of this problem. The ducks head at a appropriate angle into the wind, such that they actually fly due South. (If you rotate the figure on Sheet 16 or 16''' by 180 degrees you have the directions as given in the problem here.) The solid vectors give the velocity vectors relative to the ground, i.e. Vbg ("b" for bird) is the velocity vector pointing where the birds actually goe - not where they head which is the direction of the dashed velocity vector Vbw ("w" for wind). Identify the 2 given velocity vectors in your problem, make the sketch as on Sheet 16''' and place the velocity values given into the sketch. As is describesd on Sheet 16'' the triangle with a right angle in it allows the application of your "SOHCAHTOA" right away to get the angle, without going through the detailed setup of the vector equations on Sheet 16'. Notice that you need your calculator set to degrees in order to get the angle in degrees as asked for.

Ch3_2 #2

Part A:
See the sketch on Ch 3 Sheet 16''' on the right side of the sheet (there is no video for that sheet): in your problem Mary heads the boat straight across the river, which flows east and thus pushes Mary downstream. (The solid vectors give the velocity vectors relative to the ground, i.e. Vbg is the velocity vector pointing where Mary actually goes - not where she heads the boat which is the direction of the dashed velocity vector Vbw). Identify the 2 given velocity vectors in your problem, make the sketch as on Sheet 16''' and place the velocity values given into the sketch. Notice that d is the change of position in the y-diredtion. Thus you need to use the velocity in the y-direction to get the crossing time.
Part B:
Notice that the distance downstream is the change of position in the x-diredtion. Thus you need to use the velocity in the x-direction to get the distance downstream.

Ch3_2 #1

See Ch 3 Sheet 13,14,14'. You have to modify the equation on sheet 14' such, that the velocities, added with the appropriate relative sign , and the travel time give the travel distance. You do this twice, for the trip down the river, then the return. You obtain 2 equations with 2 unknowns, the velocity of the boat relative to the water (Vbw: Part B) and the velocity of the water relative to ground (Vwg: Part A). The latter is being asked for. Watch out for the relative sign of the velocities on the trip up the river! You solve the 2 equations for the 2 unknowns by substitution.

Ch3_1 #4

See Ch 3 Sheet 7,8,9.

Ch3_1 #3

See Ch 3 Notes Sheet 6 and 4.

Ch3_1 #2

See Ch 3 Notes Sheet 3 and 4.
Part D: Watch out where you are asked to place the angle !

Ch3_1 #1

See Ch 3 Notes Sheet 3 and 4.

Ch3_1 #4

For the vector problem:
1. Click on vector A
2. Click “add vector”
3. Select “unlabelled vector”
4. With the cross hair draw a vector on top of A
5. Repeat the same for B
6. Add Vector
7. Select D and draw it