If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

024. `query 24

Question: `qWhy was it necessary to let the string go slack at the top of the circle in order to get the desired results?

Your solution:

Confidence rating:

Given Solution:

`a** If the string goes slack just at the instant the weight reaches the 'top' of its circular path then we are assured that the centripetal acceleration is equal to the acceleration of gravity.

If there is tension in the string then the weight is being pulled downward and therefore toward the center by a force that exceeds its weight.

If the string goes slack before the weight reaches the top of its arc then the path isn't circular and our results won't apply to an object moving in a circular arc. **

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Question: `qWhy do you expect that, if the string is released exactly at the top of the circle, the initial velocity of the washer will be horizontal?

Your solution:

Confidence rating:

Given Solution:

`a** The direction of an object moving in a circular arc is perpendicular to a radial line (i.e., a line from the center to a point on the circle). When the object is at the 'top' of its arc it is directly above the center so the radial line is vertical. Its velocity, being perpendicular to this vertical, must be entirely in the horizontal direction. **

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Question: `qWhat is the centripetal acceleration of the washer at the instant of release, assuming that it is released at the top of its arc and that it goes slack exactly at this point, and what was the source of this force?

Your solution:

Confidence rating:

Given Solution:

`a** Under these conditions, with the string slack and not exerting any force on the object, the centripetal acceleration will be equal to the acceleration of gravity. **

STUDENT QUESTION:  could this answer be achieved from the equation given

INSTRUCTOR RESPONSE:  This conclusion is drawn simply because the object is traveling in a circular arc, and at this position the string is not exerting any force on it. The only force acting on it at this position is the gravitational force. Therefore its centripetal acceleration is equal to the acceleration of gravity.

Knowing the radius of the circle and v, this allows us to make a good estimate of the acceleration of gravity.

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Question: `qQuery principles of physics and general college physics 7.02. Delivery truck 18 blocks north, 10 blocks east, 16 blocks south. What is the final displacement from the origin?

Your solution:

Confidence rating:

Given Solution:

`aThe final position of the truck is 2 blocks south and 10 blocks east. This is equivalent to a displacement of +10 blocks in the x direction and -2 blocks in the y direction.

The distance is therefore sqrt( (10 blocks)^2 + (-2 blocks)^2 ) = sqrt( 100 blocks^2 + 4 blocks^2)

= sqrt(104 blocks^2)

= sqrt(104) * sqrt(blocks^2)

= 10.2 blocks.

The direction makes and angle of

theta = arcTan(-2 blocks / (10 blocks) ) = arcTan(-.2) = -12 degrees

with the positive x axis, as measured counterclockwise from that axis. This puts the displacement at an angle of 12 degrees in the clockwise direction from that axis, or 12 degrees south of east.

STUDENT QUESTION:

Why don’t we add 180 to the angle since the y is negative?

INSTRUCTOR RESPONSE:

We add 180 degrees when the x component is negative, not when the y component is negative. You that 168 degrees is in
the second quadrant, where the y component is positive.

The arctan gives us -12 degrees, which is in the fourth quadrant (where the y component is negative and the x component
positive, consistent with the given information).

We often want an angle between 0 and 360 deg; when the vector is in the fourth quadrant, so that the angle is negative, we
can always add 360 degrees to get an equivalent angle (called a 'coterminal' angle, 'coterminal' meaning 'ending at the same
point'). In this case the angle could be expressed as -12 degrees or -12 degrees = 360 degrees = 348 degrees. Either angle
specifies a vector at 12 degrees below horizontal.

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Question `prin, `gen (Optional Openstax):

Find the following for path B in the figure below:

(a) the total distance traveled, and

(b) the magnitude and direction of the displacement from start to finish.

Your solution:

Confidence rating::

Given Solution:

The path runs 4 blocks east, 3 blocks north and 3 blocks west, a total distance of (4 + 3 + 3) blocks or 10 blocks.

The displacement is a vector which has an eastward component of 1 block and a northward component of 3 blocks.  The magnitude of this vector is the distance from the initial to the terminal point.  This distance is

magnitude of displacement = sqrt( (1 block)^2 + (3 blocks)^2 ) = sqrt( 1 block^2 + 9 blocks^2) = sqrt(10 blocks^2) = 3.1 blocks^2, approx.

The angle of this vector with the eastward direction is

angle = arcTan ( 3 blocks / (1 block) ) = arcTan(3) = 70 degrees, roughly.

The displacement vector therefore has approximate magnitude 3.1 blocks and direction roughly 70 degrees.

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Question `gen (optional openstax):

Find the components of v_tot along a set of perpendicular axes rotated 30º counterclockwise relative to those in the figure below.

Your solution:

Confidence rating:

Given Solution:

All that is needed is to find the angle of v_tot with the x axis of the new system.  We can then use the sine and cosine of this angle, along with the known magnitude of v_tot, to find the requested components.

The question asks only about v_tot, but to make sure we understand the rotation of the axes we will find the angles of all three vectors.

If the axes are rotated 30 degree counterclockwise, the new x axis will be at 30 degrees with respect to the old, in the first quadrant of the old x axis. 

The new x axis will have rotated beyond the vector v_A, which was at 22.5 degrees with the original x axis, so that v_A now lies at 30 degrees - 22.5 degrees = 7.5 degrees clockwise from the new x axis.  This puts v_A in the fourth quadrant.  Since the clockwise direction is regarded as the positive direction for an angle, this means that v_A is now at angle -7.5 degrees with the new x axis.

It should be easy to see that the x axis has rotated 30 degrees toward the direction of v_tot, which was at 22.5 deg + 26.5 deg = 49 deg relative to the original x axis.  So v_tot is at angle 49 deg - 30 deg = 19 deg relative to the new x axis.

v_B originally had direction 22.5 deg + 26.5 deg + 23 deg = 72 deg relative to the x axis, so its angle with the new x axis is therefore 72 deg - 30 deg = 42 deg.

We can easily confirm, then, that the rotation of the coordinate system has rotated the x axis 30 degrees in the positive (i.e., counterclockwise) direction, which has the effect of reducing the angles of the original vectors relative to the x axis by 30 degrees. 

The original question was about the components of v_tot in the new system.  The magnitude of v_tot is 6.72 m/s, so in a system where its angle with the positive x axis is theta, its components will be

v_tot_x = 6.72 m/s * cos(theta)

and

v_tot_y = 6.72 m/s * sin(theta).

In the new system, v_tot makes angle 19 degrees relative to the x axis, so in this system its components are

v_tot_x = 6.72 m/s * cos(19 deg) = 6.4 m/s, approx.

and

v_tot_y = 6.72 m/s * sin(19 deg) = 2.2 m/s, approx..

Suggestion:

It would be beneficial to you to calculate the components of v_A and v_B relative to the new system.  If you calculate them correctly, the x components of v_A and v_B will add up to the x component of v_tot, and a similar result will occur for the y components.

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Question: `qQuery principles of physics and general college physics 7.18: Diver leaves cliff traveling in the horizontal direction at 1.8 m/s, hits the water 3.0 sec later. How high is the cliff and how far from the base does he land?

Your solution:

Confidence rating:

Given Solution:

`aThe diver's initial vertical velocity is zero, since the initial velocity is horizontal. Vertical velocity is characterized by the acceleration of gravity at 9.8 m/s^2 in the downward direction. We will choose downward as the positive direction, so the vertical motion has v0 = 0, constant acceleration 9.8 m/s^2 and time interval `dt = 3.0 seconds.

The third equation of uniformly accelerated motion tells us that the vertical displacement is therefore

vertical motion: `ds = v0 `dt + .5 a `dt^2 = 0 * 3.0 sec + .5 * 9.8 m/s^2 * (3.0 sec)^2 = 0 + 44 m = 44 m.

The cliff is therefore 44 m high.

The horizontal motion is characterized 0 net force in this direction, resulting in horizontal acceleration zero. This results in uniform horizontal velocity so in the horizontal direction v0 = vf = vAve. Since v0 = 1.8 m/s, vAve = 1.8 m/s and we have

horizontal motion: `ds = vAve * `dt = 1.8 m/s * 3.0 s = 5.4 meters.

STUDENT COMMENT/QUESTION

Why do we not calculate the magnitude for this problem, I know the number are identical but it seems that this would tell us how far from the base the diver traveled?

I understand how to calculate the magnitude using the pythagorean theorem and the directions using arc tan, but I am not quite clear on why and when it is neccessary. ?

INSTRUCTOR RESPONSE

The diver doesn't travel a straight-line path. His path is part of a parabola. It would be possible to calculate the distance traveled along his parabolic arc. However that would require calculus (beyond the scope of your course) and while it would be an interesting exercise, it wouldn't contribute much to understanding the physics of the situation.

What you did calculate using the Pythagorean theorem is the magnitude of the displacement from start to finish, i.e. the straight-line distance from start to finish.

The diver's displacement is a vector with a magnitude (which you calculated) and and angle (which you could easily have calculated using the arcTangent).  However this vector is not in the direction of any force or acceleration involved in the problem, and it's not required to answer any of the questions posed by this situation.  So in this case the displacement not particularly important for the physics of the situation.

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Question `prin, `gen (optional openstax):

A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction.

(a) At what speed does the ball hit the ground?

(b) For how long does the ball remain in the air?

(c)What maximum height is attained by the ball?

Your solution:

The ball's initial velocity is a vector with horizontal and vertical components 16 m/s and 12 m/s, respectively.  Its speed is the magnitude of this velocity, so

initial speed = sqrt( (16 m/s)^2 + (12 m/s)^2 ) = 20 m/s.

The ball lands at the vertical level as it started, so there is no net change in its gravitational potential energy.  Neglecting air resistance and other nonconservative forces, then, its kinetic energy upon landing  will be equal to its initial kinetic energy, so its speed will be the same at both points.  THus

final speed = initial speed = 20 m/s.

The ball's vertical motion for the interval between being kicked and arriving at its maximum height, assuming the upward direction to be positive, is characterized by initial velocity 12 m/s, acceleration -9.8 m/s^2 and final velocity 0. 

We could easily analyze its vertical motion using a_y = -9.8 m/s^2, v0_y = 12 m/s, vf_y = 0, using the equations of motion, and you should be able to do so.  However we gain more understanding by reasoning out the motion: 

The change in vertical velocity will therefore be -12 m/s, and the time required for this change will be `dv_y / a_y =  -12 m/s / (-9.8 m/s^2) = 1.2 seconds, approx.  Its average vertical  velocity will be v_ave_y = (12 m/s + 0) / 2 = 6 m/s, so in the 1.2 seconds it will rise to altitude (6 m/s) * (1.2 s) = 7.2 meters.

To return to the ground the ball will fall an equal and opposite displacement, starting from rest.  You should verify that this will require the same time as the ascent, so that the total time in flight will be 2 * 1.2 s = 2.4 s.

The total time of flight can also be analyzed by considering the interval between kick and landing.  For this interval the vertical displacement is zero, so we have `ds_y = 0, v0_y = 12 m/s and a_y = -9.8 m/s^2.  You may verify that the fourth equation of motion yields vf = +- 12 m/s, while the third yields `dt = 2.4 s.

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Given Solution:

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Question: `qGen phy 3.13 A 44 N at 28 deg, B 26.5 N at 56 deg, C 31.0 N at 270 deg. Give your solution to the problem.

Your solution:

Confidence rating:

Given Solution:

`a** The solution given here is for a previous edition, in which the forces are

Force A of 66 at 28 deg

Force B of B 40 at 56 deg

Force C of 46.8 at 270 deg

These forces are very close to 2/3 as great as the forces given in the current edition, and all correct results will therefore be very close to 2/3 as great as those given here.

Calculations to the nearest whole number:

A has x and y components Ax = 66 cos(28 deg) = 58 and Ay = 66 sin(28 deg) = 31

Bhas x and y components Bx = 40 cos(124 deg) = -22 and By = 40 sin(124 deg) = 33

C has x and y components Cx = 46.8 cos(270 deg) = 0 and Cy = 46.8 sin(270 deg) = -47

A - B + C therefore has components

Rx = Ax-Bx+Cx = 58 - (-22) + 0 = 80 and

Ry = Ay - By + Cy = 31-33-47=-49,

which places it is the fourth quadrant and gives it magnitude

`sqrt(Rx^2 + Ry^2) = `sqrt(80^2 + (-49)^2) = 94 at angle

tan^-1(Ry / Rx) = tan^-1(-49/53) = -32 deg or 360 deg - 32 deg = 328 deg.

Thus A - B + C has magnitude 93 at angle 328 deg.

B-2A has components

Rx = Bx - 2 Ax = -22 - 2 ( 58 ) = -139 and

Ry = By - 2 Ay = 33 - 2(31) = -29,

placing the resultant in the third quadrant and giving it magnitude

`sqrt( (-139)^2 + (-29)^2 ) = 142 at angle

tan^-1(Ry / Rx) or tan^-1(Ry / Rx) + 180 deg. Since x < 0 this gives us angle

tan^-1(-29 / -139) + 180 deg = 11 deg + 180 deg = 191 deg.

Thus B - 2 A has magnitude 142 at angle 191 deg.

Note that the 180 deg is added because the angle is in the third quadrant and the inverse tangent gives angles only in the first or fourth quandrant ( when the x coordinate is negative we'll be in the second or third quadrant and must add 180 deg). **

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Question: `qUniv. 3.58. (This problem has apparently been eliminated from recent editions, due to the now policitally incorrect nature of the device being thrown. The problem is a very good one and has been edited to eliminate politically incorrect references). Good guys in a car at 90 km/hr are following

bad guys driving their car, which at a certain instant is 15.8 m in front of them and moving at a constant 110 km/hr; an electronic jamming device is thrown by the good guys at 45 deg above horizontal, as they observe it. This device must land in the bad guy's car. With what speed must the device be thrown relative to the good guys, and with what speed relative to the ground?

Your solution:

Confidence rating:

Given Solution:

`a** The device is thrown at velocity v0 at 45 deg, giving it v0y = .71 v0 and v0x = .71 v0.

The device will return to its original vertical position so we have `dsy = 0.

Using `dsy = v0y `dt + .5 g `dt^2 with `dsy = 0 and assuming the upward direction to be positive we obtain

v0y `dt + .5 (-g) `dt^2 = 0 so that

`dt = 0 or `dt = - 2 * v0y / (-g) = 2 * 71 v0 / g.

In time `dt the horizontal displacement relative to the car will be

`dsx = v0x `dt + ax `dt; since acceleration ax in the x direction and v0x = .71 v0 is zero we have

`dsx = .71 v0 * `dt.

We also know that relative to the first car the second is moving at 20 km / hr = 20,000 m / (3600 sec) = 5.55 m/s, approx.; since its initial position is 15.8 m in front of the first car we have

`dsx = 15.8 m + 5.55 m/s * `dt.

To keep the equations symbolic we use x0Relative and vRelative for the relative initial position and velocity of the second car with respect to the first.

We thus have three equations:

`dt = 2 * .71 v0 / g = 1.42 v0 / g.

`dsx = .71 v0 * `dt

`dsx = x0Relative + v0Relative * `dt.

This gives us three equations in the variables v0, `dt and `dsx, which we reduce to two by substituting the expression -2 to obtain:

`dsx = .71 v0 * 1.42 v0 / g = v0^2 / g

`dsx = x0Relative + v0Relative * 1.42 v0 / g.

Setting the right-hand sides equal we have

v0^2 / g = x0Relative + v0Relative * 1.42 v0 / g, or

v0^2 - v0Relative * 1.42 v0 - g * x0Relative = 0.

We get

v0 = [1.42 v0Relative +-sqrt( (1.42 v0Relative)^2 - 4 * (-g * x0Relative) ) ] / 2 =

[1.42 * v0Relative +-sqrt( (1.42 * v0Relative)^2 + 4 * g * x0Relative) ] / 2.

Substituting 5.55 m/s for v0Relative and 15.8 m for x0Relative we get

[1.42 * 5.55 m/s +-sqrt( (1.42 *5.55 m/s)^2 + 4 * 9.8 m/s^2 *15.8 m) ] / 2 =

17 m/s or -9.1 m/s, approx..

We conclude that the initial velocity with respect to the first case must be 17 m/s.

Checking this we see that the device will have initial x and y velocities 7.1 * 17 m/s = 12 m/s, approx., and will therefore stay aloft for 2 * 12 m/s / (9.8 m/s^2) = 2.4 sec, approx..

It will therefore travel 2.4 sec * 12 m/s = 28 m, approx. in the horizontal direction relative to the first car.

During this time the second car will travel about 5.55 m/s * 2.4 sec = 13 m, approx., resulting in relative position 15.8 m + 13 m = 28.8 m with respect to the first. This is reasonably close to the 28 m obtained from the motion of the projectile.

Correcting for roundoff errors will result in precise agreement. **

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