For a set of physics sample tests in question-answer format:

 

You need to run the program q a prelim.  There’s a link to this program just above the large heading ‘Forms’ at http://www.vhcc.edu/dsmith/genInfo/default.htm.

 

On the Physics I homepage

 

http://www.vhcc.edu/ph1fall9/frames_pages/new_default_2.htm

 

click on Assts, go to Assignment 8 and you will see the link

 

Physics I Practice Tests

 

The link might well work from your email.

 

The following note is relevant to one of the problems on that practice test.

 

A uniform force of .06983 Newtons is applied at the rim of the disk in a direction tangent to the disk.

The moment arm runs from the axis of rotation to the line of application of the force. The force is applied at (and tangent to) the rim, so the moment arm runs from the center to the rim, a distance equal to the radius of the disk. Multiply the force by the moment arm:

torque = .06983 N * 30 cm = .06983 N * .30 m = .021 m N, approx..

To get the angular acceleration divide the torque by the moment of inertia.

 

Here are some more tests (Test 2 and final) with comments:

 

Corrected Physics I Test 1

 

Problem Number 1

t: If an object’sDs and Dv, vAve, a, DReason out the quantities v0, vf,  initial velocity is 8 cm/s, and it accelerates uniformly through 47.5 cm in 5 seconds, then what is its acceleration?

1. We can find the acceleration by using this formula: A=(vf-v0)/t

2. However, we don’t have vf. We can use this equation v=(s-s0)/t to find the vf.

V=(47.5-0)/5=9.5 cm/sec = 0.095 m/sec

`ds / `dt gives you average velocity, not final velocity.

3. Now, we can find the acceleration:

A=(9.5-8)/5=0.3 cm/sec^2

a = `dv / `dt, and `dv is final velocity - initial velocity. Your calculation shows average velocity - initial velocity.

4. Also, we can find a) vAve, which equals to vAve= (v0+vf)/2 = (8+9.5)/2=8.75 cm/sec

vAve = `ds / `dt; if acceleration is uniform then it is also the case that vAve = (vf + v0) / 2. Since acceleration is uniform it is acceptable to use the latter expression, as you have. However you have not calculated the correct final velocity. You are using average velocity where you should be using final velocity.

The reasoning requested on this problem is as follows:

average velocity is 47.5 cm / (5 sec) = 9.5 cm/s.

Initial velocity is 8 cm/s.

Ave velocity is halfway between init and final vel, since accel is uniform. Since ave vel is 1.5 cm/s greater than init vel, final vel is 1.5 cm/s greater than this, or 11 cm/s.

Change in velocity is therefore 3 cm/s, so acceleration is 3 cm/s / (5 s) = .6 cm/s^2.

t =(t1-t0)=(5-0) = 5 secDb)  

Using the equations which govern uniformly accelerated motion determine t for an object which accelerates at .6 cm/s/s through aDs and Dvf, v0, a,  distance of 47.5 cm, ending with velocity 8 cm/s.

1.Using this equation a= (vf-v0)/t, we can find the difference between two velocities:

vf-v0 = a / t = 0.6 / 5.9375 = 0.1011 cm/sec

a = `dv / `dt, so `dv = a * `dt, not a / t.

2. where t = s / v = 47.5 / 8 = 5.9375 sec

This would be `ds / v0, which is not generally an important quantity. vAve = `ds / `dt, so `dt = `ds / vAve, not `ds / v0.

3. We obtained the equation: vf-v0 = 0.1011 cm/sec

Now we can find v0: v0 = 8-0.1011 = 7.8989 cm/sec

s = (sf-s0) = (47.5 - 0) = 47.5 cmD4. Also,  

t = d0/v0 = 0/7.8989 = 0 secD5. and  

Problem Number 2

An Atwood machine consists of masses of .9 Kg and .9269999 Kg hanging from opposite sides of a pulley.

• As the system accelerates 3 meters from rest, how much work is done by gravity on the system?

1. Lets assume that the m2 is a counterweight and equals to 0.9 kg, then m1 = 0.9269999 kg. Also, g is a constant and equals to 9.8 m/sec^2.

2. Now, we can find for counter weight m1g =0.9269999*9.8 =9.08459902 N and for the first object m2g = 0.9 * 9.8 = 8.82 N

3. Now to find acceleration and Ft, we can apply F=ma to each equation, where we take upward as a positive direction: a2=a and a1= - a

4. Ft- m1g = m1a1= - m1a

Ft – m2g = m2a2= m2a

We subtract the first equation from the second one to get:

( m1-m2)g = (m1 + m2)a:

We solve this for a:

A = [(m1-m2) / (m1+m2)] g = [(0.9269999 – 0.9) / (0.9269999 + 0.9)] 9.8 = (0.0269999 /1.8269999) 9.8 = 0.0147782712 * 9.8 = 0.1448270577 m/sec^2 =0.145 m/sec^2

The object accelerates downward at a= 0.1448270577 m/sec^2

Good, but when you consider the system, it consists of both weights, one accelerating upward and the other downward. So the acceleration of the system cannot be upward or downward. The acceleration of the system will be in the direction of the greater mass; depending on how you draw the system this could be clockwise or counterclockwise.

You have not yet answered the question of how much work is done on the system by gravity.

5. The tension in the cord Ft, can be obtained from one of two equations F = ma, changing a = 0.1448270577 m/sec^2

Ft = m1g – m1a = 0.9269999 * (9.8 - 0.1448270577) = 8.950344352 = 8.95 N

Or

Ft = m2g + m2a = 0.9 (9.8 + 0.1448270577 ) = 8.950344352 = 8.95 N

very good

6. We know that W = F d cos a, where force F= mg, displacement d = (df- d0), and cos a = 180 degrees in our case, which equals to -1.

7. Now W = -F (df- d0) =- 0.1448270577 * (3 - 0) = - 0.434481173 J = - 0.43 J

You appear to have multiplied the acceleration by the displacement, not the net force by the displacement.

You are not including units in your calculations; had you included units with these quantities you would have ended up with m/s^2 * m = m^2 / s^2, which is not Joules.

Had you multiplied the acceleration by the total mass of the system you would have obtained the net force exerted by gravity on the system, and multiplying this by the displacement would have given you the work. However you do have to be careful with the signs. Is the displacement in the direction of the acceleration or opposite the direction of the acceleration?

• Assuming no friction or other dissipative forces, use the definition of KE to determine the velocity of the system after having moved through the 3 meters, assuming that the system was released from rest.

 KE. So that, we canD1. We know that KE = ½ * m * v^2, also, Wnet =  apply Newton’s second law: F net = ma, and use Eq. 2-10c, which we write as v2^2 = v1^2 +2ad.

2. Now, we can solve this equation for (v2^2 –v1^2):

(v2^2 –v1^2) = a/2d = 0.1448270577 / 2 * 3 = 0.024137843 m/sec

You haven't used the definition of KE to obtain this result, which is in any case not correct. The units of your calculation give you .024 m^2 / s^2, not .024 m/s. And this calculation does not give you velocity, but the change in the squared velocity.

From a correct calculation of work you will know the KE of the system; setting this equal to 1/2 m v^2, you can solve for v.

3. However, we know that the system was released from the rest, so the v1 = 0 m/sec

4. Now, we can determine the v2, which equals to: v2 = sqrt(0.024137843) = 0.1553635831 m/sec or = 15.54 cm/sec

This would be a correct calculation of the final velocity. It should agree with the result you obtain based on the work-energy theorem.

Problem Number 3

When masses of 30, 60 and 90 grams are hung from a certain rubber band its respective lengths are observed to be 47, 58 and 69 cm. What are the x and y components of the tension of a rubber band of length 53.03 cm if the x component of its length if 43.76755 cm?

1. According to the way I completed one of the labs: the tension would equal to 13.2575 N, because 1 N = 4 cm.

From the given information a 53 cm rubber band would support about 56 grams, which would have a weight of about .55 Newtons. You estimate the force by graphing suspended weight vs. length.

2. Using the Pythagorean theorem we can find the y-component : y^2 = 53.03^2 – 43.76755^2

Y= 29.94298694 cm

3. Now, the tension of x component is 10.9418875 N

And the tension of y component is 7.485746735 N

These components would correspond to the components of a 13.25 N force in the direction of the rubber band; had you used the correct tension you probably would have obtained the correct components.

What vertical force, when added to this force, will result in a total force of magnitude 130 grams (a gram force is the force of gravity on a one gram mass)?

1. 130 grams is 1.3 kg

130 grams is .13 kg.

However it isn't completly clear where the 130 grams comes from.

2. The vertical force will equal to F = mg = 1.30 * 9.8 = 12.74 N.

Problem Number 4

If the slope of a graph of number of paper clips needed to maintain equilibrium vs. ramp slope is 33 clips / unit of ramp slope , and if the slope of a graph of the acceleration of a cart vs. the number of paper clips attached by a string and suspended over a pulley is ( 27 cm/s2) / clip, then how many cm/s2 of acceleration should correspond to 1 unit of ramp slope? If we require 29 clips to match the mass of the cart, then if we could apply this force to the cart without the extra mass of all those clips, what would be the acceleration of the cart?

1. If the slope is 33 clips per unit and another slope is 27 cm/sec^2 / clip, than 891 cm/s2 of acceleration should correspond to 1 unit of ramp slope.

2. Since the slope is 27 cm/s^2 / clip, we can say that 29 clips would give the acceleration of 783 cm.s^2.

Problem Number 5

Explain why the slope of a position vs. time graph between two clock times is equal to the average acceleration between those two times.

1. Since the slope equals to ‘rise’ divided by ‘run’, we can say that the division position by two times will give us the average of acceleration.

2. Rise = upper position – lower position and run = right two times– left two times

The statement 'the slope of a position vs. time graph between two clock times is equal to the average acceleration between those two times' is actually incorrect. The rise of a position vs. clock time graph is the change `ds in position and the run is the change `dt in clock time. So the slope is rise / run = `ds / `dt, which is the definition of average velocity, not average acceleration.

In any case the answer to such a question always starts with 'the rise means ... ' and 'the run means ...', and ends up with 'therefore the slope means ... '. You have the right idea.

Problem Number 6

A ball of mass 8 kg rolls off the edge of a ramp with a horizontal speed of 70 cm/s.

• What is its KE as it rolls off the ramp?

1. First of all, we have to concert 70 cm/s into 0.7 m/s

2. Now, KE = ½ * m * v^2=1/2 * 8 *0.7 ^2 = 1.96 J

• How much work does gravity do on the ball as it falls 22 cm?

22 cm = 0.22 m

W = Fgrav * d * cos a = mg d cos 180 (in our case) = - 8 * 9.8 * 0.22 = - 17.248 N

Good, but in most projectile situations I recomment using a coordinate system with the x axis in the horizontal direction and the y axis in the upward vertical direction. This would mean that the gravitational force is along the negative y axis, at 270 degrees from the positive x axis as measured counterclockwise from that axis, and you would use m g cos(270 deg) = - m g as the force of gravity.

The displacement is also in the negative y direction, so the work woul be - m g cos(270 deg) * (-.22 m) = ... = 17 kg m^2/s^2 = 17 Joules. The work would be positive, since force and displacement are in the same direction, and the units are Joules, not Newtons.

• What will be its kinetic energy after falling 22 cm?

KE = ½ m v1^2 = m g y = 8 * 9.8 * 0.22 = 17.248 J

No energy is dissipated, so the KE will be 17 J greater after falling the .22 m. The KE was originally 2 J, which makes the new KE equal to 19 J.

• How much of this KE is accounted for by its horizontal velocity, and how much by its vertical velocity?

KE of the ball is 1.96 J when it is rolling down the ramp and 17.248 J when it is falling down.

• What then is its vertical velocity at this point?

Using the formula for KE, we can solve this equation for v:

V= sqrt (2KE/m) = sqrt (2 * 17.248 / 8) = 2.0765 m/s is the vertical velocity of the ball.

This is correct.

Had the question asked for the speed of the ball at this point, you would have had to use the 19 J to find the answer. I mention this because some versions of this problem ask for the speed and not just the vertical velocity.

Problem Number 7

A cart of mass 1.9 kg coasts 100 cm up an incline at 7 degrees with horizontal. Assume that frictional and other nongravitational forces parallel to the incline are negligible.

• What is the component of the cart's weight parallel to the incline?

Its displacement

The weight is directly vertically downward. If the x axis is directed along the incline, and if the incline goes up and to the right, this means that the weight is at angle 263 degrees with respect to the positive x axis. The components of the weight would therefore be W_x = W cos(263 deg) and W_y = W sin(263 deg). The x component is the component parallel to the incline and W is m g.

• How much work does this force do as the cart rolls up the incline?

1. 100 cm = 1 m

1. W = Fd cos a = mg d cos a = 1.9 * 9.8 * 1 * cos 7 = 18.481209 J

The weight does not make an angle of 7 degrees with the x axis; it makes an angle of 263 degrees if your picture is drawn as described above.

• How much work does the net force do as the cart rolls up the incline?

1. Since we already found the force, which is 1.9 N, we can say that this is the only one force which affects the cart. So that the net force equals to 1.9 N.

The net force acting on the cart includes the weight and the normal force exerted by the incline. The net force in the y direction is zero, since the cart moves and accelerates only in the x direction. So, since in this situation only weight and normal force act in the y direction, the normal force is equal and opposite to the y component of the weight. The net force is just the x component of the weight, and it acts as the cart is displaced 1 m in this direction.

2. Now, we can find work done by net force:

W = Fnet d cos a = 1.9 * 1 * cos 7 = 1. 885837688 J

The units of your quantities are kg and m, so your calculation would give you kg * m, not J. Again the cos(7 deg) is not correct here.

• Using the definition of kinetic energy determine the velocity of the cart after coasting the 100 cm, assuming its initial velocity to be zero.

1. KE = W net = 1. 885837688 J

"

See my notes, consider them carefully and let me know if you have specific questions.

 

 

Principles of Physics (Phy 121) Final Exam

 

Problem Number 1

A mass of 147 kg is raised 3.3 meters and at the same time stretches an ideal spring .56 meters from its equilibrium length, where the spring exerts a force of 658.9 Newtons.

How much PE does the system gain during this process?

How does this example illustrate the nature of a conservative force?

Assuming that this is done in the gravitational field at the surface of the Earth, the gravitational PE change is

`dPE_grav = F_grav * dy = (147 kg * 9.8 m/s^2) * 3.3 meters, which comes out around 5000 kg m^2 / s^2 = 5000 Joules.

Wherever this is done, the elastic PE change due to the spring is

`dPE_elastic = 1/2 k x^2.

We find k from the fact that a .56 meter stretch results in an elastic force of about 660 Newtons. Our force constant is therefore k = 660 N / (.56 m) = 1200 N / m, approximately. Using this we obtain

`dPE_elastic = 1/2 * 1200 N / m * (.56 m)^2 = 180 Joules, approx.

The total PE change due to gravity and the spring is therefore

`dPE = `dPE_grav + `dPE_elastic = 5000 J + 180 J = 5200 J, approx..

If the object is release to fall back to its original position as the spring returns to its equilibrium length the object will lose its 5200 J of PE and, provided no other forces act, will gain back the 5200 J in the form of KE.

Problem Number 4

9.400001 seconds after being released from rest what will be the velocity of an object which accelerates freely under the influence of the gravitational field at the surface of Earth?

Since the object is released from rest its initial velocity is 0, and it accelerates at 9.8 m/s^2 for 9.4 seconds.

Its velocity therefore increases by `dv = a `dt = 9.8 m/s^2 * 9.4 s = 92 m/s; since it started from rest this is its final velocity.

Its acceleration is uniform so its average velocity is equal to the average of its initial and final velocities:

vAve = (92 m/s + 0 m/s) / 2 = 46 m/s.

In 9.4 seconds its displacement is therefore

`ds = vAve `dt = 46 m/s * 9.4 s = 430 m, approximately.

Problem Number 5

What is the total momentum of two objects, one moving in the positive x direction at 10 m/s and the other moving at 24 m/s in the negative x direction? The mass of the first object is 10 kg and that of the second is 13 kg.

The momentum of the first object is 10 kg * (+10 m/s) = 100 kg m/s, and the momentum of the second is -24 m/s * 13 kg = -310 kg m/s, approx. The total momentum is thereofre 100 kg m/s - 310 kg m/s = -210 kg m/s, approx..

Problem Number 8

As an Earth satellite of mass `mass kg, originally in a circular orbit of radius 7.87 * 10 ^ 6 m, increases its orbital radius to 7.98 * 10 ^ 6 m, what are the KE and PE changes and the ratio of PE to KE change?

Mass is given here not as a numerical quantity but as the symbolic quantity `mass.

To get the KE of each orbit, the orbital velocity is first obtained from the radius.

To obtain velocity from orbital radius we set centripetal force equal to gravitational force:

m * v^2 / r = G M * m / r^2. Solving for v we obtain

v = sqrt(G M / r), where M is the mass of the Earth (6 + 10^24 kg, approx.). For r = 7.87 * 10^6 m we obtain

v = sqrt( 6.67 * 10^-11 N m^2 / kg^2 * 6 * 10^24 kg / (7.87 * 10^6 m) ) = 7130 m/s.

A similar calculation at 7.98 * 10^6 m yields v = 7090 m/s.

Note that since we are calculating the difference of two quantities which differ by only a small percent, we need to calculate to a sufficient level of precision; here we use 3 significant figures, consistent with the number of significant figures in the given information.

The corresponding kinetic energies are

1/2 * `mass * (7130 m/s)^2 = 2.55 * 10^7 `mass m^2 / s^2 and

1/2 * `mass * (7090 m/s)^2 = 2.52 * 10^7 `mass m^2 / s^2.

Gravitational PE is - G M m / r.

For r = 7.87 * 10^6 m this gives us -5.09 * 10^7 * `mass m^2 / s^2.

For r = 7.96 * 10^6 m this gives us -5.03 * 10^7 * `mass m^2 / s^2.

The change in KE is -3 * 10^5 `mass m^2 / s^2; the change in gravitational PE is +6 * 10^5 `mass m^2 / s^2. Note that the change in PE is double the magnitude and opposite the sign of the KE change, which is always the case when going from one circular orbit to another.

Problem Number 9

How much paint is applied per square meter if 7 gallons of paint are uniformly spread out over the surface of a sphere of radius 5.1 meters?

By what factor does the amount per square meter change in each of the following situations:

The paint is applied over a sphere of double the radius.

The paint is applied over a sphere of quadruple the radius.

The paint is applied over a sphere of radius 9 meters.

The surface area of a sphere is proportional to the square of its radius. So when the radius is doubled, the area increases by factor 2^2 = 4 and the amount per square meter is therefore 1/4 as great.

If the radius is quadrupled, the area increases by factor 4^2 = 16 and the amount per square meter is therefore 1/16 as great.

If the radius is 9 meters, then this is 9/5.1 times the original radius and the squared radius will be (9 / 5.1)^2 times as great. The amount per square meter will be 1 / (9/5.1)^2 times as great, or about .32 times as great.

Problem Number 11

If a disk has moment of inertia 2.2 kilogram meter ^ 2 and must be accelerated from 1.689 radians/second to 3.799 radians/second while rotating through 73.59 radians, what net torque must be used?

torque = moment of inertia * angular acceleration, or

tau = I * alpha.

We are given I and must find alpha in order to obtain the torque.

We know initial and final angular velocity and angular displacement so we can either use the fourth equation of motion (omegaf^2 = omega0^2 + 2 alpha `dTheta, solving the angular acceleration alpha) or reason the result out directly.

Reasoning directly, if acceleration is uniform then omega_Ave = (omegaf + omega0) / 2 = (1.69 rad/s + 3.78 rad/s) / 2 = 2.7 rad/s, approx.. To rotate through about 74 radians would therefore take time `dt = 74 rad / (2.7 rad/s) = 31 seconds, approx.. The change in angular velocity is about 3.78 rad / s- 1.69 rad/s = 2.1 rad/s, so the angular acceleration is

alpha = `dOmega / `dt = 2.1 rad/s / (31 s) = .07 rad/s^2, approx.

Thus the torque is about

tau = I * alpha = 2.2 kg m^2 * .07 rad/s^2 = .15 kg m^2 / s^2 = .15 m * N.

Problem Number 12

What velocity will be attained by the mass of 3.2 kg if, starting from rest, a rubber band does 79 Joules of work on it?

Assume that no other forces act on the object. By how much do the PE and the KE of the system change in the process?

The mass will gain 79 J of KE when the net force does 79 J of work on it, so we have

`dKE = 79 J, so

KE_f - KE_0 = 79 J. Since the object starts from rest KE_0 = 0 so

KE_f = 79 J, so

1/2 m v^2 = 79 J and

v = sqrt(2 * 79 J / m) = sqrt(2 * 79 J / (3.2 kg) ) = 7 m/s, approx..

Problem Number 14

Find the velocity of a mass of 10.7 kg, initially at rest, immediately after absorbing a bullet of mass 41 grams moving at an unknown initial velocity if the mass is attached at its equilibrium position to a spring with restoring force constant 1065 N / m, if the system is observed to displace to .103 meters amplitude with respect to the equilibrium position. Determine also the velocity of the bullet immediately before impact.

Assume that no dissipative forces act on the mass after the collision.

The .103 meter amplitude implies that after being struck by the bullet the mass had sufficient kinetic energy to allow it to gain potential energy

1/2 k A^2 = 1/2 * 1065 N/m * (.103 m)^2 = 5.4 N * m = 5.4 Joules.

During the compression of the spring it is assumed that no force other than the spring force acts on the mass, so the PE gained by the spring must have been present in the form of KE before the spring began to compress.

This means that immediately after collision the velocity of the 10.7 kg mass, plus that of the absorbed .41 gram bullet, is sufficient to give it 5.4 J of kinetic energy. Thus

1/2 m v^2 = 5.4 J and

v = sqrt( 2 * 5.4J / m) = sqrt( 2 * 5.4 J / (10.74 kg) ) = 1 m/s, approx..

During the collision of the bullet with the block, the two object exert equal and opposite forces on one another, resulting in equal and opposite impulses and therefore equal and opposite changes in momentum. Momentum is therefore conserved so that the total momentum before collision equals the total momentum after:

m1 v = (m1 + m2) v ' , where m1 and v are the velocity of the bullet before collision, m2 is the mass of the block and and v ' is the common velocity of block and bullet after collision.

We know the two masses and the after-collision velocity v ' so we solve for v. We can solve this symbolically, obtaining

v = (m1 + m2) v ' / m1 = (10.7 kg + .041 kg) * 1 m/s / (.041 kg) = 250 m/s, approx..

We could also substitute our known information into directly into the equation m1 v = (m1 + m2) v ', obtaining

.041 kg * v = (10.7 kg + .041 kg) * 1 m/s or

.041 kg * v = 10.74 kg * 1 m/s or

.041 kg * v = 10.74 kg m/s.

In this equation we can see the initial and final momentum and how the two are related. The final momentum of the system is the 10.74 kg m/s, the initial momentum is the product of the bullet's mass and velocity. Solving for v we obtain the same result as before

v = 10.74 kg m/s / (.041 kg) = 250 m/s, approx..

Problem Number 15

An object moves at constant angular velocity around a circle of radius 6.9 meters, making a revolution every 5 seconds.

If the angular position of the object is 0 radians t = 0 seconds, what are the x and y coordinates of the object's position after t seconds?

The object moves through angular displacement 2 pi radians in 5 seconds, moving at 2 pi rad / (5 s) = 1.25 rad/s, approx..

In t seconds the object will therefore move through angular displacement

`dTheta = 1.25 rad / s * t.

Since it started at position 0 radians, its angular position will therefore be just

theta = 1.25 rad / s * t.

The object's position vector will therefore have magnitude 6.9 meters and angular position 1.25 rad/s * t, and its x and y coordinates will be

r_x = r cos(theta) = 6.9 m cos(1.25 rad/s * t) and

r_y = r cos theta = 6.9 m sin(1.25 rad/s * t).

Note that these components are of the form

x(t) = A cos(omega t) and

y(t) = A sin(omega t),

with A = 6.9 m and omega = 1.25 rad / s. Either this x or y projection can be used as a model of simple harmonic motion.

bbd requests

 

General College Physics (Phy 201) Test 2

Problem Number 1

A ball has a horizontal range of 21 meters when it is projected horizontally from an altitude of 17 meters. What will be its range if it is projected at an angle of 8 degrees below horizontal with the same initial speed? Approximately how much does its horizontal range change per degree from horizontal?

When projected horizontally the initial vertical velocity is 0. Analysis of the vertical motion (v0 = 0, a = 9.8 m/s^2, `ds = 17 m) therefore yields `dt = 1.8 seconds, approximately. Since it travels 21 meters in the horizontal direction, its constant horizontal velocity is 21 meters / 1.8 sec = 12 m/s, approx..

If projected at 8 degrees below horizontal, which relative to a coordinate plane with the x direction horizontal is 352 degrees, its initial horizontal and vertical velocities would therefore be

v0_x = 12 m/s * cos(352 deg) = 11.9 m/s and

v0_y = 12 m/s * sin(352 deg) = -1.7 m/s.

To find the horizontal range we first solve the vertical motion with v0 = -1.7 m/s, `ds = -17 m and a = -9.8 m/s^2. `dt will be about 1.7 sec. Multiplying this by the initial horizontal velocity yields the horizontal range, which will again be slightly less than the original 21 meters; the result is about 20.2 meters.

This is .8 meters less than the original range. The average rate of change of range with respect to angle is therefore .8 m / (8 degress) = .1 meter / degree.

Problem Number 2

A cart of mass 2 kg coasts 95 cm down an incline at 3 degrees with horizontal. Assume that frictional and other nongravitational forces parallel to the incline are negligible.

What is the component of the cart's weight parallel to the incline?

How much work does this force do as the cart rolls down the incline?

Using the definition of kinetic energy determine the velocity of the cart after coasting the 95 cm, assuming its initial velocity to be zero.

Using the definition of kinetic energy determine the velocity of the cart after coasting the 95 cm, assuming its initial velocity to be .38 m/s.

The weight of the cart is 2 kg * 9.8 m/s^2 = 19.6 N.

Assuming that the incline goes down and to the right, a coordinate system with the x axis along the incline will have the vertically-downward weight vector at angle 273 degrees. The component of the weight parallel to the incline will therefore be

wt_parallel = 19.6 N * cos(273 deg) = 1 N, approx..

As the cart coasts .95 meters in the direction of this component, gravity does work .95 m * 1 N = .95 Joules.

Since there is no friction, and since the component of weight perpendicular to the incline is countered by the normal force exerted by the incline on the cart, this is the net force acting on the cart. So we have

`dKE = `dW_net = .95 Joules.

Since initial KE is 0, `dKE = KE_final and

KE_final = .95 Joules, so

1/2 m vf^2 = .95 J and

vf = sqrt( 2 * .95 J / (2 kg)) = .98 m/s.

Had the initial velocity been .38 m/s, the initial KE would have been 1/2 * 2 kg * (.38 m/s)^2 = .14 Joules, and

`dKE = KE_final - KE_initial so that

KE_final = `dKE + KE_initial = .95 J + .14 J = 1.09 J so

1/2 m vf^2 = 1.09 J and

vf = sqrt( 2 * 1.09 J / (2 kg) ) = 1.04 m/s.

Thus for this incline, a .38 m/s initial velocity increases the final velocity by about .06 m/s.

Problem Number 3

Sketch and label force diagrams for each of the following situations:

A mass of 45 grams is attached to a cart of mass 270 grams and suspended over a pulley of negligible mass and friction. The cart is placed on a ramp whose slope is just enough to compensate for the small frictional force acting on the cart. When the system is released, what will be the acceleration of the cart?

Since the slope of the incline is just enough to compensate for the frictional force, the weight of the cart is balanced exactly by the normal and frictional forces. The weight of the hanging mass, however, is not balanced by any other force. So the net force on the system is the weight of the hanging mass, which is .045 kg * 9.8 m/s^2 = .44 Newton.

This net force accelerates both the 45 gram mass and the 270 gram mass. Thus we have a .44 Newton net force accelerating a mass of .315 kg, giving it an acceleration of

a = F_net / m = .44 N / (.315 kg) = 1.3 m/s^2, approx..

Answer the same question if the cart is placed on a ramp making an angle of 4 degrees with horizontal, with the cart being pulled down the ramp, and if the frictional force is .024 times the normal force on the cart.

If the ramp is at 4 degrees then the component of the cart's weight parallel to the incline is

wt_parallel = 2.7 N * cos(274 deg) = .15 N.

The component of the weight perpendicular to the incline is

wt_perpendicular = 2.7 N * sin(274 deg) = -2.7 N,

which is balanced by an equal and opposite normal force exerted by the incline on the cart. The frictional force therefore has magnitude

| f_frict | = coefficient of friction * | normal force | = .024 * 2.7 N = .06 N,

and this force is in the direction opposite that of the motion. The net force is therefore

wt_parallel + wt_hanging mass + f_frict = .15 N + .44 N - .06 N = .53 N

and the acceleration of the cart is

a = F_net / m = .53 N / (.315 kg) = 1.7 m/s^2, approx..

The incline in this case is more than that required to overcome friction, so the cart has a greater acceleration.

If after adding an unknown mass to the cart on the 4 degree incline the acceleration of the system is 26.88 cm/s^2, what is the unknown mass?

If the unknown mass is designated M, the mass of the cart becomes .270 kg + M, and the components of its weight become

wt_parallel = (2.7 N + 9.8 m/s^2 * M) cos(274 deg) = .15 N + .7 m/s^2 * M and

wt_perpendicular = (2.7 N + 9.8 m/s^2 * M) sin(274 deg) = -(2.7 + 9.8 m/s^2 * M) so that the frictional force is

.024 * normal force = .06 N + .24 m/s^2 * M, opposite the direction of motion. So we have

net force = .15 N + .7 m/s^2 * M + .44 N - (.06 N + .24 m/s^2 * M); the acceleration is thus

a = F_net / mass = .15 N + .7 m/s^2 * M + .44 N - (.06 N + .24 m/s^2 * M) / (.315 kg + M).

Substituting the given value of the acceleration we get the equation

36.88 cm/s^2 = .15 N + .7 m/s^2 * M + .44 N - (.06 N + .24 m/s^2 * M) / (.315 kg + M),

which we solve for M.

It might be easier to work through the details if we use m1 for the mass of the cart, m2 for the hanging mass and M for the added mass. We have

wt_parallel = (m1 + M) g cos(274 deg) = .07 (m1 + M) g

wt_perpendicular = (m1 + M) g sin(274 deg) = -.997 (m1 + M) g and

frictional force = .024 ( .997 (m1 + M) g ) = .024 ( m1 + M) and

hanging weight = m2 g

so that

a = F_net / mass = [ .07 ( m1 + M) g + m2 g - .024 (m1 + M) g ] / (m1 + M + m2).

Our equation is

a = [ .07 ( m1 + M) g + m2 g - .024 (m1 + M) ] / (m1 + M + m2). We multiply both sides by the denominatore to get

(m1 + M + m2) * a = .07 ( m1 + M) g + m2 g - .024 (m1 + M) g, distribute the multiplication to get

m1 a + M a + m2 a = .07 m1 g + .07 M g + m2 g - .024 m1 g - .024 M g, collect terms containing M on the left-hand side to obtain

M a - .07 M g - .024 M g = .07 m1 g - m1 a - m2 a - .024 m1 g. Factoring out M we have

M (a - .07 g - .024 g) = .07 m1 g - m1 a - m2 a - .024 m1 g; dividing both sides by a - .06 g - .024 g we have

M = (.07 m1 g - m1 a - m2 a - .024 m1 g) / (a - .07 g - .024 g) . Substituting for m1, m2, g and a we obtain our final result.

Problem Number 4

An Atwood's machine consists of a disk of radius 25 cm and mass 2.5 kg, with masses of 1 kg and 1.05 kg suspended from a cord over the disk. If the cord does not slip on the disk, what is the total kinetic energy of this system when the masses are moving at 28 m/s?

If the masses are moving at 28 m/s their kinetic energy is easily calculated as

1/2 (1 kg + 1.05 kg) * (28 m/s)^2 = 800 Joules, approx.

The rotating disk also has KE = 1/2 I omega^2. We need to find I and omega:

I = 1/2 M R^2 = 1/2 ( 2.5 kg) ( .25 m)^2 = .08 kg m^2

If there is no slipping then the outer rim of the disk is moving at 28 m/s and its angular velocity is

omega = v / r = 28 m/s / (.25 m) = 112 rad/s so that

KE_rotational = 1/2 ( .08 kg m^2 ) ( 112 rad/s)^2 = 500 Joules, approx.

The total KE of the system is therefore about

800 J + 500 J = 1300 J.

Problem Number 5

What would be the orbital KE of a satellite of mass 830 kg in circular orbit about a planet of mass 99 * 10^24 kg, orbiting at a distance of 47000 km from the center of the planet?

By how much would orbital KE change as the satellite moved from this orbit to a circular orbit of radius 51700 km?

G = 6.67 * 10^-11 N m^2 / kg^2

The orbital velocity of a circular orbit is obtained by setting centripetal force equal to gravitational force:

m v^2 / r = G M m / r^2 so that

v = sqrt( G M / r). We know G, M and r, which we easily substitute to get

v = 11,900 m/s.

The orbital KE of this satellite is therefore

1/2 m v^2 = 1/2 * 830 kg * (11,900 m/s)^2 = 5.9 * 10^10 Joules.

If the radius of the orbit changes to 51700 km a similar sequence of reasoning gives us velocity about 11,300 m/s and KE about 5.3 * 10^10 Joules, a decrease of about 6 * 10^9 Joules.

Note that the gravitational PE of an orbit is

PE = - G M m / r.

For each orbit the PE is negative and has double the magnitude of the KE for that orbit, as you can easily check by substituting for G, M, m and r in the expression for PE.

When a circular orbit increases its distance from a planet, its KE decreases; its PE, which is always negative, increases by twice as much as the KE decreases. So `dKE and `dPE are not equal and opposite. Since `dW_noncons_ON = `dKE + `dPE, this implies a nonconservative force for any change from one orbit to another. This nonconservative force (hopefully) comes from a satellite's rocket engines.

General College Physics (Phy 201) Final Exam

Problem Number 1

An Atwood machine consists of masses of 1.2 Kg and 1.26 Kg hanging from opposite sides of a pulley.

As the system accelerates 3.1 meters from rest, how much work is done by gravity on the system?

Assuming no friction or other dissipative forces, use the definition of KE to determine the velocity of the system after having moved through the 3.1 meters, assuming that the system was released from rest.

The system will accelerate in the direction of the 1.26 kg mass. When the system has moved through 3.1 meters the 1.26 kg mass will be 3.1 meters lower and the 1.2 kg mass will be 3.1 meters higher; each mass will thus experience a change in gravitational PE.

The PE of the 1.26 kg mass will change by 1.26 kg * 9.8 m/s^2 * (-3.1 m), about -38 Joules.

The PE of the 1.20 kg mass will change by 1.20 kg * 9.8 m/s^2 * (+3.1 m), about +36 Joules.

If the PE changes are calculated accurately, the net change is about `dPE = -1.8 Joules.

Assuming no nonconservative forces, `dW_net_ON = `dPE + `dKE becomes `dPE + `dKE = 0 and

`dKE = - `dPE so

KEf - KE0 = -(-1.8 J). Since the system starts from rest its initial kinetic energy KE0 is 0 so

KEf = 1.8 J. Thus

1/2 m vf^2 = 1.8 J, where m is the mass of the accelerating system. Both masses are being accelerated so that mass is 1.2 kg + 1.26 kg = 2.46 kg. We obtain

vf = sqrt(2 * 1.8 J / m) = sqrt(2 * 1.8 J / (2.46 kg) ) = sqrt( (1.5 kg m^2 / s^2) / kg) = sqrt(1.5 m^2 / s^2) = 1.2 m/s, approx..

Problem Number 2

A simple harmonic oscillator of mass 40 kg has a period of .04 seconds.

If the amplitude of its motion is 16560 meters, what are the maximum magnitudes of its acceleration and velocity?

At what displacements from equilibrium can each maximum occur?

A period of a simple harmonic oscillator corresponds to a trip around its reference circle. If the period of the oscillator is .04 seconds, then the reference point is moving at omega = 2 pi radians / (.04 s) = 50 pi rad / s.

The speed of the motion of the reference point around the reference circle is r omega = 16560 meters * 50 pi rad / s = 2.5 * 10^6 m/s, approx..

The centripetal acceleration of the reference point is v^2 / r = (2.5 * 10^6 m/s)^2 / (16560 m) = 4 * 10^8 m/s^2.

Maximum acceleration occurs when the centripetal acceleration vector on the reference circle aligns with the direction of motion, which occurs when the radial vector aligns with the direction of motion. When this happens the object is at its maximum distance from equilibrium, so the max acceleration occurs at the extreme points of the motion, when the magnitude of the position of the object is equal to the amplitude of motion.

Maximum velocity occurs when the velocity vector on the reference circle aligns with the direction of motion. Since the velocity vector is perpendicular to the radial vector, this occurs when the radial vector is perpendicular to the direction of motion, which corresponds to the oscillator being at the equilibrium position.

Algebraically, let's assume that the oscillator moves along the x axis of the reference circle, so that its motion can be modeled by

x(t) = A cos(omega * t), with omega = 50 rad / s. Then its velocity and acceleration are

v(t) = -omega A sin(omega * t) and

a(t) = -omega^2 A cos(omega * t).

| v(t) | is maximized when | sin(omega * t) | = 1, which occurs when omega * t = pi / 2 or 3 pi / 2 (or either of these values plus an integer multiple of pi). When omega * t takes either of these values, cos(omega * t) = 0 so the oscillator is at equilibrium. At these times, | v(t) | = | omega * A |, since | sin(omega * t) | = 1.

| a(t) | is maximized when | cos(omega * t | = 1, which occurs when omega * t is an integer multiple of pi , and coincides with | x(t) | = A and | a(t) | = omega^2 A.

Problem Number 4

A uniform rod of mass 2.9 kg and length 97 cm is constrained to rotate on an axis about its center. A mass of .319 kg is attached to the rod at a distance of 40.74 cm from the axis of rotation. An unknown uniform torque is applied to the rod as it rotates through .13 radians from rest, which requires 1.2 seconds. The applied torque is then removed and, coasting only under the influence of friction, the rod comes to rest after rotating through 4.6 radians, which requires 20 seconds.

Find the net torque for each of the two phases of the motion.

If the applied torque is the result of a force applied at one end of the rod, and perpendicular to the rod, then what is this force?

What is the maximum KE of the system? How much of this KE resides in the .319 kg mass?

The rod has moment of inertia 1/12 M L^2 = 1/12 (2.9 kg) ( .97 m)^2 = .23 kg m^2.

The .319 kg mass has moment of inertia .319 kg * (.407 m)^2 = .05 kg m^s.

So the moment of inertia of the system is .28 kg m^2.

The first phase of motion is characterized by initial angular velocity 0, angular displacement .13 rad and time interval 1.2 seconds. Assuming uniform acceleration we find that the average and final angular velocities are .11 rad/s and .23 rad/s, so the angular acceleration is about .2 rad/s^2.

Coasting to rest the initial angular velocity is the previous final angular velocity, .23 rad/s, the time interval is 20 seconds and the final angular velocity is 0; the angular acceleration is -.23 rad/s / (20 s) = -.011 rad/s^2.

The torque for each phase is the product of moment of inertia and angular acceleration so we have

torque for first phase: `tau_1 = .28 kg m^2 * .2 rad/s^2 = .06 m * N and

torque for second phase: `tau_2 = .28 kg m^2 * -.011 rad/s^2 = -.003 m * N .

The force required to produce a given torque is equal to the force multiplied by the moment-arm; the end of the rod is .97 m / 2 = .48 m from the center. So the forces required are

force for first phase: tau_1 / r = .06 m * N / (.48 m) = .12 N and

force for second phase: tau_2 / r = -.003 m * N / (.48 m) = -.06 N.

The maximum angular velocity of the system is .23 rad / s, so its max KE is

KE_max = 1/2 I * (omega_max)^2 = 1/2 * .28 m N * (.23 rad/s)^2 = .007 Joules.

Note that there is an inconsistency in the given information. A rotating system that comes to rest in 20 sec while rotating through 4.6 radians has an average angular velocity of .23 rad/s, meaning its initial angular velocity is in fact .46 rad/s. So in this case the final velocity in the first phase does not match the initial velocity in the second; using this information the torque and force for the second phase would both be twice as great, and if .46 rad/s is used for the max angular velocity the max KE of the system would be four times that shown above.

Problem Number 5

A person is being rotated in a horizontal circle of radius 12 meters. If the person feels a centripetal force of 5.7 times her own weight, how fast is she traveling? code `t

Let her mass be m. Then her weight is m g and the centripetal force is 5.7 m g.

Let her velocity be v. Then her centripetal acceleration is v^2 / r, with r = 12 meters, and her centripetal force is m v^2 / r.

Thus m v^2 / r = 5.7 m g so that

v^2 / r = 5.7 g (note that we could actually have started with this equation by setting centripetal acceleration equal to 5.7 g).

Solving for v we have

v = sqrt(5.7 g r) = sqrt(5.7 * 9.8 m/s^2 * 12 m) = 27 m/s, approximately.

Problem Number 6

A simple pendulum of length 2.9 meters and mass .32 kg is pulled back a distance of .206 meters in the horizontal direction from its equilibrium position, which also raises it slightly. How much work must be done to accomplish this?

In the pullback position we can form a right triangle whose hypotenuse is the 2.9 m length of the pendulum, having one leg equal to the .206 meter pullback, with the other leg vertical. The length of the vertical leg is sqrt( (2.9 m)^2 - (.206 m)^2).

This vertical leg extends down from the point of attachment at the top of the pendulum, and indicates how far the pendulum is below the point of attachment at the pullback point.

The pendulum will upon release descent to its equilibrium position at a point 2.9 m below the attachment point.

Using this information, figure out how much it has to descend from its pullback position to its equilibrium position, and use this change in altitude to figure out its change in gravitational PE between the two points.

Problem Number 7

If a simple harmonic oscillator of mass 1.16 kg is subjected to a restoring force of 3.2 Newtons when displaced .0928 meters from equilibrium, what will be its its equation of motion if it is released from rest at this position?

What will be the velocity of the oscillator at clock time t = .3392 sec?

What will be its position at this clock time?

What will be its PE, its KE and its total energy?

Compare the total energy with the PE at the .0928 meter displacement.

The force constant is 3.2 N / (.0928 m) = 35 N / m, approx.

The angular frequency of the oscillation is therefore

omega = sqrt(k/m) = sqrt( (35 N/meter) / (1.16 kg) ) = 5.5 rad/s, approx..

Since it is released from its extreme point, we take this as the t = 0 point. The amplitude of motion is .0928 meters.

This corresponds to motion along the x axis of the reference circle, since the x motion starts at the extreme point. So the equation of motion is

x(t) = A cos(theta) = A cos(omega * t). Its velocity and acceleration functions are therefore

v(t) = - omega * A sin(omega * t) and

a(t) = - omega^2 * A * cos(omega * t), all with omega = 5.5 rad/s and A = .0928 m.

At clock time t = .3392 sec we have

x(.3392 sec) = .0928 m cos(5.5 rad/s * .3392 sec) = .0928 m * cos(1.8 radians) = -.03 m, very approximately, and

v(t) = -5.5 rad / s * .0928 m sin(5.5 rad/s * .3392 sec) = -.5 m/s, very approximately.

The KE at this point is 1/2 mass * v^2 = 1/2 * *1.16 kg) * (-.5 m/s)^2 = .14 Joules.

The elastic PE at this point is 1/2 k x^2 = 1/2 ( 35 N/m) (-.03 m)^2 = .017 Joules.

The total energy is thus .14 J + .17 J = .16 J.

At the .0928 m displacement, the elastic PE is

1/2 k x^2 = 1/2 * 35 N/m * (.0928 m)^2 = .16 Joules, very approximately, and since the oscillator is stationary at its endpoint the KE is zero, so the total energy at this point is .16 Joules.

Accurate calculation of these results will confirm to a higher degree of precision that PE + KE is the same at both points.

Problem Number 10

What will be the tension in the string holding a ball which is being swung in a circle of radius .7 meters, if the ball is making a complete revolution every .4 seconds? Assume that the system is in free fall (e.g., in a freely falling elevator, in orbit, etc.)?

What would be the tension in the string if the system was on and stationary with respect to the surface of the Earth, with the ball being swung in a vertical circle, when the ball is at the top of its arc?

What if the ball is at the bottom of its arc?

If the system is in free fall then it is effectively weightless and we need not consider the acceleration of gravity. In this case the net force on the ball is just the centripetal force m v^2 / r, where the velocity v of the ball is circumference / period = 2 pi * .7 m / (.4 seconds) = 11 m/s, very approx. This net force comes from the tension in the string so that

tension = centripetal force = m v^2 / r = mass * (11 meters / sec)^2 / (.7 meters) = mass * 17 m/s^2, again very approximately.

If the ball is being swung in a vertical circle in the presence of the Earth's surface gravitational field, then the centripetal force and hence the net force is the same. However the net force is now the sum of the gravitational and tension forces.

In the highest vertical position both gravity and tension act downward, as does the centripetal force, so we have

-centripetal force = - m g - tension

tension = -centripetal force + m g = -mass * 17 m/s^2 + mass * 9.8 m/s^2 = -mass * 7.2 m/s^2.

At the bottom of the arc gravity acts downward while tension and centripetal force act upward so

centripetal force = - m g + tension and

tension = centripetal force + m g = mass * 17 m/2^s + mass * 9.8 m/s^2 = mass * 26.2 m/s^2.

Problem Number 11

A uniform sphere of mass .87 kg and radius 19 cm is constrained to rotate on an axis about its center. Friction exerts a net torque of .0004104 meter Newtons on the system when it is in motion. On the disk are mounted masses of 25 grams at a distance of 16.72 cm from the axis of rotation, 6 grams at a distance of 11.21 cm from the axis and 36 grams at a distance of 7.6 cm from the axis. A uniform force of .009 Newtons is applied at the rim of the sphere at 19 cm from the axis of rotation.

As the sphere rotates through 5 radians, what will be its change in kinetic energy? Find your result by work\energy considerations only.

If the sphere starts from rest what will be its final angular velocity?

What will be the kinetic energy of each of the masses at this final angular velocity? Does the sum of these kinetic energies equal the total kinetic energy calculated earlier?

If the force was applied by a descending mass attached to the sphere by a light string around its rim, what was the mass and how far did it descend? You may assume that the mass is negligible compared to the mass of the sphere.

The moment of inertia of the sphere is 2/5 M R^2; adding the m r^2 of the individual masses gives the total moment of inertia.

The torque of the applied force is r * F = .19 m * .009 N = .004 m N, very approximately. Adding the opposing frictional torque yields the net torque.

Multiplying the net torque by the angular displacement is the rotational equivalent of multiplying force by displacement; we get the work done by the net torque:

`dW = tau_net * `dTheta

The work done by the net torque is the change in KE.

From KE = 1/2 I omega^2, now knowing I and KE we easily solve to find omega.

It is easy to find the speed of each mass at this angular velocity (multiply angular velocity by distance from axis, v = r * omega). The sum will be less than the total KE obtained earlier because the sum doesn't include the KE of the sphere itself.

A force of .009 N would be exerted by a mass of m = F / g = .009 N / (9.8 m/s^2) = .0009 kg, approx.. As the sphere rotates through 5 radians, this point being 19 cm from the axis would move through 5 rad * .19 m = .95 meters.

The descending mass is assumed negligible because if it wasn't, then its contribution to the moment of inertia would have to be considered (the contribution would be .0009 kg * (.19 m)^2 ).

Problem 12

If a simple pendulum of length 2 meters is subjected to a restoring force of 8 Newtons when displaced .194 meters from equilibrium, what is the mass of the pendulum? What will be its period of oscillation?

For small displacement, the force constant for a pendulum is m g / L so that

F = - m g / L * x, where x is the displacement from equilibrium, L is the 2 meters length and x is the .194 meter displacement. We easily solve for m to obtain

m = - F * L / (g * x) = -(-8 N) * 2 meters / (9.8 m/s^2 * .194 m) = 8 kg, very approximately.

An alternative way to look at it:

If the force is 8 Newtons at displacement .194 meters, then the force constant is 8 N / (.194 m) = 41 N / meter.

This is the force constant, so is equal to m g / L. Solving m g / L = k for m we have

m g / L = k; multiplying both sides by L / g we have

m = k * L / g = 41 N/m * 2 m / (9.8 m/s^2) = 8 N / (m *s^2) = 8 kg.