#$&*
course Phy 121
Could you please give me feedback on a few practice test 1 problems?
Problem Number 1An object of mass 3.3 kilograms is acted upon by a net force of 23.1 Newtons.
The object is initially at rest.
After the first 4.1 seconds, what will the its velocity, and how far will it have traveled?
What kinetic energy will it attain during the 4.1 seconds?
.
a = 23.1 N / 3.3 kg = 7.0 m/s^2
7.0 m/s^2 = 'dv / 4.1 s
dv = 28.7 m/s
vf = 28.7 m/s
14.35 m/s = ds / 4.1 s
ds = 58.84 m
'dKE = W = F * 'ds = 23.1 N * 58.84 m = 1359 J
.
.
.
.
.
.
.
.
.
Problem Number 2
A velocity vector has magnitude 3 and is directed at an angle with the positive x axis of 174 degrees. What are the velocities in the x and y directions of the matching motions in those directions?
Should I use the formula R = sqrt(Rx^2 + Ry^2) for this problem?
.
.
@&
You would use
R = sqrt(Rx^2 + Ry^2)
to get the magnitude R of the force, if you were given the components R_x and R_y.
Here you are given R and theta, and will use the formulas
R_x = R cos(theta)
R_y = R sin(theta)
*@
.
.
.
.
.
.
.
Problem Number 4
An object must gain 17 Joules of energy from a force of 6 Newtons being exerted on it parallel to its direction of motion.
If the object does not dissipate any of the energy added to it, then over what distance must it be pushed?
W = F * ds
17 J = 6 N * ds
ds = 2.83 m
.
.
.
.
.
.
.
.
.
.
Problem Number 5
An object of mass 12 kg experiences a variable force F(t) (here F(t) indicates function notation, not multiplication of F by t; F(t) is the force at clock time t) for .07 seconds.
If the average force is known to be Fave = 99.60001 Newtons:
Use the Impulse-Momentum Theorem to find its change in velocity.
Use Newton's Second Law and your knowledge of uniformly accelerated motion to verify your result.
F * 'dt = mvf - mv0
(99.60001) * (0.07) = 12 (vf - v0)
'dv = 0.58 m/s
a = F / m = 99.60001 N / 12 kg = 8.30 m/s^2
8.30 m/s^2 = dv / 0.07 s
dv = 0.58 m/s
.
.
.
.
.
.
.
.
.
Problem Number 6
The momentum of an object is -28 kg m/s. How fast is it moving if its mass is 7 kg?
p = m * v
-28 kg m/s = 7 kg * v
v = -4 m/s
.
.
.
.
.
.
.
.
.
Problem Number 7
A dense ball rolls off the edge of a cliff and follows a very nearly parabolic path to the floor.
At a certain instant the horizontal component of its velocity are, respectively, 4.89 meters per second and 1.27 meters per second.
Aided by a sketch, find the magnitude and angle of the velocity of the ball with horizontal.
This problem also confused me. Should I use the formula again?
.
@&
The vertical motion is characterized by
v0 = 0
a = 9.8 m/s^2 downward
`ds = 4.89 meters downward
From this you can figure out how long it takes the object to fall.
Horizontal motion for an ideal projectile is characterized by the absence of any force, so that its acceleration is zero.
The horizontal motion lasts exactly as long as the vertical motion.
What can you therefore determine?
*@
.
.
.
.
.
Problem Number 8
What are x and y the components of the vector obtained when we add force vector A, with magnitude 2.8 Newtons and angle 246.5 degrees, to the force vector B whose angle and magnitude are 98.8 degrees and 2.8 Newtons?
What are the magnitude and angle of this resultant vector?
R = sqrt (31.36 + 119232)
R = 345.3
theta = arctan (345.2 / 5.6) = 89.1 degrees
.
.
.
@&
I can't tell where most of your numbers came from.
However to solve this problem you need to first find the x and y components of the two vectors.
You then add the x components to get the x component of the resultant, and add the y components to get the y component of the resultant.
Then knowing the components of the resultant, you use them to find the magnitude and angle of the resultant.
*@
.
.
.
.
.
.
Problem Number 9
If you push an object 80 meters, using a force of 80 Newtons, then what will be its increase in kinetic energy if in the process it dissipates 2560 Joules to friction (assume that the only forces acting are your push and friction)?
W = 80 N * 80 m = 3200 J
dKE = 3200 J + 2560 J = 5760 J "
@&
80 * 80 = 6400, not 3200.
If the object gained 2560 J from friction then friction would add to its KE.
However the object dissipates 2560 J, which means that 2560 J is take from its KE.
*@
@&
In general you're in pretty good shape, but you probably need to take another good run through Introductory Problem Set 5 on vectors.
*@
course Phy 232
7/6/12 submitted around 10:21 PM
This experiment uses a cylindrical container and two lamps or other compact light sources. Fill a cylindrical container with water. The cylindrical section of a soft-drink bottle will suffice. The larger the bottle the better (e.g., a 2-liter bottle is preferable to a 20-oz bottle) but any size will suffice.
Position two lamps with bare bulbs (i.e., without the lampshades) about a foot apart and 10 feet or more from the container, with the container at the same height as the lamps. The line separating the two bulbs should be perpendicular to the line from one of the bulbs to the cylindrical container. The room should not be brightly lit by anything other than the two bulbs (e.g., don't do this in front of a picture window on a bright day).
The direction of the light from the bulbs changes as it passes into, then out of, the container in such a way that at a certain distance behind the container the light focuses. When the light focuses the images of the two bulbs will appear on a vertical screen behind the cylinder as distinct vertical lines. At the focal point the images will be sharpest and most distinct.
Using a book, a CD case or any flat container measure the distance behind the cylinder at which the sharpest image forms. Measure also the radius of the cylinder.
As explained in Index of Refraction using a Liquid and also in Class Notes #18, find the index of refraction of water.
Then using a ray-tracing analysis, as describe in Class Notes, answer the following:
1. If a ray of light parallel to the central ray strikes the cylinder at a distance equal to 1/4 of the cylinder's radius then what is its angle of incidence on the cylinder?
Radius is 5.9 cm
Central ray strikes the cylinder at distance of 5.9*1/4 =1.475 cm
Angle of incidence on the cylinder, which is measured by looking at the ray and radial line in the cylinder, is about 17 degrees
2. For the index of refraction you obtained, what therefore will be the angle of refraction for this ray?
We can use Snell’s Law to obtained the refraction for the ray: n1sin(theta1)=n2sin(theta2)
Filling in values we get: (1.0003)sin(17)=(1.33)sin(theta2)
=12.702 degrees
@&
If the incident angle is 17 degrees then this is correct.
However you haven't established how you got this result.
*@
3. If this refracted ray continued far enough along a straight-line path then how far from the 'front' of the lens would it be when it crossed the central ray?
Finding the hypotenuse: (1/4)(5.9cm)/.0436= 33.83 this should be when it crossed the central ray
@&
It isn't clear where the .0436 came from or how it's connected to the rest of your solution.
*@
4. How far from the 'front' of the lens did the sharpest image form?
From the front of the lens the sharpest image form about 6.2cm away
5. Should the answer to #3 be greater than, equal to or less than the answer to #4 and why?
The answer to #3 should be the same as the answer to #4 in an ideal case
@&
If the ray travels over 33 cm before reaching the axis, it will have passed through the 'back' of the cylinder and been refracted at that point. So it won't actually travel along that path.
*@
6. How far is the actual refracted ray from the central ray when it strikes the 'back' of the lens? What is its angle of incidence at that point? What therefore is its angle of refraction?
The actual refracted ray from the central ray when it strikes the back of the lens should be 33.83. The angle of incidence should be 17 degrees. The angle of refraction found by using Snell’s Law is 12.702 degrees.
@&
You need to throroughly explain your reasoning. I don't believe any of these values is correct.
*@
7. At what angle with the central ray does the refracted ray therefore emerge from the 'back' of the lens?
12.702 degrees
8. How far from the 'back' of the lens will the refracted ray therefore be when it crosses the central ray?
1.475 cm
"
@&
You're on the right track with much of this, but there appear to be numerous errors. You need to provide a much more documentation of your reasoning.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@