Test 1 Attempt

#$&*

course PHY 121

Time and Date Stamps (logged): 13:44:22 11-25-2011 °²Ÿ³³Ÿ±±°°Ÿ±΄Ÿ±―°° Principles of Physics (Phy 121) Test_1

Completely document your work and your reasoning.

You will be graded on your documentation, your reasoning, and the correctness of your conclusions.

Test should be printed using Internet Explorer. If printed from different browser check to be sure test items have not been cut off. If items are cut off then print in Landscape Mode (choose File, Print, click on Properties and check the box next to Landscape, etc.).

Signed by Attendant, with Current Date and Time: ______________________

If picture ID has been matched with student and name as given above, Attendant please sign here: _________

Instructions:

• Test is to be taken without reference to text or outside notes.

• Graphing Calculator is allowed, as is blank paper or testing center paper.

• No time limit but test is to be taken in one sitting.

• Please place completed test in Dave Smith's folder, OR mail to Dave Smith, Science and Engineering, Va. Highlands CC, Abingdon, Va., 24212-0828 OR email copy of document to dsmith@vhcc.edu, OR fax to 276-739-2590. Test must be returned by individual or agency supervising test. Test is not to be returned to student after it has been taken. Student may, if proctor deems it feasible, make and retain a copy of the test..

Directions for Student:

• Completely document your work.

• Numerical answers should be correct to 3 significant figures. You may round off given numerical information to a precision consistent with this standard.

• Undocumented and unjustified answers may be counted wrong, and in the case of two-choice or limited-choice answers (e.g., true-false or yes-no) will be counted wrong. Undocumented and unjustified answers, if wrong, never get partial credit. So show your work and explain your reasoning.

• Due to a scanner malfunction and other errors some test items may be hard to read, incomplete or even illegible. If this is judged by the instructor to be the case you will not be penalized for these items, but if you complete them and if they help your grade they will be counted. Therefore it is to your advantage to attempt to complete them, if necessary sensibly filling in any questionable parts.

• Please write on one side of paper only, and staple test pages together.

Test Problems:

. . . . . . . . . . . . . . . .

.

.

.

.

.

.

.

.

.

.

Problem Number 1

By how much will the kinetic energy (abbreviated KE) of an object increase as a result of a net force of 276-739-2590. Test must be returned by individual or agency supervising test. Test is not to be returned to student after it has been taken. Student may, if proctor deems it feasible, make and retain a copy of the test.exerted over a distance of 70 meters, assuming (ideally)that no energy is dissipated in the process?

.

.I’m not sure what this one is saying? It is very confusing.

.

.

.

@& There are a few problems that were 'damaged' by a search-and-replace operation, and this is one of them. You can ignore it.*@

.

.

.

.

.

Problem Number 2

A displacement vector has length 2 and is directed at an angle with the positive x axis of 75 degrees. What are the displacements in the x and y direction which would give the same net displacement?

.

.x component= 2 cos(75)=1.84

y componenet=2 sin(75)=-0.78

.

.

.

.

@& A vector at angle 75 degrees is in the first quadrant, and clearly has an x component which is less than the y component.

Your result is not consistent with this, leading me to suspect that your calculator was in radian mode.

Be sure to always check this, and check your answers to be sure then make sense.

Had your calculator been in degree mode your solution would have been correct. The x component will be somewhere in the neighborhood of .5, the y component pretty close to 2.*@

.

.

.

.

Problem Number 3

What are x and y the components of the velocity vector obtained when we add the two following velocity vectors:

• vector A, with x and y components 4.3 m/s and -5.5 m/s, and

• vector B whose x and y components are 2.2 m/s and -2.7 m/s?

What are the magnitude and angle of the resultant vector?

.Resultant of x is 4.3 m/s+2.2 m/s=6.5

Resultant of y is -5.5+-2.7=-8.2

Magnitude is sqrt(6.5^2+-8.2^2)=sqrt(42.25+67.24)=10.5

The angle is -8.2/6.5=1.26

.

.

.

@& The angle will be arcTan(-8.2 ./ 6.5), which is in the fourth quadrant. -1.26 radians might well be the angle.

You would probably be better off here putting your calculator into degree mode and specifying that the angle is in degrees.

The components have units of m/s. The magnitude will have units of m/s.*@

.

.

.

.

.

.

Problem Number 4

Two objects collide and remain stuck together after collision.

One object has mass 13 kg and is moving in the positive direction at 11 m/s and the other has mass 13 kg and and moves at 20 m/s in the negative direction.

• What is their total momentum, and what will be the velocity of this system immediately after collision?

Solution

The momentum of the first is ( 11 m/s)( 13 kg)= 143 kg m/s, and that of the second is ( - 20 m/s)( 15 kg) = -300 kg m/s.

• The total momentum is therefore 143 kg m/s + `p2 kg m/s = -157 kg m/s.

• The total mass is easily found to be 28 kg. Since the total momentum after collision is the same as that before collision, we see that after collision we have a mass of 28 kg with momentum -300 kg m/s.

• Dividing we obtain velocity ( -300 kg m/s)/( 28 kg) = -5.607143 m/s.

Generalized Solution

The total momentum of a mass m1 moving at velocity v1 and a mass m2 moving at velocity v2 is

• pTot = m1 v1 + m2 v2.

By Newton's Third Law and the Impulse-Momentum Theorem this momentum will remain unchanged during collision.

After collision we wil have one object of mass m1 + m2 and momentum m1 v1 + m2 v2. The object will therefore have velocity

velocity = momentum / mass = (m1 v1 + m2 v2) / (m1 + m2).

.

.

.

.

.

.

.

.

.

Problem Number 5

An object is pushed a distance of 13 meters by a force of 3 Newtons, with the force in the direction of the displacement.

• How many Joules of work are done by the force?

.

.dw=F*ds

dw=3 N*13 m

dw=39 J

.

.

.

.

.

.

.

.

Problem Number 6

As an unknown force is exerted on an object of constant mass 18 kg its velocity is observed to change from 7 m/s to -35 m/s is 1 seconds.

• Find the average force on the object during this time, using the Impulse-Momentum Theorem.

• Verify your results using your knowledge of uniformly accelerated motion.

.

.Fave*dt=dp

dp=m*dv

dp=18 kg*-42 m/s

Fave*1 second=-42 m/s

Fave=-42 N

vAve=-42/1=-42

.

.

.

.

@& The momentum change is 18 kg * -42 m/s = -700 kg m/s, approx.

This is the quantity that is equal to F_ave * `dt.*@

.

.

Problem Number 7

An object of mass 15 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 .05 seconds.

If the average force over this time is 144 Newtons,

• Use Newton's Second Law and your knowledge of uniformly accelerated motion to find the change in the object's velocity.

• Use the Impulse-Momentum Theorem to obtain the same result.

.

.aave=Fave/m

‘dv=aAve*’dt

aAve=144 N/15 kg=9.6 m/s

‘dv=9.6 m/s*0.05 s=0.48 m/s

144 N*0.05=7.2

7.2=15kg*’dv

‘dv=0.48 m/s

.

.

.

.

.

.

.

.

Problem Number 8

An object with mass 6 kilograms, initially at rest, is acted upon by a force of 88.2 Newtons.

• If the force acts for 7 seconds, what will be KE increase of the object, based on its velocity change?

• During the 7 seconds, based on the distance it travels, how much work is done by the net force on the object?

88.2*7 s=617.4

617.4=6 kg*’dv

‘dv=102.9

vf=102.9

‘dKE=0.5(6)(102.9)^2-0.5(6)(0)^2

‘dke=31765.23-3

‘dKE=31762.23

.’dW=88.2*’ds

‘ds=(102.9+0)/2*7 seconds

‘ds=7.35 m

‘dw=88.2*7.35=648.27 N

.

.

.

.

.

.

.

.

Problem Number 9

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, 7.87 meters per second and 3.84 meters per second.

• Aided by a sketch, find the magnitude and angle of the velocity of the ball with horizontal.

v = sqrt(vx^2 + vy^2)

theta = arctan(vy/vx)

v=sqrt(7.87^2+3.84^2)

v=sqrt(61.9+14.7)

v=8.8

The angle is arctan(3.84/7.87)=0.53

I’m not sure if I have done this one correctly? I don’t really understand figuring out the angle.

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

________________________________________

#$&*

@&

@& Check my notes and let me know if you have additional questions.*@

*@