MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_NextPart_01C5502F.77E7EBA0" This document is a Single File Web Page, also known as a Web Archive file. If you are seeing this message, your browser or editor doesn't support Web Archive files. Please download a browser that supports Web Archive, such as Microsoft Internet Explorer. ------=_NextPart_01C5502F.77E7EBA0 Content-Location: file:///C:/D9286902/Spring05.Phys1111.Exam2.SolutionWeb.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" TIME OF COMPLETION_______________

TIME OF COMPLETION_______________  &nb= sp;         NAME_________SOL= UTION____________________

 

 

<= span style=3D'font-family:"Times New Roman"'>&nb= sp;            =             &nb= sp;            =      DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1111, Exam 2            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;  Section 1

Version 1           &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =           March 30, 2005

Total Weight: 100 points

 

1.         = Check your examination for completeness prior to starting.  There are a total of ten (10)         &= nbsp;   problems on nine (9) pages.

 

2.         = Authorized references include your calculator with calculator handbook, and the Refere= nce         &= nbsp;   Data Pamphlet  (provided by your instructor).

 

3.         = You will have 75 minutes to complete the examination.

 

4.         = The total weight of the examination is 100 points.

 

5.         = There are six (6) multiple choice and four (4) calculation problems.  Do all problems.  Show   all work; partial credit will be given for correct work shown.

 

6.         = If you have any questions during the examination, see your instructor who will=

be located in the room.<= /span>

 

7.         = Start:        &= nbsp;      10:30 a.m.

Stop:            &= nbsp;   11:45 a.m.

 

 

 

&= nbsp;           &nbs= p; PROBLEM

 

&= nbsp;           &nbs= p;    POINTS

 

&= nbsp;           &nbs= p;   CREDIT

 

   &nbs= p;            &= nbsp;    1-6

 

   &nbs= p;            &= nbsp;    24

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     7

 

   &nbs= p;            &= nbsp;    19

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     8

 

   &nbs= p;            &= nbsp;    19

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     9

 

   &nbs= p;            &= nbsp;    19

 

&= nbsp;

 

&= nbsp;           &nbs= p;        10

 

&= nbsp;           &nbs= p;        19

 

 

 

&= nbsp;           &nbs= p;    TOTAL

 

&= nbsp;           &nbs= p;       100

 

 

 

 

 

&= nbsp;       POINTS EARNED

 

 

 

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS.  IN

MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR

PARTIAL CREDIT.

 

Wt.&nb= sp;     No.

 
 
04        1.         = A block of mass, m, is released from rest at point A and slides down the frictionless

surface in the figure below.  The kinetic energy of the block at= point B is equal to

 

a.&= nbsp;        (0.300)mgh.    

 

b.&= nbsp;        (0.700)mgh.

 

c.&= nbsp;        mgh.

 

d.&= nbsp;        None of the above.

 

 

 

 

04        2.         = If the work done by all the external forces acting on an object is 5.80 J, and= the object

is initially at rest, its final kin= etic energy is

 

a.&= nbsp;        zero.

 

b.&= nbsp;        2.90 J

 

c.&= nbsp;        5.80 J

 

d.&= nbsp;        cannot be determined.

 

04        5.         = A wheel is initially rotating with an angular velocity of 20.0 radians/s.  If the wheel

rotates through an angular displace= ment of 100 radians while slowing down to rest,

the constant angular acceleration w= as

 

a.&= nbsp;        -2.00 radians/s2.

w2 =3D wo2 + 2 a<= span style=3D'font-family:"Times New Roman"'> (q - q0)<= /o:p>

b.&= nbsp;        0.200 radians/s2.

(w2 - wo2)/ (2(q - q0)) =3D a     a =3D (0 - (20.0 rad/s)2)/(2 x 100 rad)  

`        &= nbsp;           &nbs= p;  c.         = 5.00 radians/s2.

 

d.&= nbsp;        -200 radians/s2.

Wt.&nb= sp;     No.

 

04        4.        A person riding a bicycle is traveling due North, and decreases her linear velocity

from 10.0 m/s to 3.00 m/s.  The direction of the angular acceleration vector of the

front wheel points

 

a.&= nbsp;        East.

 

b.&= nbsp;        West.

 

c.&= nbsp;        North.

 

d.&= nbsp;        downward.=

 

 

 

04        5.         = The curve shown below represents the magnitude of the force exerted on a ball during

a collision with a wall plotted aga= inst time.  The area under the curve represents:

 

a.&= nbsp;        the force on the ball.    &= nbsp;           &nbs= p;                   &= nbsp;  

 

b.&= nbsp;        the work done on the ball.

 

c.&= nbsp;        the change in linear momentum of the ball.<= /o:p>

 

d.&= nbsp;        the power delivered by the wall to the ball.

 

 

 

 

 

Wt.&nb= sp;     No.

 

19        7.         = An applied force of magnitude A =3D 50.0 N, oriented at an angle of  35.0° with

respect to the vertical as shown in= the diagram below, pushes a block of mass,

 
m =3D 3.00 k= g, up a wall a distance of 1.80 m.  The coefficient of kinetic friction

between the block and the wall is 0= .230.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(04)        &= nbsp;        a.         = Draw a free body diagram for the block.

 

 

 

(04)        &= nbsp;        b.         = Calculate the work done by the applied force.

 

WA =3D A d cos(q) =3D  (50.0 N)(1.80 m)cos (35.0o)

=3D73.7J

 

 

(04)        &= nbsp;        c.         = Calculate the work done by the weight of the block.

 

 

Ww =3D w d cos(q) =3D (3.00 kg)(9.80 m/s2)(1.80 m)cos (180o) =3D -52.9 J

 

 

 

(07)        &= nbsp;        d.         = Calculate the work done by the kinetic friction force.

 

 

Wf =3D fk d cos(q) =3D mk n d cos(q)

 

n =3D A sin (35.0o) =3D 28.7 N

 

Wf =3D (0.230)(28.7 N)(1.80 m)cos (180o) =3D -11.9 J

 

 

 
Wt.&nb= sp;     No.

 

19        9.        A bullet of mass, m1 =3D 0.0200 kg, is fired horizontally into a b= lock of wood of

mass, m2 =3D 8.00 kg.  The initial speed of the bullet is= 300 m/s.  The block rests on<= /o:p>

a horizontal surface.  

 

 

 

 

 

 

 

 

(05)        &= nbsp;        a.         = If the collision between the bullet and the block is perfectly inelastic, calculate

the final speed of the masses immed= iately after the collision.

 

 

m1V1i =3D (m1 + m2)Vf

 

Vf =3D m1V1i /(m1 + m2) =3D 0.748 m/s

 

 
 
(14)        &= nbsp;        b.         = A wooden block, m2 =3D 8.00 kg, as in part a above is now attached= to an

unextended horizontal spring of spr= ing constant of 300 N/m.  A bullet= ,

m1 =3D 0.0200 kg, is aga= in fired horizontally into the block and again undergoes

 &nb= sp;            =             &nb= sp;         a perfectly inelastic collision.  Assume the speed of the masses immediately

<= span style=3D'font-family:"Times New Roman"'>after the collision is 0.748 m/s.

 

 

 

 

 

 

 

 

 

If the maximum compression of the s= pring is 0.0668 m, calculate the

coefficient of kinetic friction bet= ween the block with the bullet and the

horizontal surface.

 

 

 

 

 

 

 

WNC =3D TEf - TEi =3D ½ k x2 – &f= rac12; (m1 + m2) Vf2

 

WNC =3D fk x cos (180o) =3D - (m1 + m2) g mk x

 

- (m1 + m2) g mk x  =3D &fr= ac12; k x2 – ½ (m1 + m2) Vf2<= /sup>

 

mk  =3D - (&f= rac12; k x2 – ½ (m1 + m2) Vf2)/[(m1 + m2) g x]

 

mk  =3D 0.300

 

 

Wt.&nb= sp;     No.
 
<= /span>

19        10.       A coi= n is located 0.250 m from the center of a rotating turntable.  The coefficient<= /p>

of static friction between the coin= and the turntable is 0.500.  =

 

 

 

 

 

 

 

 

 

 

 

(06)        &= nbsp;        a.         = Draw a complete free body diagram for the coin.

 

 

 

 

 

 

 

 

(13)        &= nbsp;        b.         = Calculate the magnitude of the maximum tangential velocity of the coin so<= /span>

that it does not slip off of the turntable.  Show your work.

 

 

FC =3D m aC =3D mV2/R

 

FC =3D fS =3D mS n =3D mS mg

 

mV2/R =3D mS mg

 

V2 =3D mS g R

 

V =3D ( mS g R)½

 

V =3D 1.11 m/s

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PHYS 1111 Exam 2, Version 1

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Spring 2005

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