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TIME OF COMPLETION_______________  &nb= sp;         NAME___SOLUTION_= _________________________

 

 

<= 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;            =             &nb= sp; July 16, 2007

Total Weight: 150 points

 

1.         = Check your examination for completeness prior to starting.  There are a total of ten (10) prob= lems on seven (7) pages.

 

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

 

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

 

4.         = The total weight of the examination is 150 points.

 

5.         = There are six (6) multiple choice and four (4) calculation problems. Work all cal= culation problems and 5 (five) multiple choice.&nbs= p; Show all work; partial credit will be given for correct work shown.<= o:p>

 

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

be located in the classroom.

 

7.         = Start:        &= nbsp;      1:30 p.m.

Stop:            &= nbsp;   2:45 p.m

 

 

 

&= nbsp;           &nbs= p; PROBLEM

 

&= nbsp;           &nbs= p;    POINTS

 

&= nbsp;           &nbs= p;   CREDIT

 

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

 

   &nbs= p;            &= nbsp;    30

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     7

 

   &nbs= p;            &= nbsp;    30

 

&= nbsp;

 

8

 

   &nbs= p;            &= nbsp;    30

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     9

 

   &nbs= p;            &= nbsp;    30

 

&= nbsp;

 

&= nbsp;           &nbs= p;        10

 

&= nbsp;           &nbs= p;        30

 

 

 

&= nbsp;           &nbs= p;    TOTAL

 

&= nbsp;           &nbs= p;       150

 

 

 

 

 

&= nbsp;          PERCENTA= GE

 

 

Note: All multiple choice problems below are wort= h 6 points each.

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT.

 

1. The quantity mgy is

 

            = (A) The kinetic energy of the object.

            = (B) The elastic potential energy of the object.

            = (C) The gravitational potential energy of the object.

            = (D) The work done on the object by a force.

            = (E) The power supplied to the object by the force.

 

 

2. An object is held at a heigh= t h above the ground. A second objec= t with four times the mass of the first is held at the same height. The potential energy of the second object compared to the first is

            = (A) One-forth as much.

            = (B) One-half as much.

            = (C) Twice as much.

            = (D) Four times as muc= h.

            = (E) Eight times as much.

 

 

  3. An acorn falls from a tree.

 

            = (A) Its kinetic energy increases and its gravitational potential energy decreases.

            = (B) Its kinetic energy increases and its gravitational potential energy increas= es.

            = (C) Its kinetic energy decreases and its gravitational potential energy increas= es.

            = (D) Its kinetic energy decreases and its gravitational potential energy decreas= es.

            = (E) Both kinetic and gravitational potential energies remain constant.

 

 

4. Two children are riding on a merry-go-round. = Child A is at a greater distance from the axis of rotation than child B. Which ch= ild has the larger angular displacement?

 

            = (A) Child A.

            = (B) Child B.

            = (C) They have the same zero angular displacement.

            = (D) They have the same non-zero angular displacement.

            = (E) There is not enough information given to answer the question.   

 

 

 

5. Identical forces act for the same time interv= al on two different masses. The change in momentum of the smaller mass is

 

      (A) Smaller than the change in momentum of the larger mass, but not zero.<= /o:p>

      (B) Larger= than the change in momentum of the larger mass.

      (C) Equal to the chan= ge in momentum of the larger mass.

      (D) Zero.<= o:p>

      (E) There = is not enough information to answer the question.

 

 

6. It is more difficult to start moving a heavy = box from rest than it is to keep pushing it with constant velocity, because

 

 

(A) The normal force is greater= when the box is at rest.

(B) m= s < mk=

      (C) Initially, t= he normal force is not perpendicular to the normal force.

(<= /span>D) = ms > m<= /span>k

      (E) m= s =3D mk

&nb= sp;

 

&nb= sp;

7. A 6.00-kg block slides down a frictionless incline making an angle of 60.0o with the horizontal. 

 

a. List all the for= ces acting on the block, and find the work done by each force when the block sl= ides 2.00 m (measured along the incline).

 

Gravitational force and normal forc= e

 

Ww =3D (6.00 kg)(9.81 m/s2)(2.00 m)cos(30.0o) =3D 102 J

 

Wn =3D 0 J

b.= What is the total work done on a block?

Wne= t =3D Ww + Wn =3D 102 J

&nb= sp;

&nb= sp;

`       c. W= hat is the speed of the block after it slid 2.00 m if it starts with an initial sp= eed of 1.50 m/s?

&nb= sp;

&nb= sp;

Wnet =3D Kf &= #8211; Ki =3D ½ mVf2 - ½ mVi2

 

½ mVf2= =3D Wnet + ½ mVi2

 

Vf2 =3D 2 Wnet /m + Vi2

 

Vf =3D (2 Wnet /m + Vi2) 1/2

 

        &= nbsp;   V f =3D 6.02 m/s

&nb= sp;

&nb= sp;

8.   &n= bsp; A 5.00 –kg o= bject moving to the right with  a sp= eed of 4.00 m/s collides head-on with a 10.0-kg object moving toward it with a spe= ed of 3.00 m/s. The 10.0-kg object stops dead after the collision. =

a.   &n= bsp;  What is the final velocity of the 5.00-kg object? Please specify both magniude and direction.

&= nbsp;

m1V1i + m2V2i =3D m1V1f + m2 V2f

 

V2f =3D 0

m1V1i + m2V2i =3D m1 V1f

 

V1f   =3D (m1V1i + m2V2i)/ m1 <= /span>

 

V1f   =3D ((5.00 kg)(4.00 m/s= ) + (10.0 kg)(-3.00 m/s))/(5.00 kg) =3D -2.00 m/s (af= ter collision the 5.00-kg object is moving with a speed of 2.00 m/s to the left= )

 

b.   &n= bsp; Is this collision = elastic or inelastic?

Ki =3D ½ m1V1i2 + ½ m2V2i2 =3D ½ (5.00 kg) (4.00 m/s)2 + ½ (10.0 kg) (3.00 m/= s)2 =3D 85.0 J

 

Kf =3D ½ m1<= /sub>V1f2 =3D ½ (5.00 kg) (2.00 m/s)2 =3D 10.0 J

 

Kinetic energy is not conserved, collision is inelastic.<= o:p>

 

&= nbsp;

9.     A compact disc rot= ates from rest to 500 rev/min in 5.50 s.   

a.      What is the angular acceleration of the disc assuming it is constant?

(500 rev/min)(2p rad/rev)(1 min/60s) =3D 52.4 rad/s

 

w =3D  <= /span>w0 + a t

 

a =3D (w -  w0) / t =3D ((52.4 rad/s) = – (0 rad/s))/(5.50 s) =3D 9.52 rad/s2

 

&= nbsp;

b.     How many revolutio= ns does it make in 5.50 s?

&= nbsp;

w2 =3D  w02= + 2 a (q - q0)

 

(q - q0) =3D (= w2 -  w02)/(2 a)

 

(q - q0) =3D (= (52.4 rad/s)2 -  (0 rad/s)2)/(2 (9.52 rad/s2)) =3D 144 rad =3D 23.0 rev=

 

 

c.      What is the centri= petal acceleration of the point on the rim of the disc (6.00 cm from the center) = at the moment the disc reaches angular velocity of 500 rev/min?

&= nbsp;

a= cp =3D w2 R =3D (52.4 rad/s)2(0.0600 m) =3D 165 m/s2

 <= /o:p>

  1. You= drop your physics textbook (of 4.00 kg mass) from the second floor window 7= .00 m above the ground.
    1. N= eglecting the air resistance, how fast do you expect it to move as it reaches t= he ground?

Ei =3D m g yi= + ½ mVi2   =3D m g yi

 

Ef =3D m g yf= +  ½ mVf2  =3D ½ mVf2  

 

Ei =3D Ef

m g yi =3D ½ mVf2  

 

g yi =3D ½ V= f2  

 

Vf  =3D (2 g yi)1= /2  

 

Vf  =3D (2 (9.81 m/s2)= (7.00 m))1/2  

 

Vf  =3D 11.7 m/s

 

    1. Y= ou measure the speed of the book just before it hits the ground and find= it to be 9.00 m/s. You correctly assume that the air resistance slows the book down.  How much wor= k is done on the book by air resistance?

 

Wf =3D Ef = 211; Ei =3D  ½ m Vf = 2  - m g yi

 

Wf =3D Ef = 211; Ei =3D  ½ (4.00 kg) (9.00 = m/s) 2  - (4.00 kg= ) (9.81 m/s2)(7.00 m) =3D -113 J

 

    1. W= hat is the average air resistance force that acts on the book during the fall?   

 

Wf =3D f d cos (180o) =3D - f d

f =3D - Wf /d =3D - (- 113 J)/(7.00 m) =3D 16.1 N

 

 

 

 

 

 

 

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= 2

PHYS 1111 Exam 2, Version 1

Summer 2007

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