MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_NextPart_01C8AE1C.BC2BE610" 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_01C8AE1C.BC2BE610 Content-Location: file:///C:/D9286902/Spring08.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__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;            =              April 2, 2008

Total Weight: 100 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 100 points.

 

5.         = There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice.  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 classroom.

 

7.         = Start:        &= nbsp;      6:00 p.m.

Stop:            &= nbsp;   7:15 p.m

 

 

 

&= nbsp;           &nbs= p; PROBLEM

 

&= nbsp;           &nbs= p;    POINTS

 

&= nbsp;           &nbs= p;   CREDIT

 

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

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     7

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

8

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     9

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

&= nbsp;           &nbs= p;        10

 

&= nbsp;           &nbs= p;        20

 

 

 

&= nbsp;           &nbs= p;    TOTAL

 

&= nbsp;           &nbs= p;       100

 

 

 

 

 

&= nbsp;          PERCENTA= GE

 

 

 

 

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

 

A ma= ss is attached to one end of a string. The other end is attached to a rigid support. The mass is released from rest at A and swings in a vertical arc to points B, C, and D, as shown below. Neglect air= resistance.
 

1. At what point does the mass have the least gravitational potential energy?<= o:p>

(A) A.

(B) B.

(C) C.

(D) D.

(E) None= of the given points.

 

2. At what point does the mass have the least kinetic energy?

(A) A.

(B) B.

(C) C.

(D) D.

(E) None= of the given points.

&nb= sp;

&nb= sp;

3. At what point does the mass have negative gravitational potential energy?

(A) A.

(B) B.

(C) C.

(D) D.

&nb= sp;

4.  In an inelastic colli= sion, if the momentum is conserved, then which of the following statements is true about kinetic energy?

 

(A) Kinetic energy is also cons= erved.

(B) Kinetic energy is gained.

(C) Kinetic energy is lost.

(D) Kinetic energy can either be gained or lost.

(E) None of the above is correc= t.

&n= bsp;

&n= bsp;

&n= bsp;

&n= bsp;

5. A per= son applies a constant force on an object of mass 20.0 kg that causes the objec= t to move horizontally at a constant speed of 0.200 m/s through a distance of 0.800 m along a rough surface. Wha= t is the net work done on the object?

&nb= sp;

(A) 160 J.

 

(B) 10.0 J.

 

(C) 16.0 J.

 

(D) 0 J.

 

(E) Cannot be determined without knowing the magnitude of the applied force.

 

 

 

 

6. A bal= l drops some distance and loses 30.0 J of gravitational potential energy. Do NOT ig= nore air resistance.  How much kine= tic energy did the ball gain?

&nb= sp;

(A) More than 30.0 J.

 

(B) Exactly 30.0 J.

 

(C) Less than 30.0 J.

 

(D) Exactly 0.0 J.

 

(E) Exactly 60.0 J.

 

&nb= sp;

&nb= sp;

7. A 100= kg warthog (Phaco= choerus aethiopics) squeals with delight as it slides from rest down the greased (frictionless) slope a= nd plane as pictured below.

a. How f= ast does the hog move as it just reaches the bottom of the slope?<= /p>

&nb= sp;

Ei =3D m g yi + &fra= c12; m Vi 2 =3D m g yi

 

Ef =3D m g yf + ½ m Vf<= /span> 2 =3D ½ m Vf= 2

 

Ei =3D Ef=

&nb= sp;

m g yi =  =3D ½ m Vf 2

 

g yi  =3D ½ V= f 2

 

Vf 2 =3D 2 g yi =

 

Vf&n= bsp; =3D 9.90 m/s=

&nb= sp;

&nb= sp;

b. As th= e hog continues along the plane, it encounters another hog of 150 kg mass, initia= lly at rest. The hogs collide and as a result of the collision the 150-kg hog a= cquires 5.20 m/s velocity to the right. How fast does the 100-kg hog move after the collision and in what direction? 

 <= /span>

m1 V1i  + m2 V2i  =3D m1 V1f  + m2 V2f 

m1 V1f  =3D m1 V1i  + m2 V2i - m= 2 V2f 

V1f  =3D (m1 V1i<= span style=3D'mso-spacerun:yes'>  + m2 V2i - m= 2 V2f )/m1

V1f  =3D ((100 kg)(9.90 m/s)  - (150 kg)(5.20 m/s))/(100 kg) =3D= 2.10 m/s, to the right

&nb= sp;

&nb= sp;

8.   &n= bsp; You are pushing a = 7.00 kg box along a rough ceiling with a force of 180 N directed at 35.0o  to the horizontal. The magnitude of= the friction force acting on the box is 15.0 N.

a.<= span style=3D'font:7.0pt "Times New Roman"'>      Calculate the work done by each for= ce on the box after you pushed it through 0.750 m. 

&nb= sp;

&nb= sp;

W =3D F = Dx cos (= q)

 

WA<= /sub> =3D A Dx cos(q) =3D (180 N) (0.750 m) cos(35.0o)  =3D 147 = J

 

Wn =3D n Dx cos(q) =3D n (0.750) cos(90.0o) =3D 0 J

 

Ww =3D w Dx cos(q) =3D (7.00 kg)(9.81 m/s2) (0.750 m) cos(90.0o)=3D 0 J

 

Wf =3D f Dx cos(q) =3D (15.0 N)(0.750 m) cos(180o)=3D - 11.3 J

 

&nb= sp;

b.<= span style=3D'font:7.0pt "Times New Roman"'>     Calculate the total work done on th= e box.

&nb= sp;

WNE= T =3D WA + Wn + Ww + Wf =3D 147 J + 0 J + 0 J - 11.3 J =3D 136 J

&nb= sp;

&nb= sp;

&nb= sp;

c.<= span style=3D'font:7.0pt "Times New Roman"'>      If initially the box was at rest, w= hat is the final speed of the box after you pushed it through 0.750 m?<= /span>

&nb= sp;

WNE= T =3D ½ m Vf 2 – ½ mV= i 2

&nb= sp;

Vi =3D 0 m/s

&nb= sp;

½= mVf 2 =3D WNET<= /sub>

&nb= sp;

Vf  =3D= √( 2WNET/m) =3D <= span style=3D'color:red'>6.23 m/s

 

9.   &n= bsp; The force shown in= the force-time diagram below acts on a 1.50 kg object. The force is directed al= ong the x-axis. Find

 

a.   &n= bsp;  The impulse of the force.

I =3D (2.00 N)(3.0= 0 s) + ½ (2.00 N)(2.00 s) =3D 8.00 N-s

 

b.   &n= bsp; The final velocity= of the object if it is initially at rest.

I <= /sub>=3D m Vf  = 211; mVi

&nb= sp;

Vi =3D 0 m/s

&nb= sp;

Vf    =3D I/m  =3D 5.33 m/s

c.   &n= bsp;  The final velocity of the object if it is initially moving along the x axis with a velocity of -2.00 m/s.

I <= /sub>=3D m Vf  = 211; mVi

&nb= sp;

Vi =3D -2.00 m/s

&nb= sp;

Vf    =3D (I + mV= i )/m  =3D 3.33 m/s

 

  1. The launching mechanism of a t= oy gun consists of a spring of unknown spring constant, as shown below. If the spring is compressed a distance of 0.120 m and the gun is fired vertic= ally as shown, the gun launches a 20.0-g projectile from rest to a maximum height of 20.0 m above the starting point. Neglecting all resistive forces, determine

&nb= sp;

    1. The spring constant,

&nb= sp;

&nb= sp;

Ei =3D m g yi + &fra= c12; m Vi 2 + ½ k xi 2 =3D ½ k xi 2

 

Ef =3D m g yf + ½ m Vf<= /span> 2+ ½ k xf 2 =3D m g yf

 

Ei =3D Ef=

&nb= sp;

½ k xi 2 =3D m g yf  

 

k =3D (2 m g yf )/xi 2   

&nb= sp;

k <= /sub>=3D 545 N/m

&nb= sp;

    1. The speed of the projectile a= s it moves through the equilibrium position of the spring. 

 

Ei =3D m g yi + &fra= c12; m Vi 2 + ½ k xi 2 =3D ½ k xi 2

 

Ef =3D m g yf + ½ m Vf<= /span> 2+ ½ k xf 2 =3D m g yf + ½ m <= span class=3DSpellE>Vf 2

 

Ei =3D Ef=

 

½ k xi 2 =3D m g yf + ½ m Vf 2

 

k xi 2 =3D 2 m g yf + m Vf 2=

 

Vf 2 =3D (k xi 2 - 2 m g yf )<= /span>/m

 

Vf  =3D√ (k xi 2 - 2 m g yf )/m

&nb= sp;

Vf  =3D= 19.7 m/s=

&nb= sp;

&nb= sp;

&nb= sp;

&nb= sp;

 

 

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8

PHYS 1111 Exam 2, Version 1

Spring 2008

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