TIME OF COMPLETION_______________ NAME    SOLUTION

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1111, Exam 1                                                                                                      Section 1

Version 1                                                                                                                June 12, 2003

Total Weight: 100 points

 

5.               Check your examination for completeness prior to starting.  There are a total of ten (10) problems on eight (8) pages.

 

5.               Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet  (provided by your instructor).

 

5.               You will have 80 minutes to complete the examination.

 

5.               The total weight of the examination is 100 points.

 

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

 

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

be located in the classroom.

 

7.         Start:               11:00 a.m.

Stop:                12:20 p.m

 

 

 

              PROBLEM

 

                 POINTS

 

                CREDIT

 

                     1-6

 

                     30

 

 

 

                      7

 

                     15

 

 

 

8

 

                     15

 

 

 

                      9

 

                     20

 

 

 

                     10

 

                     20

 

 

 

                 TOTAL

 

                    100

 

 

 

 

 

           PERCENTAGE

 

 

 

     

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

 

  1. Jeff throws a ball straight up. For which situation is the vertical velocity negative?

 

a.      On the way up.

(5)

b.     At the top.

 

c.      On the way back down.

 

 

d.     None of the above.

 

 

2.Take a look at the position vs. time graph representing one-dimensional motion of an object. At what point is the instantaneous velocity of the object negative? 

 

a.      A.

 

b.     B.

 

c.      C. (The slope is negative at point C)

 

d.     None of these.

 

3.  Which one of these expressions is NOT valid?

a.      (1.00 m ) / (2.00 s).

(5)

b.     (3.50 m/s) – (2.67 m/s).

 

c.      (4.56 N) (10.5 m).

 

d.     (2.45 m) + (100 m/s2).

 

 

4. The speed of a body traveling in a straight line with a constant positive acceleration increases linearly with:

 

a.      Time.  V = V0 + a t

(5)

b.     Displacement.

 

c.      Distance.

 

d.     Time squared.

 

 

 

 

  5.   Why does it take more force to start walking than to continue walking?

 

a.      Is does not.

(5)

b.     Because kinetic friction is less that static friction.

 

c.      Because air drag at walking speeds is much less than the needed internal force, ma.

 

d.     People get tired walking and prefer to stand still.

 

 

 

6.Is it possible to devise a technique to push on a table without it pushing back on you?

 

a.      Yes, out in space.

(5)

b.     Yes, if someone else also pushes on it.

 

c.      A table never pushes in the first place.

 

d.     No.

 

 

 

7. A Jaguar in an auto accident in England in 1960 left the longest recorded skid marks on a public road: an incredible 290-m long. Assuming a constant acceleration of –3.90 m/s2, calculate: 

 

(15)

a.      The Jag’s speed when the brakes locked.

 

    V2 = V02 + 2 a ( x- x0)                                        V = 0 m/s;  x0 = 0 m; x = 290 m

                              

 

    (0 m/s)2 = V02 + 2 (-3.90 m/s2)( 290 m - 0 m)                                                                  

 

    V02 = 2262 m2/s2

   

    V0 = 47.6 m/s

 

b.     How much time did it take to come to the full stop?

 

    V = V0 + a t

 

    t = (V - V0)/a

 

    t = (0 m/s – 47.6 m/s) / (-3.90 m/s2) = 12.2 s

 

 

c.      What was the average speed of the Jaguar during that time?

(Hint: use the definition of average acceleration here.)

 

     Vaverage = (x - x0)/(t - t0) = (290 m)/(12.2 s) = 23.8 m/s

 

 

8.  The cat named Feynman is prowling its neighborhood. It travels 24.5 m at the direction of 25o to the tall oak tree, sits there for some time, and then runs after a bird traveling another 56.0 m at the direction of –60o. What is the magnitude and direction of Feynman’s total displacement?

(All directions are given from the positive x-axis of the conventional rectangular coordinate system.)

 

(15)

 

  A: 24.5 m @ 25o

  B: 56.0 m @ -60.0o

 

             Ax = A cos(qA) = (24.5 m) cos(25o) = 22.2 m

            Ay = A sin(qA) = (24.5 m) sin(25o) = 10.4 m

         

            Bx = B cos(qB) = (56.0 m) cos(-60.0o) =  28.0 m

            By = B sin(qB) = (56.0 m) sin(-60.0o) = -48.5 m

 

           Rx = Ax - Bx  = 22.2 + 28.0 m = 50.2 m

           Ry = Ay - By = 10.4 m – 48.5 m = -38.2 m

 

           R = (Rx2 + Ry2)1/2 = ( (50.2 m)2 + (-38.2 m)2)1/2 = 63.1 m

qR  = tan-1 (Ry /Rx) = tan-1 (-38.2 m /50.2 m) = -37.2o

 

 

9. The cat named Feynman sits on a window sill. Here comes a bird, and Feynman is off to catch it.

It jumps at a speed of 2.00 m/s making 40o angle with the horizontal, misses the bird, and lands with a meow 1.50 s later. 

 

a.      Find the components of Feynman’s initial velocity.

 

V0: 2.00 m/s @ 40.0o

 

                        Vx0 = V0 cos(q0) = 2.00m/s cos(40.0o) = 1.53 m/s

                        Vy0 = V0 sin(q0) = 2.00m/s sin(40.0o) = 1.29 m/s

 

b.     How far in horizontal direction does Feynman travel?

 

                      x = x0 + Vx0 (t) = 0 m + (1.53 m/s) (1.50 s) = 2.30 m

(20)

 

c.      Find how high the window sill is.

 

y = y0 + V0yt – ½ g t2

 

(0 m) = y0 + (1.29 m/s)(1.50 s) – ½ (9.80 m/s2) (1.50 s)2

 

y0  = 9.09 m

 

d.     With what speed does Feynman lands?

 

Vx  = V0x  = 1.53 m/s

 

Vy  = V0y – gt = (1.29 m/s) – (9.80 m/s2) (1.50 s) = -13.4 m/s

 

                        V = sqrt (Vx2  + Vy2) = 13.5 m/s

 

10.A frustrated parent of mass 70.0 kg is dragging a child of mass 20.0 kg having a temper tantrum across level but rough ground. The parent exerts a force of magnitude 80.0 N on the child; the force is upward and forward at the angle of 45o to the horizontal. The parent and child are moving with a constant speed of 1.50 m/s homeward from a video arcade.

 

 

 

a.Draw a free body diagram indicating all forces on the child.

 

 

(20)

 

 

 

            Ax = A cos(45.0o) = (80.0 N) cos(45.0o) = 56.6 N

            Ay = A sin(45.0o) = (80.0 N) sin(45.0o) = 56.6 N

 

         

            nx = n cos(90.0o) = 0

            ny = n sin(90.0o) = n

 

 fx = f cos(180o) = -f

            fy = f sin(180o) = 0

 

            wx = w cos(270o) = 0 

            wy = w sin(270o) = – (20.0 kg) (9.80 m/s2) = -196 N

 

          

 

b. Calculate the frictional force of the ground on the child.

 

S Fx = m ax         

           

            56.6 N –f = (20.0 kg) (0 m/s2)

 

f = 56.6 N

 

c.Calculate the normal force of the surface on the child.

 

S Fy = m ay         

           

            56.6 N  + n – 196 N = (20.0 kg) (0 m/s2)

 

            n = 139 N

 

d.Find the coefficient of friction for the child on the ground.

 

            f = m n

 

            m = f/n = (56.6 N)/ (139 N) = 0.406

 

e. Is it kinetic or static friction?

 

            Kinetic: the child is in motion.