Practice Exam I: Physics 141, Chapters 1-4

1. False This describes constant speed but not constant velocity which must also include the requirement of constant direction.
              Moving equal distances in equal intervals of time around a circular race track (for example) would not be constant velocity
              (the direction is constantly changing).

2. False     "distance traveled" applies to one specific kind of vector: displacement.

3. b             vf = vo + at = 0 + (9.8 m/sec2)(5 sec)

4. d             DELTAy = vot + ½at2 = 0 + ½(9.8 m/sec2)(10 sec)2

5. c             DELTA x = xf - xo = 120 miles - 40 miles

6. B             (A had the headstart!)

7. c             120 mi /4 hour

8. c             Notice line is straight (constant slope) here. ~30 mi /36 min or 10 mi /0.2 hour

9. A

10. c

11. d         It slows down after tB but continues to move with positive speed (i.e., forward) until tD (where it has slowed to zero...and then
                 starts going backwards!).

12. d         Look for where the slopes are parallel!

13. A

14. c         vf2 = vo2 + 2ax   ->  0 = (29 m/sec)2 + 2a(58m) for a , and then use F=ma to find the force.

15. e         This is the sort of question I like to ask...what if the mass doubled? What if the distance halved?
                 What if we increased the angle? Such questions test not whether you can just plug into an equation,
                 but understand the relationship between the variables.

Here, vf2 = vo2 + 2ax     ->     DELTA x = vo2/2a. Notice if we replace vo with 2vo the square means that DELTA x would become (22) times as great. PROBLEMS
  1. DELTA v = vf -vo = 300 m/sec - 400 m/sec = -100 m/sec


    2. vavg = (400 m/sec + 300 m/sec)/2 = 350 m/sec

    t = DELTA x / vavg = (0.10 m)/350 m/sec = 2.86´ 10-4 sec = 0.000286 sec

    a = DELTA v/ t = (-100 m/sec)/ 2.86´ 10-4 sec = -350000 m/sec2

    F = ma = (0.006 kg)(-3500000 m/sec2) = -2100 N

    Assuming a uniformly solid wall (so there is a constant deceleration or constant stopping force)
    to bring the bullet to rest (vf =0) means vf2 = vo2 + 2 a DELTAx
    0 = (400 m/sec)2 + 2(-350000 m/sec2) DELTA x

    or DELTA x= (-16000 m2/sec2) / (-2´ x 350000 m/sec2) = 0.23 m = 23 cm

  2. Vertically: Dy = vot + ½at2

    3. 75 m = 0 + ½(9.8 m/sec2) t 2
    so the time to drop 75 m is t 2 = 2(75 m)/(9.8 m/sec2) = 15.3 sec2

    t = 3.9 sec
     

  3. Horizontally: DELTA x = vot + ½at2 = (115 m/sec)( 3.9 sec ) + 0 = 449 m