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Physics 215 Fall 2008 Exam 1 Version C (707777)
Instructions
Be sure to answer every question. Follow the rules shown on the screen for filling in the Scantron form. Each problem is worth
10% of the exam. When you are finished, check with Dr. Mike or his TA to be sure you have finished the scantron correctly.
1. Algebra and Constant Acceleration [1021151]
The x-component of the first equation for constant acceleration, can be solved for time. Which of these is the right equation?
*5*c*
*10*a*
*7*b*
*2*e* There is no way to know.
*2*d*
Solution or Explanation
If we start with
and rearrange, we will find that
which is a quadratic equation that has the solution
.
2. Vectors [1021207]
Given that
*5*b*
*2*e*
*10*a*
*3*c*
*3*d*
, what is the unit vector that points in the same direction as
?
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Solution or Explanation
The way to find a unit vector that has the same direction as another vector is to divide that vector by its magnitude. The
magnitude of is
. Thus, the new unit vector is
.
3. Mathematics in Physics [711489]
There are 1.057 quarts in a liter and 4 quarts in a gallon. A barrel equals 42 gallons. How many liters are there in 5.5 barrrel(s)?
*4*d*0.87 liters
*1*e*61.04 liters
*10*a*874.17 liters
*2*b*0.06 liters
*2*c*8.74 x 105 liters
Solution or Explanation
We begin with 5.5 barrels. This is the same as
(5.5 barrels)(42 gallons / 1 barrel)(4 quarts / 1 gallon) (1 liter / 1.057 quarts) = 874.17 liters.
4. Constant Acceleration [711502]
At t = 0, a stone is dropped from a cliff above a lake; 1.8 seconds later another stone is thrown downward from the same point
with an initial speed of 33 m/s. Both stones hit the water at the same instant. Find the height of the cliff.
*3*b*15.89 m
*1*e*3.97 m
*1*d*55.50 m
*10*a*39.45 m
*5*c*31.78 m
Solution or Explanation
Let us call T0 the time when the first stone was thrown. Then the second stone is thrown at a time T0 + 1.8 s. Let us also call
the time at which both stones hit the water Tf. The equation to use for both stones is
y = y0 + v0yt + 1/2 ay t2.
For the first stone, we have
tI = Tf - T0
and for the second,
tII = Tf - (T0 + 1.8 s).
Thus, tII = tI - 1.8 s.
Remember this. I will call it equation (1). We will need it later.
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Now, the height of the cliff is h = y0 - y for both stones. Thus,
-h = v0yI tI + 1/2 ayI tI2 = v0yII tII + 1/2 ayII tII2.
For the first stone, v0yI = 0 and ayI = -g. So that,
-h = - 1/2 g tI2.
This, I will call equation (2). For the second stone, v0yII = 33 m/s and ayII = -g. So that,
-h = (33 m/s) tII - 1/2 g tII2.
This is equation (3). Equations (1), (2) and (3) are three equations with three unknown variables tI, tII and h. You should be able
to solve this for any of the three unknown variables.
Here is how you do it. We can eliminate tII by putting equation (1) into equation (3) to get
-h = (33 m/s) (tI - 1.8 s) - 1/2 g (tI - 1.8 s)2.
We can also eliminate tI be solving equation (2) for tI and putting this into the last equation. Thus, we are left with
-h = (33 m/s) ( (2h/g) - 1.8 s) - 1/2 g (
(2h/g) - 1.8 s)2.
Now all we need to do is solve for h. After using some algebra, you will find that
(2h/g) = [(33 m/s) (1.8 s) - 1/2 g (1.8 s)2] / [(33 m/s) + g (1.8 s)]
which gives us
h = 39.45 m.
5. Method of Susbtitution [1021204]
Given that both of the following equations are both true. Which of the answers is a possible solution for .
*5*c* 3.12
*2*d* 4.52
*10*a* 3
*2*e* 3.22
*5*b* 4
Solution or Explanation
Start with the first equation
and plug this into the second equation
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.
We can now use the quadratic equation to solve for
.
6. Algebra, Trigonometry and Geometry [711497]
Calculate the following with your calculator. Given A=11, B=44, C=4, and D=4. What is the value of B-(2D+3/16A2)/C?
*5*b*42.00
*10*a*36.33
*4*d*9.00
*1*e*41.67
*4*c*3.33
Solution or Explanation
Remember to use the PEMDAS order (parentheses, exponents, multiplication/division, addition/subtraction) going from left to
right. If we had put the parentheses in explicitly, we would have had
B-(((2D)+((3/16)(A2)))/C) = 36.33.
7. Constant Acceleration [716875]
A particle moves in the xy plane with constant acceleration. At t = 2 s, the particle is at x = 286 m, y = 171 m, and has velocity at
time zero of v = 7 m/s i + 9 m/s j. The acceleration is given by the vector a = 10 m/s2 i + 5 m/s2 j. Find the position vector at time
zero.
*7*b*(320 m) i + (199 m) j
*4*c*(289.75 m)
*7*e*(252 m) i + (199 m) j
*10*a*(252 m) i + (143 m) j
*2*d*(376.83 m)
Solution or Explanation
The equation we use is
r = r0 + v0 (t - t0) + 1/2 a (t - t0).
Solving the components for r0, we get
x0 = x - vx0 (t - t0) - 1/2 ax (t - t0) = (286 m) - (7 m/s) (2 s) - 1/2 (10 m/s2) (2 s)2 = 252 m.
Similary,
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y0 = y - vy0 (t - t0) - 1/2 ay (t - t0) = (171 m) - (9 m/s) (2 s) - 1/2 (5 m/s2) (2 s)2 = 143 m.
8. Circular Motion [713781]
Bailey D. Wonderdog is chasing her tail at a constant speed of v = 6 m/s. The circle she moves in has a diameter of d = 85 cm.
What is the magnitude and direction of her acceleration?
*7*b*0.85 m/s2 toward the center of the circle
*10*a*84.71 m/s2 toward the center of the circle
*2*e*0.85 m/s2 opposite to the direction of her motion
*2*d*0.85 m/s2 in the direction of her motion
*5*c*84.71 m/s2 in the direction of her motion
Solution or Explanation
The acceleration of any object going in a circle is the centripetal acceleration, which has a magnitude of
a = v2 / r = v2 / (d/2) = [6 m/s]2 / ([85 cm]/2) = [6 m/s]2 / ([85 cm]/2) (100 cm / 1 m) = 84.71 m/s2.
9. Vectors [771679]
Find the cross product of A = 9 cos(30o) i and B = 3 sin(52o)j.
*5*c*26.41 i
*10*a*18.43 k
*1*e*10.16
*3*d*18.43
*5*b*27.00 j
Solution or Explanation
The magnitude of the cross product is |A X B| = A B cos , where is the angle between the vectors. In this case = 90o and
we have |A X B| = [9 cos(30o)] [3 sin(52o)] cos 90o = 18.43. The direction is given by the right-hand rule. Point your fingers to
the right and bend their tips upward. Your thumb will be point toward you. This is the positive z-direction (k).
10. Preview of Calculus and Velocity [711490]
Figure 2-35 shows the x-component of position of a car plotted as a function of time.
Figure 2.35.
At which time t0 to t7 is the x-component of velocity zero?
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*4*c*t1
*4*e*t7
*10*a*t2
*4*b*t0
*4*d*t3
Solution or Explanation
The velocity is zero when the slope is zero. This only happens at t2.
Assignment Details
Name (AID): Physics 215 Fall 2008 Exam 1 Version C (707777)
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