CSULA
I)
(SOLUTION)
MIDTERM
PHYSICS 211
Fall 2007
CONCEPTUAL QUESTIONS
1. Consider the following controls in an automobile: gas pedal, brakes, and steering wheel. The
controls in this list that cause an acceleration of the car are:
(e) All three controls
Gas pedal causes the car to speed up. The brakes cause the car to slow down. The steering wheel
changes the direction of the velocity vector.
2.
If vector A is added to vector B which two conditions must be satisfied in order for the resultant
vector to be equal zero?
(a) A and B are parallel and in opposite direction
(b) A and B have the same magnitude.
3.
After a ball is thrown upward and is in the air, its acceleration
(e) Remains the same
For the entire time interval that the ball is in the air, the acceleration is that due to gravity.
4.
You push an object, initially at rest, across a frictionless floor with a constant force for a time interval t
resulting in a final speed of v for the object. You repeat the experiment, but with a force that is twice as
large. What interval is now required to reach the same final speed v?
(d) t /2
With twice the force, the object will experience twice the acceleration. Because the force is
constant, the acceleration is constant and the speed of the object (starting from the rest) is given by
v = at. With twice acceleration the object will arrive at speed v at half the time.
II) Solve the following problems
1.
(10 pts). A person walks first at constant speed of 5.00 m/s along a straight line from point A to point B
and then back along the line from B to A at a constant speed of 3.00 m/s. Find the average speed over the
entire trip.
Let d represent the distance between A and B. Let t1 be the time for which the walker has the higher
d
speed in 5.00 m s  .
t1
d
Let t2 represent the longer time for the return trip in 3.00 m s   .
t2
d
d
Then the times are t1 
and t2 
.
 5.00 m s
 3.00 m s
The average speed is:
v
v
Totaldistance

Totaltim e

2 15.0 m
2
2
s
8.00 m s

d d
 3.00dm
d
 5.00 m s


s
2d
 8.00 m s d
15.0 m
2
s2

3.75 m s
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2. (10 pts). A ball is dropped from rest from a height h above the ground. How long does the ball take to
reach a height h/2?
Position at any time for ball:
y f  yi  vi t  1 2 gt 2   1 2 gt 2
Replacing: y f   1 2 h 
h
g
 1 2 h   1 2 gt 2  t 
3. (10 pts). Given the following displacements with rectangular coordinates:
(-3, 2) m (-5, 3) m and (6, 1) m
A) Obtain a unit-vector expression for the resultant displacement of the three vectors.

d   3, 2m   5, 3m  (6,1)m  (2, 6)m .

In unit-vector notation: d   2iˆ  6 ˆj m


B) Determine the magnitude and direction of the resultant displacement.
Magnitude:
Direction:
d  22  62 m  40m  2 10m
6
  tan 1    72  II Quadrant. (  = 108o)
2
4. (10 pts). A rock is thrown upward from the level ground in such way that the maximum height h of its
flight is equal to the horizontal range R. At what angle  was the rock thrown?
vi2  sin 2 i
v2 sin2  i
; R
.
Since R = h and sin(2) = 2sincos 
h i
g
2g
vi2 sin2  i 2vi2 sin  icos i

2g
g
sin  i
 tan  i  4
cos i
 i  76.0
5. (20 pts). Boxes A and B (masses of 10 kg and 12 kg respectively) are connected by a cord (mass
neglected). Boxes are resting on a frictionless table. The force applied FP = 40.0 N.
A) Sketch a free-body y diagram for each box (show motion direction as well as forces acting
ON each box and label them correctly). (4 pts).
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B) Using second Newton’s law, write the equation of motion for mA. (3 pts).
and
ΣFx = FP – FT = mAa
FAN  m A g
C) Using second Newton’s law, write the equation of motion for mB. (3 pts).
and
ΣFx = FT = mBa
FBN  mB g
D) Using parts B and C, find the acceleration of each box. (5 pts).
FP – FT = mAa  40 N – FT = 10a
FT = mBa 
FT = 12a
----------------------------40 N = 22a  a = 40/22 m/s2 = 1.82 m/s2
E) Using parts B and C, find the tension in the cord connecting the boxes. (5 pts).
FT = mBa  FT = 12(1.82) N = 21.8 N
6. (20 pts). The systems shown in the figure are in equilibrium. If the spring scales are calibrated in newtons
(N). What do the spring scales read? (Assume frictionless, and neglect the masses of the pulleys)
F  n  T  m g  0
Take the component along the incline
nx  Tx  m gx  0 
0  T  m g sin 30.0  0
T  m g sin 30.0 
m g 5.00 9.80

2
2
 24.5 N .
T  m g  m a 0
Isolate either mass
T  mg.
The scale reads the tension T. 


T  m g  5.00 kg 9.80 m s2  49.0 N .
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