Download Professor Epstein SP07

MIDTERM
Page 1 of 5
NAME__________KEY________________RECITATION INST________ KEY _____________
GRADE_______ KEY _______________
Physics 111
Professor Epstein
48 minutes
2:30 section
SP07
May 2, 2007
Midterm Equations
Motion with Constant Acceleration
1 2
xx
v
t a
t
0
0
2
(v  v)
x  0
t
2
v  v0  at
v2  v02  2a(x  x0 )
Quadratic Equation
2
If ax bxc0, then
2

b b

4
a
c
x
2
a
Newton’s 2nd Law
F
Fm
a
n
e
t

Means
Fx max and Fy may
Force
fs sFN
fk kFN
Gmm
1 2
r2
G6.671011Nm2 / kg2
wmg
F
g 9.80 m/s2, near surface of Earth
Conditions for Equilibrium
Fnet 
F0
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Page 2 of 5
NAME_________ KEY _______________RECITATION INST_______ KEY __________________
Physics 111
Professor Epstein
2:30 section
SP07
May 2, 2007
48 minutes
Questions 1 through 10 are multiple choice, 5 points each.
1. You step on a hot stone path with your bare feet. A nerve impulse, generated in your foot, travels
through your nervous system at an average speed of 110 m/s. How much time does it take for the
impulse, which travels a distance of 1.71 m, to reach your brain? (5 points)
(a)
64.3 s
(b)
7.73 x 10-3 s
**(c) 1.55 x 10-2 s
(d) 1.60 x 10-2 s
2. A baseball is thrown straight (horizontally) off the edge of a cliff that is 45 m high. A tourist observed that the
point of impact is 99 m from the base of the cliff. How fast was the baseball traveling when it went over the
cliff? (5 points)
(a) 10.0 m/s
**(b) 32.7 m/s
(c) 14.9 m/s
(d) 4.59 m/s
3. A park ranger being chased by an angry wolf is running in a straight line towards her car at a speed of
3.25 m/s. The car is a distance d away. The wolf is 30.0 m behind the park ranger and running at 6.50
m/s. The park ranger reaches her car safely. What is the maximum possible value for d? (5 points)
(a)
60.0 m
(b)
20.0 m
(c)
40.0 m
**(d) 30.0 m
4. The left ventricle of the heart accelerates blood from rest to a velocity of 0.250 m/s. The
displacement of the blood during the acceleration is 0.0215 m. The acceleration of the blood is? (5
points)
**(a) 1.45 m/s2
(b) 0.116 m/s2
(c) 0.581 m/s2
(d) 2.91 m/s2
5. The acceleration of a projectile equals zero when the projectile reaches the top of its trajectory. (5
points)
(a) True
** (b) False
MIDTERM
Page 3 of 5
NAME________ KEY _______________RECITATION INST______ KEY _________________
Physics 111
2:30 section
Professor Epstein
SP07
May 2, 2007
48 minutes
6. A baseball is through upward at a 45° angle and then falls back to the ground. Does the baseball, in
any part of its trajectory, ever have a speed that is one-half of its initial value? (5 points)
(a) Yes
**(b) No
7. A trucks and a car are traveling in the same direction, although the car is 245 m behind the truck.
The speed of the car is 34.3 m/s and the speed of the truck is 19.2 m/s. How much time does it take for
the car to catch up with the truck? (5 points)
(a) 17.8 s
(b) 33.6 s
(c) 12.6 s
**(d) 16.2 s
8. A baseball is hit upward and travels along a parabolic arc before it strikes the ground. Which one
of the following statements is necessarily true? (5 points)
(a) The acceleration of the ball decreases as the ball moves upward.
(b) The velocity of the ball is zero m/s when the ball is at the highest point in the arc.
**(c) The x-component of the velocity of the ball is the same throughout the ball's flight.
(d) The velocity of the ball is a maximum when the ball is at the highest point in the arc.
9. What is the magnitude of the gravitational force acting on a 79.5-kg student due to a 60.0-kg student
sitting 2.25 m away in the lecture hall? (5 points)
(a) 3.14  109 N **(b) 6.29  108 N
(c) 7.91  1010 N
(d) 1.41  107 N
10. The drawing at right shows a wire tooth brace
used by orthodondists. The topmost tooth is
protruding slightly, and the tension in the wire
exerts two forces T’ and T each at an angle of
16.0° below the dashed line to the left and right
respectively. If T’ and T each have the same
magnitude of 23.2 N, what is the magnitude of the
net force exerted on the tooth by these forces? (5
points)
(a)
46.2 N **(b) 12.8 N
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(c)
44.6 N
(d) 22.3 N
Page 4 of 5
NAME_________ KEY __________________RECITATION INST_____ KEY ______________
Physics 111
Professor Epstein
48 minutes
2:30 section
SP07
May 2, 2007
PART II: "SHOW WORK" PROBLEMS
(25 points each question, 50 points total)
Show all work and equations to receive full credit. Partial credit is given.
11. On a windless day an airplane travels with a speed of 900 km/h with respect to the air at an altitude
of 7.00 km. The pilot then has the airplane climb at an angle of 35.0 degrees to the horizontal while
maintaining the same constant speed. The pilot releases a package when the plane reaches an altitude of
8.00 km (while still maintaining the same speed and at the same 35.0 degree angle). Assume that the air
resistance is zero. Note points for each portion are not the same.
(a) What is the magnitude of the horizontal velocity of the package after release? (5 points)
voh = cos35.0o x 900km/h = 0.819 x 900km/h = 737.2km/h = 204.79m/s= 205m/s
Direction of horizontal velocity is same direction of airplane’s horizontal direction.
(b) What is the direction (1 point) and magnitude magnitude (4 points) of the initial vertical velocity of
the package after release? (total 5 points)
Initial direction of vertical velocity is upward.
vov = sin35.0o x 900km/h = 0.574 x 900km/h = 516.2km/h = 143.4m/s=143 m/s
(c) How long does it take for the package to reach the ground after its release? (5 points)
h = vot + ½ at2
-8000 m = 143.4m/s t - ½ x 9.8m/s2 x t2
t = 57.61s = 57.6s
(d) How far does the package travel in the horizontal direction before striking the ground? (5 points)
voh x t = 204.79m/s x 57.61s = 11797 m = 11800 m
(e) What is the speed of the package the instant before its impact with the ground? (5 points)
vv = vov +at = 143.4m/s - 9.8m/s2 x 57.61s = - 421.14m/s = - 421m/s
Because there is nonzero horizontal speed, total speed is vector sum of horizontal and vertical speeds.
Speed = ((- 421.14m/s )2 + (voh)2)1/2 = 468.3m/s = 468m/s
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Page 5 of 5
NAME__________ KEY __________________RECITATION INST_______ KEY ______________
Physics 111
2:30 section
Professor Epstein
SP07
48 minutes
May 2, 2007
12. A 6.00 kg box is stationary on an incline plane that is at an angle of 30.0 degrees to the horizontal.
Note points for each portion are not the same.
(a) Draw the free body diagram showing all forces acting on the box. (5 points)
N
f
mg
(b) Calculate the magnitude of the frictional force on the box. (4 points). In what direction does the
frictional force point?(2 points) (total 6 points)
f = mg sin30.0o = 6.00kg x 9.8m/s2 x 0.5 = 29.4 N
upward with an angle of 30.0 degree to the horizontal
(c) Calculate the minimum value for the static coefficient of friction for the box on the incline plane.. (6
points)
f = 29.4N = N = mg cos30.0o =  6.00kg x 9.8m/s2 x 0.866
 = 0.577
(d) The box is pushed to give it an initial velocity of 10.0 m/s going up the incline plane. During the
ascent of the box, in what direction does the frictional force between the box and the incline plane
point? (3 points)
downward with an angle of 30.0 degree to the horizontal
(e) Assume the 30.0 degree incline plane in part (d) has a kinetic coefficient of friction of 0.200. What
is the direction (2 points) and magnitude (3 points) of the acceleration along the incline plane of the
box after it is pushed to achieve an initial velocity of 10.0 m/s going up the incline plane. (5 points)
Direction of acceleration is down the incline plane (or equivalent statement)
N – mgsin30.0o = – mgcos30.0o – mgsin30.0o
a = –( cos30.0o + sin30.0o )g = – (0.2 x 0.866 + 0.5) x 9.8 m/s2 = - 6.60 m/s2