ID CODE: A Physics 201 Midterm Exam 1 Oct. 3 , 2011

Physics 201 Midterm Exam 1
Oct. 3 rd , 2011
ID CODE: A
Name: .............Yibin Pan................................... Student ID: .........................
Section: .........................
TA (please circle):
Nick Brewer Todd Garon Zhen Liu Sean O’Connor Nicole Vassh
Instructions:
1. Don’t forget to write down your name, student ID#, and section number. You need do
this on (this page of) your test book and on your Scantron sheet as well.
2. Answer all multiple choice questions in this test book by indicating the best answer
among choices. You must do this both on your test book and on your Scantron sheet.
Follow instructions on the Scantron sheet on how to mark valid answers.
3. When you finish, you need to turn in both this test book and the Scantron sheet.
4: Use the blank side of question pages as additional draft spaces. An extra blank sheet
is provided at the end of the test book.
5: Only one answer is allowed per problem/question. All problems have equal weight.
Constants: g=9.8 m/s 2
(Note: Neglect air friction for all projectile motion).
Please be very careful with the first question even though the answer will not
count towards your grade:
1. ENTER THE ID CODE ABOVE IN THE UPPER RIGHT CORNER
B. ID Code B
C. ID Code C
D. ID Code D
E. ID Code E

2. A rectangle has a length of 1.323 m and a width of 4.16 m. Using significant figure
rules, what is the area of this rectangle?
a. 5.50368 m 2
b. 5.5037 m 2
c. 5.504 m 2
d. 5.50 m 2
e. 5.5 m 2
3. The position of a particle moving along the x axis is given by x = (21 + 22t-6.0t 2 ) m
where t is in s. What is the average velocity during the time interval t =1.0 s to t= 3.0
s?
a. −6.0 m/s
Δx=x(3.0s) – x(1.0s) = 22*(3-1) – 6.0*(3
b. −4.0 m/s
c. −2.0 m/s
d. −8.0 m/s
e. 8.0 m/s
2 -1 2 ) = -4 m
vavg = Δx/Δt = -2.0m/s
4. An automobile moving along a straight track changes its velocity at a constant
acceleration from 40 m/s to 80 m/s in a distance of 200 m. What is the acceleration of
the vehicle during this time?
a. 8.0 m/s 2
b. 9.6 m/s 2
c. 12 m/s 2
d. 6.9 m/s 2
e. 0.20 m/s 2
vf 2 =vi 2 +2aΔx 2 � a = (vf 2 -vi 2 )/(2Δx 2 )= 12 m/s 2
5. A car travels around an oval racetrack at constant speed. The car has a non-zero
acceleration
a. at all points except B and D.
b. at all points except A and C.
c. at all points except A, B, C, and D.
d. everywhere, including points A, B, C, and D.
e. nowhere, because it is traveling at constant speed.
per sig. figure rule, result of multiplication takes the smallest
number of sig. figures among operands.
In this case, the smallest number of sig. figures is 3 (of 4.16)
Direction keeps changing along the path � a!=0.

6. A rock is thrown downward from an unknown height above the ground with an initial
speed of 10 m/s. It strikes the ground 3.0 s later. Determine the initial height of the
rock above the ground.
a. 44 m
b. 14 m
c. 74 m
d. 30 m
e. 60 m
7. The position of an object (moving to the right) at equal time intervals is shown
below:
Which graph below correctly represents position versus time for this object?
8. At t=0.0s, a particle is at position (x,y) = (3.0,2.0) m, at t=1.0, 2.0, and 3.0s, the
particle is at (x,y)= (4.0,5.0), (3.0,2.0), (4.0,4.0) m, respectively. What is the
particle’s displacement during the time interval between 0.0s and 3.0s?
a. (1.0i + 2.0j )m
b. 1.7 m
c. 8.1 m
d. (4.0i + 4.0j )m
e. none of above
9. The speed of an object is given by v=5.00t 2 Displacement is determined by initial and final position, all middle
steps are irrelevant.
ri= (3.0i + 2.0j ), ri= (4.0i + 4.0j ), Δr=rf-ri = (1.0i + 2.0j )m
note: displacement is a vector, so all scalar answers can be ruled out
+4.00t where v is in m/s and t is in s.
What is the acceleration of the object at t = 2.00 s?
a. 5.00 m/s 2
b. 14.0 m/s 2
c. 20.0 m/s 2
d. 24.0 m/s 2
e. 28.0 m/s 2
Set y=0 at where the rock is thrown. yi=0, yf=-h, a=-g, v0=-10 m/s.
yf= yi + v0t +1/2 at 2 = 0 + (-10)*3 +0.5*(-9.8)*3 2 = -74.1 m
� h=74m
a= dv/dt = 5.00(2t) +4.00 = 4.00 + 10.00t
at t=2.00, a= 24.0 m/s 2
Draw it carefully and you will
see only answer “e.” matches the
motion pattern.
Note: this problem requires knowledge of simple derivative which is a
requirement of this course.

10. A particle starts from the origin at t = 0 with a velocity of 6.0i m/s and moves in the
xy plane with a constant acceleration of (�-2.0i+ 4.0j) m/s 2 . At the instant the particle
achieves its maximum positive x coordinate, how far is it from the origin?
a. 36 m
b. 20 m
c. 45 m
d. 27 m
e. 37 m
11. A particle moves in the xy plane with a constant acceleration given by a= -4.0j m/s 2 .
At t = 0, its position and velocity are 10i m and (-2.0i + 8.0j) m/s, respectively. What
is the distance from the origin to the particle at t = 2.0 s?
a. 6.4 m
b. 10 m
c. 8.9 m
d. 2.0 m
e. 6.2 m
12. Which of the following quantities is directly proportional to the time interval after a
projectile has left the barrel that shot it out? The x axis is horizontal; the y axis is
vertically upward.
a. Δ|v| in y direction, the acceleration is a constant. ( ay=-g)
b. Δay
c. Δy Δvy= ayt
d. Δ|r|
e. Δvy
13. A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes
the ground at a point 65 m horizontally away from and below the point of release.
What is the speed of the ball just before it strikes the ground?
a. 43 m/s
b. 47 m/s
c. 39 m/s
d. 36 m/s
e. 14 m/s
when it achieves maximum positive x position, vx=0 (=vxf)
vxf=vxi+axt � t= (vxf-vxi)/ax = (0-6)/(-2) =3s.
xf= xi + vxit+0.5axt 2 = 0 + 6*3 + 0.5*(-2.0)*3 2 = 9 m
yf= yi + vyit+0.5ayt 2 = 0 + 0*3 + 0.5*4.0*3 2 = 18 m
r=sqrt(Δx 2 + Δy 2 ) = 20.1m
xf= xi + vxit+0.5axt 2 = 10 + (-2)*2 + 0.5*(0)*3 2 = 6m
yf= yi + vyit+0.5ayt 2 = 0 + 8.0*2 + 0.5*(-4.0)*2 2 = 8 m
r=sqrt(xf 2 + yf 2 ) = 10 m
in y direction: -100 = 0 +0*t +0.5*(-g)t 2 � t=4.52s
vyf= 0 –g*4.52 = -44.27 m/s
now in x direction, vxf=vx= 65/4.52= 14.38 m/s
vf= sqrt(vyf 2 +vxf 2 )=47 m/s

14. A particle is in a counter-clockwise uniform circular motion of period T and radius
R. At t=0, its position is (x,y)=(0,0), and at t=T/2, is position is (x,y) = (2R, 0). What
is the magnitude of its average velocity in time interval between t=0 and T/2?
a. 0;
b. πR/T
c. -πR/T
d. 4R/T
e. none of above
15. For the same setting as in the previous problem. What is the magnitude of its average
acceleration in time interval between t=0 and T/2?
a. 0;
b. 4π 2 R/T 2
c. 8πR/T 2
d. 8π 2 R/T 2
e. none of above
16. Still in the same setting as above. What is the direction of its average acceleration in
time interval between t=0 and T/2?
a. No direction as the average acceleration is zero;
b. +y
c. -y
see solution to the previous problem
d. +x
e. -x
17. Define the x axis as horizontal and the y axis vertically upward. A projectile is thrown
up at an angle (and falling down later). When the projectile reaches the maximum
height, its
a. vx = 0 and vy= 0
b. vy = 0 and ay= 0
c. vx = 0 and ax= 0
d. ax = 0 and ay= 0
e. none of above
18. A baseball is hit at ground level. The ball is observed to reach its maximum height
above ground level 3.0 s after being hit. And 2.5 s after reaching this maximum
height, the ball is observed to barely clear a fence that is 97.5 m from where it was
hit. How high is the fence?
a. 8.2 m
b. 15.8 m
c. 13.5 m
d. 11.0 m
e. 4.9 m
Between t=0 and t=T/2, the magnitude of displacement is 2R.
vavg = |Δr|/dt = 2R/(T/2) =4RT
v(t=0) = -2πR/T j, v(t=T/2) = 2πR/T j,
a = (v(t=T/2)- v(t=0)) /(T/2) = 8πR/T 2 j � a=8πR/T 2
at maximum height, vy=0. but vx is not necessarily zero
at t=3.0s, vy=0 ( maximum height) � vyi=0-(-g)(3.0) = 3x9.8 (m/s)
at t=3.0+2.5=5.5s, y-yi=vyi*(5.5) +0.5*(-g)(5.5) 2 =13.47m

19. Two balls, projected at different times so they don't collide, have trajectories A and B,
as shown below.
Which statement is correct?
a. v0B must be greater than v0A.
b. Ball A is in the air for a longer time than ball B.
c. Ball B is in the air for a longer time than ball A.
d. Ball B has a greater acceleration than ball A.
e. Ball A has a greater acceleration than ball B.
20. A river has a uniform flow speed of 3.0 m/s toward the east. A boat with a speed of
8.0 m/s relative to the water leaves the south bank and heads in such a way that it
crosses to a point directly north of its departure point. Take notations vrE as the
relative velocity of the water flow w.r.t to the earth, vbr as the relative velocity of the
boat w.r.t to the river’s water flow, vbE as the relative velocity of the boat w.r.t to the
earth. Which of the following vector graphs represents the relationship among these
velocities?
e. none of above
21. In the above setting, if the river is 0.20 km wide, how long does it take the boat to
cross the river?
a. 29 s
b. 23 s
c. 25 s
d. 27 s
e. 17 s
vbE=vrE + vbr � c. is correct.
vbE=vrE + vbr �Take vector relationship as shown in c. above,
vbE = sqrt(vbr 2 -vrE 2 ) = 7.4 m/s
t=200/7.4= 27 s
The total flight time is twice of that from the top to the ground.
Now think of only vertical dimension, trajectory B has higher
maximum height, it take longer for B to fall down.