Instructor: Hacker Course: Physics 210 Sample Exam 3 Solutions (Kinematics)

Instructor: Hacker
Course: Physics 210
Sample Exam 3 Solutions (Kinematics)
Print your name neatly. If you forget to write your name, or if the grader can’t read your
writing, you can lose up to 100 points. Answer all the questions that you can.
This exam will consist of 21 multiple-choice problems. You may not use calculators or
other electronic devices on this exam. The use of such a device will be regarded as an
attempt to cheat, and will be pursued accordingly. All diagrams and figures on this exam
are rough sketches: they are not generally drawn to scale.
No partial credit will be given for these problems. However, you can miss one of the 21
problems without penalty. Your grade will be based on your best 20 problems. You will
not receive extra credit for getting all 21 right.
Your grade on the exam will be based entirely on the answers that you circle on this
sheet. If you have no answer or a wrong answer there, the grader will not look at the
page with the problem to see if the right answer appears there. Illegible or ambiguous
answers will be graded as wrong. You are responsible for copying your answers clearly,
correctly, and in the right place.
Although there is no partial credit on this exam, you must show your work in the space
provided on the exam. There is additional scratch paper at the end of the exam: do
not use it unless you have filled all the scratch space provided on the page with the
problem. If you answer a difficult problem without doing any written work, the grader
will assume that you got the answer by guessing or by copying from someone else, and
will not give you credit for the problem even though you’ve indicated the correct solution
on the answer sheet.
Circle your answers here. Do not detach this sheet from the test.
1.
a
b c
○
d
e
8.
a
b c
○
d
e
a b
15. ○
c
d
2.
a
b c
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d
e
9.
a
b
c d
○
e
16.
b
c
d e
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d
e
a b
17. ○
c
d
e
e
18.
a
b c
○
d
e
d e
○
19.
a
b
c
d e
○
d
e
20.
a
b
c d
○
d
e
21.
a
b
c
a b
3. ○
c
d
e
10.
a
b c
○
a b
4. ○
c
d
e
11.
a
b
c d
○
5.
b
c
d e
○
12.
a
b
c
a b
6. ○
c
d
e
13.
a
b c
○
b c
○
d
e
a b
14. ○
7.
a
a
c
a
e
e
d e
○
Physics 210 sample exam 3 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 2
One-dimensional kinematics
x
6
s
D
s
s
E
F
s
GA
s
A
A
C
A
A
As
s
B
s
H
A
-
t
Figure 1: A car is being driven erratically along a straight stretch of highway. The graph
shows its position x as a function of the time t.
Problem 1. For the interval between points F and G in figure 1, the car’s velocity is:
(a) positive
*(b) negative
(c) zero
Solution: The slope of the curve x = x(t) is negative between F and G.
Problem 2. In this problem, use downward as the positive direction. An elevator is
going down from the 39th floor to the 3rd. It is now at the level of the fourth floor,
and is slowing down as it approaches the third. Which of the following describes the
elevator’s situation?
(a) positive velocity; positive acceleration
*(b) positive velocity; negative acceleration
(c) negative velocity; negative acceleration
(d) negative velocity; positive acceleration
(e) none of these
Solution: The elevator is moving downward (the positive direction), so the velocity is
positive. It is slowing down as it moves downward, so the acceleration is negative.
Fundamental formulas for motion in 1-D
Equation 1:
Equation 2:
Equation 3:
v = v0 + at
(use when: position not in problem statement)
2
2
v = v0 + 2a(x − x0 )
(use when: time not in problem statement)
1
x = x0 + v0 t + at2
(use when: want position as a fcn of time)
2
Physics 210 sample exam 3 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 3
Problem 3. A ship is initially moving northward at 15 m/s. It accelerates southward
at 2 m/s2 for 4 s. What is its final speed?
*(a)
(c)
(e)
7 m/s
34 m/s
None of these
(b) 23 m/s
(d) 62 m/s
Solution: Careful! Notice that the initial velocity is northward, but that the acceleration is southward. We will take the northward direction to be positive, so the acceleration
is negative. We know initial velocity, acceleration, and time; we want to know final velocity.
v = v0 + at = 15 m/s + (−2 m/s2 )(4 s) = 7 m/s
The final velocity is positive, which means that the ship is moving northward. However,
the problem only asks for speed, which doesn’t include the direction.
Problem 4. A car is initially 30 m south of an intersection and moving northward at 6
m/s. It accelerates northward at 2 m/s2 for 3 s. At the end of this time, where is it with
respect to the intersection? Round your answer to the nearest meter.
*(a)
(c)
(e)
3 m south
3 m north
None of these
(b) 57 m south
(d) 57 m north
Solution: We will take northward to be the positive direction, so the initial position
is negative and the initial velocity and acceleration are positive. We know the initial
position, initial velocity, acceleration, and time; we want to know the final position.
1
1
x = x0 + v0 t + at2 = −30 m + (6 m/s)(3 s) + (2 m/s2 )(3 s)2 = −3 m
2
2
Since x is negative, it is south of the intersection.
Problem 5. A physics student on Planet X shoots a ball straight upward from a spring
cannon, with an initial speed of 10 m/s. The ball reaches the highest point in its flight
after 100 s. What is the magnitude of the Planet’s surface gravity?
1
1
(a)
m/s2
(b)
m/s2
2
20
1
(c) 10 m/s2
*(d)
m/s2
10
(e) None of these
Solution: We will take the positive direction to be upward in the direction of motion,
so the initial velocity will be positive and the acceleration will be negative. We know the
initial velocity, the final velocity (zero), and the time; we want to know the acceleration.
v = v0 + at
⇒
a=
v − v0
−v0
10 m/s
1
=
=
=
m/s2 .
t
t
100 s
10
Physics 210 sample exam 3 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 4
Problem 6. (Similarity Problems) An object dropped from a building with height
Y strikes the ground with speed V . If an object is dropped from a building with height
4Y , how fast is it moving when it hits the ground?
*(a)
(c)
(e)
2V
8V
None of these
(b)
(d)
4V
16V
Solution: We will take the positive direction as downward, and the top of the building
as y0 = 0. The initial velocity is v0 = 0. Since time isn’t in the statement of the problem,
we use the fundamental formula (2): v 2 = v02 + 2a(y − y0 ); in this case, v 2 = 2gy. Taking
the square
root of this equation gives v as a function of y. The governing equation is
√
from the governing
equation,
we
v = 2gy.√In the first case, let y1 = Y and v1 = V ; then √
p
√
have v1 = 2gy1 . In the second case, y2 = 4y1 ; so v2 = 2gy2 = 2g(4y1 ) = 2 2gy1 =
2v1 = 2V .
Problem 7. (This problem requires solving a system of equations)
A policeman is lurking behind a billboard beside the highway when someone speeds by
him at 30 m/s. The policeman immediately accelerates at a constant rate of 6 m/s2 until
he has caught up with the speeder. How long does it take for the policeman to catch up?
(a) 5 s
(c) 15 s
(e) None of these
*(b) 10 s
(d) 20 s
Solution: We will use the formula for position as a function of time: x = x0 +v0 t+ 12 at2 .
We will use the position of the billboard as x0 = 0. We need to use the formula twice:
once for the policeman, once for the speeder. For the speeder, x0 = 0, v0,s = 38 m/s, and
as = 0. For the policeman, x0 = 0, v0,p = 0, and ap = 6.7 m/s2 . Hence their positions
are:
1
xs = x0 + v0,s t + as t2 = v0,s t
2
1 2 1 2
xp = x0 + v0,p t + ap t = ap t
2
2
When the policeman catches up with the speeder, their positions will be the same: so
xs = xp . Equating the expressions:
1
v0,s t = ap t2
2
⇒
ap t2 − v0,s t = 0
⇒
t (ap t − 2v0,s ) = 0
There are two solutions:
t = 0 and t =
2v0,s
2(30 m/s)
=
= 10 s
ap
6 m/s2
The first solution refers to the fact that both were at the billboard at t = 0; the second
solution is the time at which the policeman catches the speeder.
Note: In general, the velocity of the police officer and the speeder at the position
xfinal = xs = xp will not be the same vs,f 6= vp,f .
Physics 210 sample exam 3 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 5
One-dimensional calc-based problems
Problem 8. An object is moving along the x-axis, with its position as a function of time
given by x = x(t). The origin O is located at x = 0. The object is definitely moving
towards the origin if
d −x2 d −x2 e
<0
*(b)
e
>0
(a)
dt
dt
(c)
d x
(e ) < 0
dt
(d)
d(x3 )
(e)
<0
dt
d x
(e ) > 0
dt
(f) None of these
Solution: The correct answer is (b).
2
If e−x is increasing, then −x2 is increasing; so x2 is decreasing; so x is getting closer to
the origin.
2
(a) can’t be right: if e−x is decreasing, then −x2 is decreasing; so x2 is increasing; so x
is moving away from the origin.
(c) can’t be right: if ex is decreasing, then x is decreasing; but x could be negative, and
moving away from the origin.
(d) can’t be right: if ex is increasing, then x is increasing; but x could be positive, and
moving away from the origin.
(e) can’t be right: if x3 is decreasing, then x is decreasing; but x could be negative, and
moving away from the origin.
Problem 9. The acceleration of gravity at the surface of Planet X is gX . The planet
has no atmosphere, so there is no air drag on falling objects. An object initially at rest
is dropped from a large height at t = 0. Find the time T for which the distance that the
object falls between t = 0 and t = T is the same as the distance that it falls between
t = T and t = T + 1. Which of the following statements about T is true?
(a) 0 < T ≤ 1
*(c) 2 < T ≤ 3
(e) 4 < T
(b) 1 < T ≤ 2
(d) 3 < T ≤ 4
Solution: The equation governing the fall is y = 12 gX t2 . Want the time T :
y(T ) = ∆y = y(T + 1) − y(T )
+ y(T )
−−−−−−−→ y(T + 1) = 2y(T )
Let y= 21 gX t2
1
1
2
2
−−−−−−−→
g (T + 1) = 2
g T
2 X
2 X
÷
1
g
2 X
−−−
−→ (T + 1)2 = 2T 2
complete the square
−−−−−−−−−−−→ T 2 − 2T + 1 = 2 ⇒ (T − 1)2 = 2
√
√
−−−−−−→ T = 1 ± 2 ≈ 2.14 (since T > 0)
Physics 210 sample exam 3 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 6
One-dimensional rotational kinematics
Problem 10. (Similarity Problem) Two bugs are clinging to a spoke of a turning
bicycle wheel. The first bug is 10 cm from the center of the wheel; the second bug is 20
cm from the center. If the first bug is experiencing an angular speed of ω1 , what is the
second bug’s angular speed?
(a) √
ω1 /2
2 ω1
(c)
(e) None of these
*(b) ω1
(d) 2ω1
Solution: Both bugs are moving at the angular speed of the wheel as a whole: ω1 .
Problem 11. Starting at rest, a wheel experiences an angular acceleration of 2 rad/s2 .
How much time elapses before the wheel has turned through 16 rad?
√
8s
(a) 2 s
(b)
*(c) 4 s
(d) 8 s
(e) None of these
Solution: We know ω0 = 0, α, θ0 = 0, and θ. We want to know t. We have a formula
we can use in this situation:
1
1
θ = θ0 + ω0 t + αt2 = αt2
2
2
2θ
·2/α
−−→ t2 =
α
r
1/2
√
2θ
2(16 rad)
−→ t =
=
=4s
α
2 rad/s2
Problem 12. A wheel is turning at a rate of 12 rad/s. When a brake is applied to it, it
experiences an angular acceleration of α. If the wheel turns through 9 rad before coming
to a halt, what is α?
(a) −1/3 rad/s2
(c) −2 rad/s2
(e) None of these
(b) −4/3 rad/s2
*(d) −8 rad/s2
Solution: We know θ0 = 0, θ, ω0 , and ω = 0. We want to know α. We have an equation
we can use in this situation.
ω 2 = ω02 + 2α(θ − θ0 )
⇒ 0 = ω02 + 2αθ
⇒ 2αθ = −ω02
ω2
(12 rad/s)2
= −8 rad/s
⇒ α=− 0 =−
2θ
2 · 9 rad
Physics 210 sample exam 3 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 7
Two-dimensional kinematics
Graphical Understanding
Problem 13. The graph at right
shows the top view of the path of a
particle moving along a horizontal
surface. The particle is moving
from the lower left to the upper
right. Which of the labelled vectors
is the velocity vector at point P ?
~
(a) A
*(b)
~
(c) C
(d)
(e) None of these
y6
~
B
~
A
@
I
@
P
@s
@
@
~
B
~
D
~
D
R
@
~
C
-
x
Solution: The velocity vector is tangent to the curve in the direction of motion.
Uniform circular motion
Problem 14. A windmill has three blades, each 25 m long; the blades make a complete
circle every 5 s. A bug is clinging to the tip of one of the blades. How much radial
acceleration does the bug experience? Ignore the effect of gravity.
*(a)
(c)
(e)
4π 2 m/s2
20π m/s2
None of these
(b) 4π m/s2
(d) 20π 2 m/s2
Solution: The radius of the circle traced by the bug is r = 25 m. The bug makes a
complete circle in t = 5 s. Hence the bug’s speed is v = 2πr/t. Its centripetal acceleration
is
4π 2 r2
4π 2 r
4π 2 (25 m)
v2
=
=
=
= 4π 2 m/s2
ac =
r
rt2
t2
(5 s)2
Physics 210 sample exam 3 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 8
Problem 15. (Similarity Problem) A highway has a curve with a radius of 100 m.
Large trucks must drive the curve at a speed of V or less to keep from skidding off the
road. If the highway were rebuilt so that the curve had a radius of 200 m, how fast could
trucks safely drive around the curve?
√
(b) 2V
*(a) √2 V
(c)
2 2V
(d) 4V
(e)
None of these
Solution: Let R = 100 m be the radius of a curve that a truck can drive at a speed of
V . Then the centripetal acceleration experienced by such a truck is A = V 2 /R. If we
increase the radius to 2R, we want to know what speed v will cause trucks to experience
the same acceleration A.
A=
v2
V2
=
R
2R
⇒
v2 =
2RV 2
= 2V 2
R
⇒
√
v=
2V 2 =
√
2V
Trajectories
Problem 16. The following four diagrams show the trajectory of a cannonball. Which
diagram correctly shows the acceleration vectors at three points along the trajectory?
r
y6
6
~0
r
(a)
?
r
r -
y6
r
r
Q
7
-
-
x
r
y6
?
r
Z
~
Z
r
(c)
(b)
s
Q
x
r
y6
r
*(d)
?
r
?
?
-
x
-
x
(e) None of these
Solution: At every point, the acceleration is 9.8 m/s2 , directed straight down. Your
vectors should all be of the same length.
Physics 210 sample exam 3 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 9
Problem 17. A movie stuntman wants to ride a motorcycle off a flat roof, sail across
a gap X meters wide, and land on another roof 5 m below the first one. How fast does
he need to be moving when he leaves the first roof? For ease of computation take the
magnitude of gravity to be 10 m/s2 .
√
*(a) X m/s
(b)
X
p 2 m/s
√
(c)
X/ 2 m/s
(d)
X/2 m/s
(e)
None of these
Solution: The stuntman is going off a flat roof, with no mention of a ramp; so we
assume that the elevation is θ0 = 0. There is no vertical component of initial velocity,
and v0 = v0x .
p
The time it takes for the motorcycle to fall y = 5 m is t = 2y/g. During this time,
the motorcycle needs to cover a horizontal distance of x = X m. Hence the initial
(horizontal) velocity must be
s
r
x
x
g
10 m/s2
v= =p
=
x=
(X m) = X m/s
t
2y
2(5 m)
2y/g
Problem 18. Your potato gun has a muzzle velocity of v0 . You want to shoot a potato
from ground level and have it strike the ground txmax seconds later. At what elevation
do you have to fire the gun?
2v0
−1
−1 gtxmax
(a) θ0 = sin
*(b) θ0 = sin
gtxmax
2v0
gtxmax
−1 2ymax
(c) θ0 = sin
(d) θ0 = sin
g
2v0
(e) None of these
Solution: We know the initial velocity, the gravitational acceleration, and the time of
flight. We want to know the angle of elevation. Use the formula for flight time:
gtxmax
2v0 sin θ0
−1 gtxmax
⇒ sin θ0 =
⇒ θ0 = sin
txmax =
.
g
2v0
2v0
Physics 210 sample exam 3 solns
Copyright ©Wayne Hacker 2010. All rights reserved.10
Problem 19. (Lab Similarity Problem) You fire a spring cannon from ground level
at a fixed elevation. The spring cannon uses a compressed spring to launch the ball.
The cannon has three settings with the first being the lowest and the third being the
strongest. Using the first setting (one click) you find that the ball emerges from the
muzzle with a speed of V ; it strikes the ground at a distance X from the cannon. You
then increase the energy stored in the spring by using the second setting (two clicks).
The muzzle velocity of the ball is now α V , where α is a positive constant. What is the
new range of the ball?
√
αX
(b) α X
(a)
√
*(d) α2 X
(c) α α X
(e) None of these
Solution: Use the formula (the governing equation) for range:
xmax =
v02
sin(2θ0 )
g
⇒
xmax,1 =
V2
sin(2θ0 )
g
If we increase v0 from V to α V , we get
xmax,2
(α V )2
sin(2θ0 ) = α2
=
g
V2
sin(2θ0 ) = α2 xmax,1 = α2 X
g
Thus, if we double the initial speed we quadruple the distance traveled by the ball.
Actually, in the real world the faster the ball travels, the larger the air drag.
Two-dimensional calc-based problem
Problem 20. Consider a position vector ~r in the xy-plane with tail at the origin O
= (0, 0) and tip at the point P = (x, y). An object’s position as a function of time is
~r(t) = x(t)ˆı + y(t)ˆ
. In vector notation, the origin is expressed as ~r = ~0. In which of the
following situations is the object definitely moving closer to the origin?
(a) vx > 0 and vy > 0
*(c) xvx + yvy < 0
(e) None of these
(b) vx < 0 and vy < 0
(d) xvx + yvy > 0
Solution: The object is moving closer to the origin if its distance from the origin
is decreasing with time. Since distance is always positive (actually non-negative) and
f (r) = r2 is a monotonic increasing function, it follows that the distance between the
object and the origin is decreasing if and only if the distance squared is decreasing. It
therefore suffices to show that the distance squared is decreasing in time. Let ~r(t) =
hx(t), y(t)i be the position vector of the particle. Then the magnitude of the position
vector is r = k~rk. In symbols we need the following condition to be satisfied:
0>
d
d 2
r =
(~r · ~r) = 2(~r · ~v ) = 2(xvx + yvy )
dt
dt
⇒
xvx + yvy < 0 ,
where reading from left-to-right the second equality follows from the dot product for
vectors and the last equality follows from the component definition of the dot product.
The answer is (c).
Physics 210 sample exam 3 solns
Copyright ©Wayne Hacker 2010. All rights reserved.11
Problem 21. A particle is moving in two dimensions. Its position vector as a function
of time is ~r(t) = h3t2 + 5t − 1, −t − 4i. Find the acceleration vector ~a(t) as a function
of time.
(a) h3t2 + 4t − 5i
(b) h6t + 5, −1i
3 5 2
1 2
*(d) h6, 0i
(c)
t + 2 t − t, − 2 t − 4t
(e) None of these
Solution: To find the acceleration we take two time derivatives of the position vector.
d~r
d 2
(t) =
3t + 5t − 1, −t − 4 = h6t + 5, −1i
dt
dt
d~v
~a = (t) = h6, 0i
dt
~v =