Download Kinematics Practice Packet 2021 KEY

Time Weighted Practice.notebook
Tips:
a. this is equal DISTANCE, not equal
time! Once you know times, you
can use time weighted average.
b. this is equal TIME, you can use
mathematical average.
c. one motion, same direction
d. there's a turn around! Consider
direction! Displacement and
Velocity depend on that direction
change!!
i) d>a=b=c
ii) all same, a=b=c=d
iii) d>b>c>a
iv) b>c=d>a
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1) A light plane must reach a speed of 33 m/s for takeoff. How long a runway ls needed if the
(constant) acceleration is 3 rn/s1 .
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2) A marble is tossed up from a height of 20 m with a velocity of 100 m/s. (let g=10m/s2J
a. What is the marble's velocity after 5 seconds?
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1
27
NATIONAL
MATH + SCIENCE
INITIATIVE
Science
To NMSI Instructors Presenting This Session
This topic-based session consists of a pre-assessment piece, a content piece (for student reference), multiplechoice questions (including a multi-mark question) and four free responses. It is advised that you ask students
to consider the pre-assessment piece during the period of time after they have walked into your room but before
the session officially starts. Asking students probing questions about the pre-assessment will allow you to
determine to what degree students remember (or were correctly taught) motion principles.
Do not spend much time (if any at all) on the content piece. It is there as a reference for students. You are
encouraged to launch directly into the multiple-choice questions or the free-response questions as you feel is
appropriate from your experience with the pre-assessment.
When going over multiple-choice questions or free-response parts, give students some think time and allow
them to work as partners (or groups no bigger than 3). Let them prepare remarks to make relating to each
question. Encourage students to say something relevant to the problem even if they cannot arrive all the way to
a right answer. For example, for problem M2, a student might contribute that displacement is the area of a
velocity graph even if the student can’t make it all the way to the right answer.
The multiple-choice questions are designed to be a trivial mental task for those who have mastered the concepts
related to motion, and an impossible ordeal for those who have shallow understanding. This reflects the
question style on the Physics 1 exam, which questions are easy only after complete understanding is attained.
When going over multiple-choice questions, focus on the basic principles required to answer the question
correctly. Say principles repeatedly with different wording. Draw diagrams and graphs if that will help
students understand the approach to the problem better.
There are four free-response questions, which are related to motion but follow this pattern:
• The first is a question that heavily relies on one or more representations.
• The second is a laboratory or experiment-related question.
• The third is a qualitative-quantitative translation.
• The fourth involves writing a paragraph-length-response.
Avoid being the “sage on the stage.” The session should be a mixture of you interacting with students and
students interacting with each other. Segments that involve you talking and the students listening should be
short—no more than 5 minutes at the beginning and sporadic 2-to-3 minute talks after that. Be positive at all
times, having a “can-do” attitude. If you think that the students are “doomed” or “too far behind”, do not let
this on. Do not allow students (by your actions or the actions of other students) to feel bad because they don’t
know something. Don’t say “your teacher didn’t teach you this?” or anything of that sort—neither they nor you
can control how their teacher conducts their AP Physics classroom. Emphasize continually that the purpose of
Saturday Study Sessions is to leave the session better and more capable than you came.
Pre-Assessment Answers
(a-i)
The object slows down between 0 < t < 1 s and between 2 < t < 3 s. This is because the slope of the
position graph (which is velocity) becomes less steep during those intervals.
(a-ii)
Already given
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NATIONAL
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(b-i)
Science
The object slows down between 1 < t < 2 s and between 3 < t < 4 s. This is because the height of the
velocity graph is velocity, and during these intervals the graph is going toward zero velocity.
(b-ii)
(c-i)
The object slows down between 2 < t < 4 s. This is because the object sped up during the first half of
the interval (when the acceleration was positive), so the negative acceleration must mean that the object
is slowing down.
(c-ii)
Multiple-Choice Answers
M1.
Answer: B
Instantaneous speed is a tiny change in position over a tiny time. The last two dots (for 1.4 s and 1.5 s) are
separated by a displacement of 0.2 m and a time interval of 0.1 s.
M2.
Answer: D
The person has the same position as at the beginning when their net displacement is zero. Displacement is the
area of a velocity vs. time graph, so the time at which the positive area of the graph (from 0 to 5 seconds) equals
the negative area (from 5 to the-time-we-are-trying-to-find seconds) must be equal. The triangle between 5 s
and 8 s is not enough negative area to cancel the area of the positive trapezoid between 0 and 5 s, so the time we
are trying to find must be beyond 8 s.
M3.
Answer: B
If the car has an acceleration of 2 m/s2 to reach 4 m/s, then it traveled for 2 seconds according to Δv = aΔt.
Using either x = ½at2 (a = 2 m/s2, t = 2 s) or v2 = 2ax (v = 4 m/s, a = 2 m/s2), we can find the distance through
which the acceleration took place to be 4 m. That means that the object travels the remaining 12 m (of the 16 m
distance) going a constant 4 m/s, which takes 3 s according to Δx = vΔt. That makes for a total time of (2 s
accelerating) + (3 s constant speed) = 5 seconds.
M4.
Answer: B
A falling object has a falling distance that is related to its falling time by the equation d = ½at2. Rearrange to
get d/t2 = ½a. The d is on the top, so it is the “rise”, and t2 on the bottom is the “run”. The slope would then
equal half the acceleration of the falling object.
M5.
Answer: B
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NATIONAL
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Science
Answer choices A, B, and C are illustrated below. If the object traveled a distance 3D, then it traveled a
distance D while accelerating (area of a triangle on the velocity vs. time graph) and then a distance of 2D while
traveling with constant speed (the area of the rectangle under the constant speed on the velocity graph).
Car accelerates for T
seconds, then keeps
accelerating with
acceleration a for another
T seconds.
M6.
Car accelerates for T
seconds, then stops
accelerating and continues
with constant velocity.
Car accelerates for T
seconds, then decelerates
to rest with the same
magnitude acceleration it
started with.
Answer: B&C
Constant acceleration specifically requires that the distance traveled by related to time by the equation d = ½at2
(if the object starts from rest), so checking to see if distance is proportional to the square of the falling time
would verify that free-fall is constant acceleration. Constant acceleration also specifically requires the velocity
to increase as a function of time by v = at (if the object starts from rest), so a velocity vs. time graph would be
linear if the motion is constant acceleration.
Merely the fact that all objects released from rest move together does not guarantee constant acceleration;
pendulums released from the same string length and initial angle move together but are not uniformly
accelerating. Merely the fact that objects speed up does not mean that the acceleration is uniform; cars speed up
from rest, but a car’s acceleration decreases as the car reaches top speed.
Free-Response Answers
F1.
(7 points)
(a)
2 points
(b)
(1)
At time t = 5 seconds, the elevator is moving upward (it had only upward acceleration before that)
at 6 m/s (area of the velocity vs. time graph between zero and 5 s).
(2)
At time t = 10 seconds, the elevator is at rest (net area of the velocity graph between 0 s and 10 s is
zero).
1 points
(3)
The elevator is above its initial position because the elevator only moved upward and never
downward.
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NATIONAL
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(c)
Science
4 points
(4)
The velocity graph is continuous and horizontal for 0 < t < 2 s, 4 < t < 6 s, and 8 < t < 10 s.
(5)
The velocity graph is a straight line with positive slope for 2 < t < 4 s and a straight line with
negative slope for 6 < t < 8 s
(6)
The position graph is horizontal for 0 < t < 2 s and 8 < t < 10 s, and a reasonably straight line with
positive slope for 4 < t < 6 s.
(7)
The position graph has positive slope and is part of an upward-opening parabola for 2 < t < 4 s and
has positive slope and is part of a downward-opening parabola for 6 < t < 8 s
F2.
(12 points)
(a)
4 points
(1)
An appropriate diagram shows the yo-yo falling on its string and some sort of distance and time
measuring equipment.
(2)
The student explains how distances can be measured, including equipment to be used.
(3)
The student explains how times can be measured, including equipment to be used.
(4)
The student indicates that multiple trials must be taken, or explains a method that strongly implies
multiple distances and times will be measured.
Example Response:
Hold the yo-yo a certain distance above a table (distance measured by a
meterstick). Release the yo-yo so it unrolls until it hits the table, timing its fall
with a stopwatch. Repeat for several different distances above the table.
Example Response:
Hold the yo-yo above a motion sensor. Release the yo-yo while the motion
sensor takes data of distance and time.
Example Response:
Hold the yo-yo next to a meterstick and video the yo-yo falling next to the
meterstick. Use video analysis software to take data of distance fallen at different
times.
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4
NATIONAL
MATH + SCIENCE
INITIATIVE
(b)
Science
3 points
(5)
EITHER a graph of velocity vs. time is shown that is a line with slope less than 10 (numbers on the
axes would need to be shown to indicate this), OR a table of velocity vs. time is shown where the
velocities are proportional to the time, but the ratio of velocity to time is less than 10.
Time (s)
1
2
3
4
(c)
(6)
The student indicates that A is correct because EITHER the slope of the velocity vs. time graph is
constant, OR the ratio of velocity to time is the same.
(7)
The student indicates that B is incorrect because EITHER the slope of the velocity vs. time graph
is less than 9.8 or 10 m/s2 OR the ratio of velocity to time is less than 9.8 or 10 m/s2.
1 points
(8)
(d)
Velocity (m/s)
2
4
6
8
Drop the yo-yo without holding the string (so that the only force acting is the weight force).
4 points
(9)
Measure the height of the ball at A and at B, OR measure one of
those heights and measure the length of the string. The student also
indicates equipment to be used.
(10) Measure the speed of the ball at the instant it passes through B.
This point is not granted if the student attempts to use a stopwatch,
but only if the student includes equipment such as a photogate or
video analysis that can measure small amounts of time.
(11) Use mgy to find potential energy and ½mv2 to find kinetic energy.
(12) The student indicates that point A has only potential energy, while B has both potential and kinetic
energy and that the total energies at A and B need to be compared to see if they are the same.
F3.
(10 points)
(a)
2 points
(1)
The student uses the equation x = ½at2 and plugs in a = 6 m/s2 and t = 5 s.
(2)
The student states that D = 75 m.
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5
NATIONAL
MATH + SCIENCE
INITIATIVE
Science
(b&c) 5 points
(d)
(3)
For 0 < t < 5 s, the graph is a line.
(4)
For 0 < t < 5 s, the graph has positive slope and
passes through the origin.
(5)
The graph is continuous at t = 5 s and changes slope
(6)
For 5 < t < 10 s, the graph is a line with different
slope than for 0 < t < 5 s
(7)
For 5 < t < 10 s, the graph is a line with negative
slope that passes through (10 s, 0 m/s).
3 points
Student A is correct.
(8)
The student states that the area of velocity vs. time is distance
(9)
The student indicates that the second half of the graph needs to have area equal to the first half of
the graph.
(10) The student needs to connect these two statements to the fact that student A is correct by
explaining that the second half of the graph shows the car slowing down the same way it sped up
during the first half.
F4.
(7 points)
(a)
2 points
(b)
(1)
The rocket is momentarily at rest.
(2)
The rocket is at its highest point.
3 points
The greatest speed is attained going downward, and the greatest amount of time is spent moving upward.
(3)
The area of a velocity vs. time graph is displacement (or distance)
(4)
The area above the time axis “Aup” equals the area below the time axis “Adown”
(5)
Aup has a wider base and shorter height (Adown has shorter base and higher height)
(6)
Time moving upward is greater because of the greater base of Aup
(7)
Greatest speed is attained moving downward because of higher height of Adown.
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6
NATIONAL
MATH + SCIENCE
INITIATIVE
Example Response:
Science
The area of the positive part of the graph represents upward motion and the area
of the negative part downward motion, and these areas must equal. The positive
triangle has a wider base representing more time moving upward, while the
negative triangle has a higher height representing greater speed attained.
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7