11 The frog leaps from its resting position at the
lake's bank onto a lily pad. If the frog has a
mass of 0.5 kg and the acceleration of the leap
is 3 m/s^{2}, what is the force the frog exerts on
the lake's bank when leaping?
A 0.2 N
B 0.8 N
C 1.5 N
D 6.0 N
A formula chart on a beginning page of the test provides the formula
Force = mass x acceleration F = ma
and a constants/conversions table on the same beginning page provides the value
g = acceleration due to gravity = 9.8 m/s^{2}
and provides the unit definition
newton (N) = kg m/s^{2}.
Out of the 187214 students who took the test, the response frequencies for answers A, B, C, D, and no answer were, respectively, 7258, 15018, 154840, 9370, and 728.
Will Ken Mercer or Rebecca BellMetereau defeat Physicist Mark Loewe?
Solution
When resting, the frog already exerts a force straight down on the lake's bank equal to the frog's gravitational weight,
W = mg = (0.5 kg)(9.8 m/s^{2}) = 4.9 N.
To accelerate in some direction at 3 m/s^{2}, the frog must add a force of magnitude
F = ma = (0.5 kg)(3 m/s^{2}) = 1.5 N
in the direction opposite to the direction of the frog's acceleration. The force that the frog exerts on the lake's bank when leaping is the (vector) sum of the force that the frog exerts when resting with the additional force. This sum of forces depends on the direction of the frog's acceleration but must have a magnitude no greater than 4.9 N + 1.5 N = 6.4 N (corresponding to an acceleration of the frog straight up) and no less than 4.9 N  1.5 N = 3.4 N (corresponding to an acceleration of the frog straight down). Since answers A, B, and C are outside of this range, they are incorrect.
Answer D is acceptable as a correct answer because 6.0 N is well located within the range of values that are consistent with the given information. This is shown below along a force magnitude scale. [The force magnitude 5.1 N is shown because, for any direction of the frog's acceleration above the horizontal, the magnitude of the force that the frog exerts on the lake's bank when leaping must be greater than [(4.9 N)^{2} + (1.5 N)^{2}]^{1/2} = 5.1... N.]
Texas State Board of Education, District 5
Scoring mistakes
The Texas Education Agency made the mistakes of giving credit for the incorrect answer C and of denying credit for the correct answer D. The TEA wrongly increased the scores of those 154840 students who chose answer C and wrongly decreased the scores of those 9370 students who chose answer D. The remedy is to remove credit for answer C and to give credit for answer D.
Texas SBOE, District 5
Failures to correct the scoring mistakes
Mark Loewe informed the Texas Education Agency of the scoring mistakes on 4 June 2003. On 4 August 2003, the TEA issued the response
TAKS Released Test Items and issued, in "Additional Information Regarding Released Science Items", the statements
Item 11 asked students to calculate the force a frog exerts when jumping. The
frog has a mass of 0.5 kg and leaps with an acceleration of 3m/s^{2}. This item was
intended to measure the Integrated Physics and Chemistry (IPC) TEKS 4(B),
which required students to apply Newton's Law. The question raised about the
item is whether the force exerted by the frog must include both the weight of the
frog and the force needed to accelerate. That force is a vector sum and varies
with the angle of the leap. There is not enough information given to solve the
more advanced physics components of this problem. When viewed as an
introductorylevel problem for students first learning the connection between
force and acceleration, the item has only one correct answer and is a valid
question for all students tested at this level.
The TEA states that "However, upon review, these items were determined to be correct and no action was taken by the agency." This TEA statement refers to questions that include question 11. By "determined to be correct" the TEA means that answer C is correct and that answer D is incorrect. The TEA's claim that answer C is correct is false. The TEA's claim that answer D is incorrect is false.
The TEA states that "The question raised about the item is whether the force exerted by the frog must include both the weight of the frog and the force needed to accelerate." The need to account for the weight of the frog is basic. The TEA has not explained why whoever raised this question should have any doubt about the need to account for the weight of the frog. The TEA's statement is additionally surprising given the fact that, in an earlier (14 July 2003) letter to Dr. Loewe, then Associate Commissioner Smisko acknowledged that "In their analysis of item 11, the reviewers noted that the force exerted by the frog must include the weight of the frog plus the force needed to accelerate."
The TEA states that "There is not enough information given to solve the more advanced physics components of this problem." Only basic considerations of physics are used to show that answers A, B, and C are incorrect. Only basic considerations of physics are used to show that answer D is well located within the range of values that are consistent with the given information. Answer D is acceptable as a correct answer.
The TEA states that "When viewed as an introductorylevel problem for students first learning the connection between force and acceleration, the item has only one correct answer and is a valid question for all students tested at this level." By "the connection between force and acceleration" the TEA probably means Newton's Second Law, F = ma, which, when applied to the frog, gives 1.5 N for the total force on the frog. The TEA did not present the method by which it obtained the incorrect answer 1.5 N for the force the frog exerts on the lake's bank. Question 11 requires students to apply Newton's Second Law and additional basic considerations of physics. By "only one correct answer" the TEA means answer C. The TEA's claim that answer C is correct is false. Answer C has been shown to be incorrect using nothing more than a few of the most basic considerations of physics, which were undoubtedly applied by some students who took the test and which should readily have been applied by many students who took the test.
In the earlier letter to Dr. Loewe, then Associate Commissioner Smisko states that "Both fieldtest and livetest data have also been reexamined for each of these items; on both of the items in question, students performed well." With regard to question 11, Dr. Smisko's statement that "students performed well" is false.
The Texas Education Agency would have Texans believe that over 82% of the 187214 students who took the test chose a correct answer for question 11 even though only 5% chose a correct answer. That only 5% of Texas 11thGraders chose a correct answer (and some students may simply have guessed) demonstrates a deficiency in basic physics education in Texas that should be corrected, not hidden.
The TEA's false claims, which serve to hide a miserable performance, are due to incompetence or dishonesty. Texans are illserved by such incompetence or dishonesty.
The TEA has continued to propagate the scoring mistakes and false claims without correction.
On 15 July 2005,
Mark Loewe gave written and oral testimony to the
State Board of Education on the topic "TEA failed to correct TAKS mistakes"; question
11 was discussed in some detail. The SBOE has failed to acknowledge or to correct any of the TEA's scoring mistakes or false claims. [Then SBOE member Don McLeroy (District 9  College Station) deserves credit for recognizing the scoring mistakes on question
11.]
Newspaper articles
Question 11 has been discussed in several newspaper articles and letters to the editor:
"Physicist challenges 2 TAKS answers", Fort Worth StarTelegram, 6 June 2003,
"Disputed TAKS answers correct, state agency says", Fort Worth StarTelegram, 7 June 2003,
"Ambiguous questions", letter, Fort Worth StarTelegram, 10 June 2003,
"Physics, testing and a frog", letter by Charles Reidl and letter by Mike Stepp, Fort Worth StarTelegram, 14 June 2003,
Force magnitude
6.4 N  Leaping straight up (maximum possible force)



6.0 N  D < This answer is very well located within
 the range of possible correct answers.
 It corresponds to leaping in a direction
 41 degrees above the horizontal.

 Answer D is acceptable as the correct
 answer. The TEA denied credit for it.


5.1 N  Leaping horizontally

4.9 N  Resting














3.4 N  Leaping straight down (minimum possible force, even if "leaps"
 below the horizontal are considered)

















1.5 N  C < The TEA gave away credit for this incorrect answer






0.8 N  B





0.2 N  A

0.0 N 