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Descriptive Statements:
- Analyze one-dimensional motion of objects in a variety of situations using multiple representations (e.g., words, graphs, equations, mathematical models), including solving problems with constant acceleration.
- Analyze two-dimensional motion of objects in a variety of situations using multiple representations (e.g., words, graphs, equations, mathematical models, vectors), including solving problems with constant acceleration.
- Analyze different forces (e.g., gravity, spring, friction, air resistance, buoyancy, tension, normal) and resultant net forces (e.g., centripetal) acting on objects in a variety of situations (e.g., inclined planes, springs, ropes, pulleys, uniform circular motion) in one or two dimensions, including the use of free body diagrams.
- Apply knowledge of Newton's laws of motion of objects and systems for a variety of situations, including solving problems in one or two dimensions.
- Demonstrate knowledge of scientific and engineering practices, crosscutting concepts, safety procedures and the proper use of equipment, and the engineering design process related to kinematics and the laws of motion.
Sample Item:
A 4.0 kg4 point 0 kilogram ball travels in the x-y coordinate plane with initial velocity (v0x, v0y) = (4.0, 0.0) m/s(v subscript zero x comma v subscript zero y) equals (4.0 comma 0.0) meters per second and initial position (s0x, s0y) = (0.0, 0.0) m. The ball experiences a single, constant force given by (Fx, Fy) = (0.0, 10.) N(F subscript x comma F subscript y) equals (0.0 comma 10) Newtons. Which of the following values represents the magnitude of the velocity of the ball at t = 2.0 sseconds?
- 2.5 m/smeters per second
- 5.0 m/smeters per second
- 6.4 m/smeters per second
- 9.0 m/smeters per second
Correct Response and Explanation (Show Correct ResponseHide Correct Response)
C. The force and mass information allows one to calculate the acceleration vector using the second law, (ax, ay) = (0.0, 2.5) m/s2(a subscript x comma a subscript y) equals (0.0 comma 2.5) meters per second squared. The horizontal velocity component remains constant, vx = 4.0 m/sv subscript x equals 4.0 meters per second, and the vertical velocity component is calculated using vy = v0y +ayt = (0.0)+(2.5 m/s2)(2.0 s) = 5.0 m/sv subscript y equals v subscript zero y plus a subscript y times t equals 0.0 plus 2.5 meters per second squared times 2.0 seconds equals 5.0 meters per second. The magnitude of velocity is then calculated using 
= 6.4 m/smagnitude of vector v equals square root of quantity v subscript x squared plus v subscript y squared equals root 41 = 6.4 meters per second.
Descriptive Statements:
- Analyze changes in energy of objects in open and closed systems in a variety of situations.
- Apply knowledge of work, energy, and power principles to objects and systems in a variety of situations.
- Apply the work-energy theorem to open and closed systems with conservative and nonconservative forces in a variety of situations, including solving problems.
- Apply the impulse-momentum theorem to systems in a variety of situations, including solving problems.
- Analyze elastic and inelastic interactions using momentum conservation in one or two dimensions, including solving problems.
- Demonstrate knowledge of scientific and engineering practices, crosscutting concepts, safety procedures and the proper use of equipment, and the engineering design process related to energy and momentum conservation.
Sample Item:
A physicist is analyzing a martial arts demonstration in which a thick piece of wood is split in two by a person's hand. A motion sensor measures the speed of the hand before impact to be 12 m/s12 meters per second and a video recorder measures the time it is in contact with the board to be 1.5 × 10–2 s1.5 times 10 to the power negative 2 seconds. What is the average force acting on the board from the hand? (Assume that the effective mass of the hand is 1.1 kg1.1 kilograms and that the hand comes to rest at the end of the impact.)
- 8.8 × 101 N8.8 times 10 to the power 1 Newtons
- 8.0 × 102 N8.0 times 10 to the power 2 Newtons
- 8.8 × 102 N8.8 times 10 to the power 2 Newtons
- 1.8 × 103 N1.8 times 10 to the power 3 Newtons
Correct Response and Explanation (Show Correct ResponseHide Correct Response)
C. Using the impact-momentum relationship, the average force is calculated: Favg = △p/△t = m(△v)/△t = (1.1 kg) (12 m/s) /(1.5 × 10–2 s) = 880 NF subscript average equals delta p divided by delta t equals m times delta v divided by delta t equals 1.1 kilograms times 12 meters per second divided by 1.5 times 10 to the power negative 2 seconds equals 880 Newtons. The third law of motion shows that the force on the board is identical to the force on the hand.
Descriptive Statements:
- Apply knowledge of kinematics for systems in rotational motion using multiple representations (e.g., words, graphs, equations, mathematical models), including solving problems with constant angular acceleration.
- Analyze different torques acting on objects in a variety of situations, including solving statics and dynamics problems using the laws of motion.
- Apply knowledge of conservation of energy and angular momentum to closed systems in rotational motion, including solving problems.
- Apply knowledge of the characteristics of simple harmonic motion (e.g., amplitude, frequency, period, energy, kinematic quantities) in a variety of situations using multiple representations (e.g., words, graphs, equations, mathematical models).
- Analyze objects and systems in simple harmonic motion in a variety of situations (e.g., pendula, springs).
- Demonstrate knowledge of scientific and engineering practices, crosscutting concepts, safety procedures and the proper use of equipment, and the engineering design process related to rotational motion and simple harmonic motion.
Sample Item:
A sphere with moment of inertia 3.0 kg•m23.0 kilogram meters squared is spinning with an initial angular velocity of 10. rad/s10. Radians per second. A constant torque of 5.0 N•m5.0 Newton meters squared is applied to the sphere, causing the sphere to spin faster. Which of the following values is closest to the angular acceleration of the sphere?
- 1.7 rad/s21.7 Radians per second
- 2.5 rad/s22.5 Radians per second
- 3.3 rad/s23.3 Radians per second
- 5.9 rad/s25.9 Radians per second
Correct Response and Explanation (Show Correct ResponseHide Correct Response)
A. Applying the second law in rotational form: α = τ / I = 5.0 N•m/3.0 kg•m2 = 1.7 rad/s2alpha equals tau divided by I equals 5.0 Newton meters divided by 3.0 kilogram meters squared equals 1.7 radians per second squared. The initial angular velocity value is superfluous information.
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