Newton's First Law Quiz
Authored by ELIZABETH WOODARD
Physics
11th Grade
NGSS covered
Used 4+ times
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This quiz thoroughly assesses Newton's First Law of Motion and fundamental force analysis concepts appropriate for 11th-grade physics students. The questions require students to understand equilibrium conditions, analyze force components in multiple directions, apply the concept of net force to predict motion changes, and work with mathematical relationships between forces. Students must demonstrate mastery of key principles including that objects at rest or moving at constant velocity have zero net force, that unbalanced forces cause acceleration, and that forces must be analyzed as vectors with both magnitude and direction. The problems demand sophisticated reasoning about real-world scenarios such as projectile motion on the Moon, terminal velocity calculations, and force decomposition with trigonometric relationships. Students need to distinguish between instantaneous conditions and overall motion patterns, understand the relationship between force and acceleration through F=ma, and recognize that equilibrium requires balanced forces in all directions. Created by Elizabeth Woodard, a Physics teacher in US who teaches grade 11. This comprehensive assessment serves multiple instructional purposes, functioning effectively as a unit review before major examinations, homework assignments to reinforce classroom learning, or formative assessment to gauge student understanding of foundational mechanics concepts. The quiz works particularly well for identifying student misconceptions about force and motion relationships, making it valuable for targeted remediation and reteaching opportunities. Teachers can use individual questions as warm-up problems to activate prior knowledge or as discussion starters for deeper conceptual exploration. The variety of question formats, including multiple correct answers and mathematical derivations, allows for differentiated assessment that meets diverse learning needs while maintaining rigorous academic standards. This assessment aligns with Next Generation Science Standards HS-PS2-1, which requires students to analyze data to support the claim that Newton's second law accurately describes the relationship between net force, mass, and acceleration.
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13 questions
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1.
MULTIPLE SELECT QUESTION
15 mins • 1 pt
An astronaut jumps from the surface of the Moon and follows the path shown by the dashed line. At Point P, the astronaut is at the highest point in the path. Air resistance is negligible. Is the astronaut in equilibrium? Why or why not?
Yes, because the astronaut only has a constant horizontal velocity.
Yes, because the astronaut’s vertical velocity is instantaneously zero.
No, because the astronaut has a component of velocity in the horizontal direction.
No, because there is a downward net force on the astronaut.
Answer explanation
Tags
NGSS.HS-PS2-1
2.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
A person holds a string that is attached to a box. The person allows the box to move straight downward at a constant speed v, as shown. The weight of the box is 5 N. Which of the following claims about the tension in the string is correct?
The tension must be zero.
The tension must be less than 5 N but greater than zero.
The tension must be equal to 5 N.
The tension must be greater than 5 N.
Answer explanation
Tags
NGSS.HS-PS2-1
3.
MULTIPLE SELECT QUESTION
15 mins • 1 pt
Two applied forces are exerted on a box that slides along a flat surface with negligible friction. The force on the right is at an angle θ to the horizontal, as shown. The box is initially moving to the right at a constant velocity. The angle θ is then slowly decreased, but the magnitudes of the applied forces remain constant. Which of the following correctly describes the motion of the box as the angle is decreased?
The velocity of the box remains constant because the sum of all the forces on the box remains constant.
The velocity of the box remains constant because the horizontal components of the forces do not change.
The velocity of the box changes because the sum of the vertical components of the forces no longer equals zero.
The velocity of the box changes because the sum of the horizontal components of the forces no longer equals zero.
Answer explanation
Tags
NGSS.HS-PS2-1
4.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
At the instant shown, a paper airplane is moving horizontally to the right. The air exerts two independent forces on the plane: an upward lift force with magnitude less than the plane’s weight and a resistive drag force. Which of the following correctly describes the change in velocity, if any, of the plane in the horizontal and vertical directions at this instant?
No change | No change
No change | Increasing downward
Decreasing | No change
Decreasing | Increasing downward
Answer explanation
Tags
NGSS.HS-PS2-1
5.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
A puck slides freely across the horizontal surface of a frozen pond with negligible friction. Which of the following is true of the motion of the puck?
The puck will slide with a constant nonzero acceleration.
The puck will slide with a decreasing acceleration.
The puck will slide with an increasing acceleration.
The puck will slide with a constant velocity.
Answer explanation
Tags
NGSS.HS-PS2-1
6.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
A box moves with constant speed to the right along a horizontal surface with negligible friction. Two forces, with magnitudes F1 and F2, are exerted on the box at angles θ1 and θ2, respectively, as shown. Which of the following is a correct expression for F1 in terms of F2?
F1 = F2 (sin θ2 / cos θ1)
F1 = F2 (cos θ2 / sin θ1)
F1 = F2 (sin θ2 / sin θ1)
F1 = F2 (cos θ2 / cos θ1)
Answer explanation
Tags
NGSS.HS-PS2-1
7.
MULTIPLE CHOICE QUESTION
15 mins • 1 pt
A string of negligible mass is attached to a box of mass m. With the string taut, the box sits at rest on top of a scale, as shown. The scale has a force reading of FN. Which of the following must be a correct equation for the magnitude FT of the tension in the string?
FT = FN + mg
FT = FN
FT = mg
FT = mg - FN
Answer explanation
Tags
NGSS.HS-PS2-1
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