Physics — Semester A
Free Practice · 10 Questions · 20 min
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Question 1 of 10
TEKS 3Medium Calc Word Diagram

A 5.0-kg block on a horizontal surface is pushed with a horizontal applied force of 30 N. A frictional force of 12 N opposes motion. What is the block’s acceleration?

Horizontal push with friction (free-body view)m = 5.0 kgF = 30 Nf = 12 NNW = mg
A3.6 m/s²
B6.0 m/s²
C8.4 m/s²
D2.4 m/s²
Explanation
📌 Net force = 30 − 12 = 18 N right. a = F_net/m = 18/5.0 = 3.6 m/s². Distractor A ignored applied force; C ignored friction; D added forces instead of subtracting (sign error — friction opposes motion).
Question 2 of 10
TEKS 2Medium Calc Word Diagram

The position-vs-time graph shows an object's motion. Between t = 2 s and t = 6 s, which best describes the object's motion?

Position vs. time0246810t (s)0102030x (m)(2, 10)(6, 10)focus interval
AThe object is at rest.
BThe object is moving backward.
CThe object is accelerating.
DThe object is moving at constant velocity.
Explanation
📌 A horizontal line on a position-vs-time graph means position is not changing — the object is stationary (at rest). Distractor A confuses horizontal line with acceleration; B confuses horizontal with constant motion; D confuses horizontal (no motion) with negative slope (backward motion).
Question 3 of 10
TEKS 2Hard Calc Word Diagram

A projectile is launched from ground level with an initial speed of 20 m/s at 30° above horizontal. What is the maximum height reached? Use g = 9.8 m/s² and ignore air resistance.

Projectile launch: v0 = 20 m/s, θ = 30°v0 = 20 m/sθ = 30°apex (max height)
A20.4 m
B5.10 m
C15.3 m
D2.55 m
Explanation
📌 Vertical component v_iy = v sin θ = 20 × sin(30°) = 10 m/s. At max height v_y = 0. Use v_fy² − v_iy² = −2g·h → 0 − 100 = −19.6h → h = 5.10 m. Distractor A used v_iy/2 (halved velocity); C used cos(30°) instead of sin(30°) — the classic sin/cos swap; D ignored angle entirely, used full v² / 2g.
Question 4 of 10
TEKS 2Hard Calc Word Diagram

The velocity-vs-time graph shows the velocity of an object over a 10-second interval. The velocity rises linearly from 0 to 20 m/s over the first 4 seconds, holds constant at 20 m/s from 4 to 7 seconds, then decreases linearly to 0 by t = 10 s. What is the total displacement of the object from t = 0 to t = 10 s?

Velocity vs. time0246810t (s)05101520v (m/s)(4, 20)(7, 20)
A100 m
B140 m
C160 m
D200 m
Explanation
📌 Area under v-t graph = displacement. Segment 1 (0-4s): triangle ½ × 4 × 20 = 40 m. Segment 2 (4-7s): rectangle 3 × 20 = 60 m. Segment 3 (7-10s): triangle ½ × 3 × 20 = 30 m. Total = 40 + 60 + 30 = 140 m. Distractor A misses one segment; C treats segment 1 as rectangle; D treats all as rectangles (very common graph error).
Question 5 of 10
TEKS 1Easy Word

A student measures the mass of a sample as 0.00284 kg. Which expression correctly represents this mass in scientific notation with proper SI units?

A2.84 × 10⁻³ kg
B2.84 × 10⁻² g
C28.4 × 10⁻⁴ kg
D2.84 × 10³ kg
Explanation
📌 Move decimal 3 places right: 0.00284 → 2.84 × 10⁻³. Coefficient must satisfy 1 ≤ a < 10, and SI mass unit is kg. Distractor B has wrong exponent sign; C violates scientific notation convention; D uses wrong unit (2.84 × 10⁻² g = 0.0000284 kg, two orders off).
Question 6 of 10
TEKS 5Hard Calc Word Diagram

A 2.0-kg cart moving at 4.0 m/s collides with a stationary 3.0-kg cart. The two carts stick together after the collision. What is the velocity of the combined mass immediately after the collision?

Two carts collide and stick together on a horizontal trackBeforem1 = 2 kgv1 = 4 m/sm2 = 3 kg(at rest)After (stuck together)m1 + m2v′ = ?
A6.7 m/s
B1.6 m/s
C2.0 m/s
D4.0 m/s
Explanation
📌 Conservation of momentum (perfectly inelastic): m₁v₁ + m₂v₂ = (m₁+m₂)v_f. (2.0)(4.0) + 0 = (5.0)v_f → v_f = 1.6 m/s. Distractor B averaged velocities without mass weighting; C ignored mass conservation; D used wrong combined mass. Note: kinetic energy is NOT conserved here — 15 J is lost to heat, sound, and deformation.
Question 7 of 10
TEKS 3Easy Word

A force of 24 N is applied to a 6.0-kg object. Assuming no other forces act on it, what is the object’s acceleration?

A18 m/s²
B4.0 m/s²
C0.25 m/s²
D144 m/s²
Explanation
📌 F = ma → a = F/m = 24 / 6.0 = 4.0 m/s². Distractor A divided m by F (reciprocal); C subtracted; D multiplied.
Question 8 of 10
TEKS 2Easy Word

A cyclist travels 240 m in 30 s at constant velocity. What is her average velocity?

A7,200 m/s
B0.125 m/s
C8.0 m
D8.0 m/s
Explanation
📌 Average velocity v = d/t = 240 m / 30 s = 8.0 m/s. Distractor A multiplied (÷/× swap); C took reciprocal; D forgot velocity units (m instead of m/s).
Question 9 of 10
TEKS 4Easy Word

According to Newton’s law of universal gravitation, if the distance between two objects is doubled while their masses remain the same, the gravitational force between them becomes:

Two objects — distance changesBeforem₁m₂dAfter (distance doubled)m₁m₂2d
Adoubled
Bhalved
Cfour times the original
Done-fourth of the original
Explanation
📌 F = GmM/d². Force is inversely proportional to d². Double d → d² becomes 4d² → F becomes F/4. Distractor A treats it as direct proportion; B applies 1/d instead of 1/d² (forgot the square); D squares wrong direction (direct instead of inverse).
Question 10 of 10
TEKS 3Medium Calc Word Diagram

A 4.0-kg block slides down a frictionless ramp inclined at 30° to the horizontal. What is the magnitude of the block’s acceleration along the ramp? Use g = 9.8 m/s².

Block on frictionless incline (θ = 30°)30°m = 4 kgW = mgN
A8.5 m/s²
B4.9 m/s²
C9.8 m/s²
D2.0 m/s²
Explanation
📌 On a frictionless incline, acceleration along the ramp = g·sin θ = 9.8 × sin(30°) = 9.8 × 0.5 = 4.9 m/s². Mass cancels — same rate for any object at same angle. Distractor A unnecessarily divided by mass; C used cos(30°) instead of sin(30°) — the sin/cos swap; D used full g without decomposition.

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