Physics — Semester A Practice

Question 1 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.

A20.4 m

B2.55 m

C5.10 m

D15.3 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 2 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

B8.0 m/s

C0.125 m/s

D8.0 m

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 3 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?

A2.0 m/s

B6.7 m/s

C1.6 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 4 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.

A8.5 m/s²

B2.0 m/s²

C9.8 m/s²

D4.9 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.

Question 5 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:

Afour times the original

Bone-fourth of the original

Cdoubled

Dhalved

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 6 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?

A144 m/s²

B18 m/s²

C4.0 m/s²

D0.25 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 7 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 × 10³ kg

B2.84 × 10⁻² g

C28.4 × 10⁻⁴ kg

D2.84 × 10⁻³ 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 8 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?

A2.4 m/s²

B3.6 m/s²

C8.4 m/s²

D6.0 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 9 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?

A100 m

B200 m

C160 m

D140 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 10 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?

AThe object is moving at constant velocity.

BThe object is moving backward.

CThe object is accelerating.

DThe object is at rest.

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).

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