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100+ Free Physics Level 4 Practice Questions
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Key Facts: Physics Level 4 Exam
Master TASC Physics Level 4 (PHY415115) with 100 syllabus-aligned practice questions and worked physics calculations covering classical mechanics, circular motion & gravitation, electromagnetism, waves & optics, and modern physics. These practice questions are an English-language multiple-choice study aid for revising course knowledge and are not an official TASC paper or a simulation of the written external examination format.
Sample Physics Level 4 Practice Questions
Try these sample questions to test your Physics Level 4 exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A projectile is launched horizontally from a cliff of height h = 19.6 m with an initial horizontal velocity of v_x = 15.0 m/s. Assuming g = 9.80 m/s² and negligible air resistance, what is the horizontal distance (range) traveled by the projectile before impact?
A.15.0 m
B.30.0 m
C.45.0 m
D.60.0 m
Explanation: The vertical motion is governed by h = 0.5 * g * t². Solving for time: t = sqrt(2h / g) = sqrt((2 * 19.6) / 9.80) = sqrt(4.00) = 2.00 s. The horizontal distance traveled is x = v_x * t = 15.0 m/s * 2.00 s = 30.0 m.
2A cannonball is launched with an initial speed of v_0 = 40.0 m/s at an angle of 30.0° above the horizontal over level ground. Taking g = 9.80 m/s², calculate the maximum height reached above the launch point.
A.10.2 m
B.20.4 m
C.40.8 m
D.81.6 m
Explanation: The initial vertical component of velocity is v_0y = v_0 * sin(30.0°) = 40.0 * 0.500 = 20.0 m/s. At maximum height, v_y = 0. Using v_y² = v_0y² - 2g * h_max gives h_max = v_0y² / (2g) = (20.0)² / (2 * 9.80) = 400 / 19.6 = 20.4 m.
3A soccer ball is kicked with initial velocity v_0 = 28.3 m/s at 45.0° to horizontal on level ground. What is the total time of flight before the ball lands (g = 9.80 m/s²)?
A.2.04 s
B.2.89 s
C.4.08 s
D.5.77 s
Explanation: The initial vertical velocity is v_0y = 28.3 * sin(45.0°) = 20.0 m/s. Time to reach maximum height is t_up = v_0y / g = 20.0 / 9.80 = 2.04 s. Total time of flight is T = 2 * t_up = 2 * 2.04 = 4.08 s.
4A heavy sign of mass m = 12.0 kg hangs in static equilibrium suspended by two identical light cables, each making an angle of 35.0° with the horizontal ceiling. What is the tension force T in each cable (g = 9.80 m/s²)?
A.58.8 N
B.71.8 N
C.102.5 N
D.205.0 N
Explanation: For vertical equilibrium, the sum of vertical tension forces must equal weight: 2 * T * sin(35.0°) = m * g = 12.0 * 9.80 = 117.6 N. Solving for T: T = 117.6 / (2 * sin(35.0°)) = 117.6 / (2 * 0.5736) = 117.6 / 1.147 = 102.5 N.
5A block of mass m = 5.0 kg slides down a frictionless incline inclined at 30.0° to horizontal. What is the magnitude of the block's acceleration down the ramp (g = 9.80 m/s²)?
A.2.45 m/s²
B.4.90 m/s²
C.8.49 m/s²
D.9.80 m/s²
Explanation: The component of gravitational force parallel to the incline is F_parallel = m * g * sin(30.0°). By Newton's second law, a = F_parallel / m = g * sin(30.0°) = 9.80 * 0.500 = 4.90 m/s².
6An 8.0 kg wooden block slides down a ramp angled at 25.0° above horizontal. The coefficient of kinetic friction between the block and ramp is μ_k = 0.20. Calculate the net acceleration of the block down the ramp (g = 9.80 m/s²).
A.1.78 m/s²
B.2.36 m/s²
C.4.14 m/s²
D.5.92 m/s²
Explanation: Parallel force F_g = m * g * sin(25.0°) = 8.0 * 9.80 * 0.4226 = 33.13 N. Normal force N = m * g * cos(25.0°) = 8.0 * 9.80 * 0.9063 = 71.05 N. Friction force f_k = μ_k * N = 0.20 * 71.05 = 14.21 N. Net force F_net = 33.13 - 14.21 = 18.92 N. Acceleration a = 18.92 / 8.0 = 2.36 m/s².
7A force F = 50.0 N is applied to pull a sled across a horizontal floor at an angle of 60.0° above the horizontal over a displacement of d = 10.0 m. How much work is done by the applied force on the sled?
A.250 J
B.433 J
C.500 J
D.866 J
Explanation: Work is defined as W = F * d * cos(θ). Here W = 50.0 N * 10.0 m * cos(60.0°) = 500 * 0.500 = 250 J.
8A 2.0 kg block starts from rest and slides down a frictionless curve from a height h = 5.0 m onto a horizontal frictionless track, where it strikes a spring of spring constant k = 400 N/m. What is the maximum compression x of the spring (g = 9.80 m/s²)?
A.0.35 m
B.0.49 m
C.0.70 m
D.0.98 m
Explanation: By conservation of mechanical energy, potential energy at top equals elastic potential energy at maximum compression: m * g * h = 0.5 * k * x². Substituting values: 2.0 * 9.80 * 5.0 = 98.0 J. Thus 0.5 * 400 * x² = 98.0 => 200 * x² = 98.0 => x² = 0.49 => x = 0.70 m.
9A 0.15 kg baseball traveling horizontally east at 30.0 m/s is struck by a bat and leaves in the opposite direction (west) at 40.0 m/s. If the impact duration is Δt = 0.0050 s, what average force does the bat exert on the ball?
A.300 N
B.1500 N
C.2100 N
D.4200 N
Explanation: Choosing east as positive: v_i = +30.0 m/s and v_f = -40.0 m/s. Change in momentum Δp = m(v_f - v_i) = 0.15 * (-40.0 - 30.0) = 0.15 * (-70.0) = -10.5 kg m/s. Magnitude of average force F_avg = |Δp| / Δt = 10.5 / 0.0050 = 2100 N.
10Car A (mass 1000 kg) is traveling east at 20.0 m/s when it collides at an intersection with Car B (mass 1500 kg) traveling north at 15.0 m/s. The two vehicles lock together upon impact. What is the speed of the combined wreck immediately after collision?
A.8.00 m/s
B.12.0 m/s
C.17.0 m/s
D.25.0 m/s
Explanation: Conservation of momentum in 2D: p_x = m_A * v_A = 1000 * 20.0 = 20000 kg m/s (East). p_y = m_B * v_B = 1500 * 15.0 = 22500 kg m/s (North). Total momentum p_total = sqrt(p_x² + p_y²) = sqrt(20000² + 22500²) = sqrt(400000000 + 506250000) = sqrt(906250000) = 30104 kg m/s. Combined mass = 1000 + 1500 = 2500 kg. Final speed v_f = 30104 / 2500 = 12.04 m/s ≈ 12.0 m/s.
About the Physics Level 4 Practice Questions
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