MCAT Physics Exam 2025/2026 – Complete Questions with Detailed
Answers, Explanations, and Calculations | Graded A+ | Guaranteed
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1. A car of mass 1200 kg accelerates from rest to 25 m/s in 10 s. What is the average net
force on the car?
A) 2500 N
B) 3000 N
C) 3500 N
D) 4000 N
Answer: B) 3000 N
Calculation: a = Δv/Δt = 25/10 = 2.5 m/s². F = ma = 1200×2.5 = 3000 N.
2. A 0.5 kg ball is thrown upward at 20 m/s. What is the maximum height? (g = 9.8 m/s²)
A) 15 m
B) 20 m
C) 25 m
D) 30 m
Answer: B) 20 m (≈20.4 m)
Calculation: v² = v0² − 2gh → h = v0²/(2g) = 20²/(2×9.8) = 400/19.6 ≈ 20.41 m.
3. Spring with k = 200 N/m compressed by 0.1 m. Potential energy stored?
A) 0.5 J
B) 1.0 J
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C) 2.0 J
D) 4.0 J
Answer: B) 1.0 J
Calculation: U = 1/2 k x² = 0.5×200×0.1² = 1.0 J.
4. Two resistors 6 Ω and 12 Ω in parallel. Equivalent resistance?
A) 4 Ω
B) 6 Ω
C) 8 Ω
D) 18 Ω
Answer: A) 4 Ω
Calculation: 1/R = 1/6 + 1/12 = 2/12 +1/12 =3/12 → R = 12/3 =4 Ω.
5. Light from air (n=1.00) enters glass (n=1.50) at 30°. Angle of refraction?
A) 19.5°
B) 20°
C) 25°
D) 30°
Answer: A) 19.5°
Calculation: n1 sinθ1 = n2 sinθ2 → sinθ2 = (1.00/1.50) sin30° = (2/3)(0.5)=1/3 → θ2 =
arcsin(0.333)=19.47°.
6. A 2 kg block sliding on frictionless surface collides elastically with stationary 1 kg block.
After collision, speed of 2 kg block?
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A) 0 m/s
B) 1 m/s
C) 2/3 m/s
D) 4/3 m/s
Answer: C) 2/3 m/s
Calculation: For 1D elastic: v1' = (m1−m2)/(m1+m2) v1 = (2−1)/(3) v = (1/3)×v. If v initial =1
(assume 1 m/s standard MCAT setup) then v1' =1/3; but question missing initial speed. Assume
v0=1 m/s → v1' = (2−1)/(2+1)×1 =1/3 m/s. Note: If initial speed not given, treat symbolically. (If
v0 unspecified, answer depends on v0.)
7. A car rounds a curve of radius 50 m at 20 m/s. What is centripetal acceleration?
A) 4 m/s²
B) 6 m/s²
C) 8 m/s²
D) 10 m/s²
Answer: A) 8 m/s²
Calculation: ac = v²/r = 20²/50 = 400/50 = 8 m/s².
8. 0.1 A flows through 100 Ω resistor. Power dissipated?
A) 0.1 W
B) 1 W
C) 10 W
D) 100 W
Answer: B) 1 W
Calculation: P = I²R = 0.1²×100 = 0.01×100 =1 W.
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9. A gas undergoes isothermal expansion at 300 K, absorbing 500 J heat. Work done by
gas?
A) 0 J
B) 250 J
C) 500 J
D) 1000 J
Answer: C) 500 J
Calculation: For isothermal ideal gas, ΔU = 0 so Q = W. Thus W = Q = 500 J.
10. A proton (q=+e) moves through uniform E-field 1×10^4 N/C over 0.02 m. Change in
potential energy?
A) 3.2×10^-17 J
B) 1.6×10^-19 J
C) 3.2×10^-19 J
D) 6.4×10^-19 J
Answer: A) 3.2×10^-17 J
Calculation: ΔV = E·d =1×10^4×0.02 =200 V. ΔU = qΔV = (1.602×10^-19 C)×200 ≈3.204×10^-17 J.
11. Two waves of same frequency interfere; path difference = λ/2. Resulting amplitude?
A) Constructive
B) Destructive
C) Partially constructive
D) Beats
Answer: B) Destructive
Calculation: Path difference λ/2 → phase difference π → destructive interference (minimum
amplitude).