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Relative velocity, motion in a plane, projectile and uniform circular motion: NEET MCQs with answers

89 questions on this topic of Kinematics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A particle moves in a circle of radius 5 m with a constant speed of 10 m/s. The magnitude of change in its velocity after it completes one-fourth of a revolution is:

  1. 0 m/s
  2. 10 m/s
  3. 10√2 m/s
  4. 20 m/s
Show answer and explanation

Answer: (C) 10√2 m/s

In one-fourth of a revolution, the velocity vector turns through 90°. The magnitude of change in velocity is |Δv⃗| = 2v sin(θ/2) = 2(10) sin(45°) = 20(1/√2) = 10√2 m/s.

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2. A ball is projected with speed 20 m/s at 90° to the horizontal. Taking g = 10 m/s², the maximum height it reaches is:

  1. 30 m
  2. 20 m
  3. 10 m
  4. 40 m
Show answer and explanation

Answer: (B) 20 m

H = u² sin²θ / 2g = 20² × (sin 90°)² / 20 = 20 m.

hard

3. A particle is projected from the ground with an initial speed u at an angle θ to the horizontal. The angle subtended by the projectile at the point of projection when it is at the highest point of its trajectory is α. The relationship between tan α and tan θ is:

  1. tan α = 2 tan θ
  2. tan α = tan θ
  3. tan α = (1/4) tan θ
  4. tan α = (1/2) tan θ
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Answer: (D) tan α = (1/2) tan θ

At the highest point, the horizontal distance from the point of projection is x = R / 2 and the vertical height is y = H. The angle of elevation α is given by tan α = H / (R / 2) = 2H / R. Since H = (u² sin² θ) / (2g) and R = (2u² sin θ cos θ) / g, we have H / R = (1/4) tan θ. Thus, tan α = 2 × (1/4) tan θ = (1/2) tan θ.

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4. A projectile is fired with speed 30 m/s at 30° above the horizontal from level ground. Taking g = 10 m/s², its horizontal range is:

  1. 45 m
  2. 77.9 m
  3. 39 m
  4. 90 m
Show answer and explanation

Answer: (B) 77.9 m

R = u² sin 2θ / g = 30² × sin 60° / 10 = 77.9 m.

easy

5. Given below are two statements: Statement I: In uniform circular motion, the tangential acceleration of the particle is zero. Statement II: In uniform circular motion, the total acceleration of the particle is always directed towards the center of the circle. In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

In uniform circular motion, speed is constant, so tangential acceleration a_t = dv/dt = 0. Consequently, the only acceleration acting on the particle is the radial (centripetal) acceleration, which is always directed towards the center. Thus, both statements are correct.

hard

6. A stone is thrown horizontally with velocity 15 m/s from the top of a tower. Taking g = 10 m/s², the radius of curvature of its path at t = 2 s after launch is:

  1. 125 m
  2. 104.2 m
  3. 62.5 m
  4. 83.3 m
Show answer and explanation

Answer: (B) 104.2 m

At t = 2 s, v_x = 15 m/s, v_y = gt = 20 m/s. The speed is v = √(15² + 20²) = 25 m/s. The angle with horizontal is tan θ = 20/15 = 4/3, so cos θ = 3/5. The normal component of acceleration is a_n = g cos θ = 10 × (3/5) = 6 m/s². The radius of curvature is ρ = v² / a_n = 25² / 6 = 625 / 6 ≈ 104.2 m.

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7. Two particles A and B are located at points (0, 0) and (0, 40 m) respectively at t = 0. Particle A moves along the positive x-axis with a constant speed of 8 m/s, and particle B moves along the negative y-axis with a constant speed of 6 m/s. The minimum distance between them during their motion is:

  1. 24 m
  2. 32 m
  3. 20 m
  4. 16 m
Show answer and explanation

Answer: (B) 32 m

At time t, the position vectors are r_A = 8t î and r_B = (40 - 6t) ĵ. The square of distance is d² = (8t)² + (40 - 6t)² = 64t² + 1600 - 480t + 36t² = 100t² - 480t + 1600. Minimizing with respect to t: d(d²)/dt = 200t - 480 = 0, giving t = 2.4 s. At t = 2.4 s, d² = 100(5.76) - 480(2.4) + 1600 = 576 - 1152 + 1600 = 1024, so d_min = √1024 = 32 m.

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8. A ball is projected with speed 30 m/s at 60° to the horizontal. Taking g = 10 m/s², the maximum height it reaches is:

  1. 45 m
  2. 33.7 m
  3. 39 m
  4. 67.5 m
Show answer and explanation

Answer: (B) 33.7 m

H = u² sin²θ / 2g = 30² × (sin 60°)² / 20 = 33.7 m.

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9. A projectile is fired with speed 50 m/s at 45° above the horizontal from level ground. Taking g = 10 m/s², its horizontal range is:

  1. 500 m
  2. 177 m
  3. 125 m
  4. 250 m
Show answer and explanation

Answer: (D) 250 m

R = u² sin 2θ / g = 50² × sin 90° / 10 = 250 m.

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10. A projectile is fired with speed 50 m/s at 60° above the horizontal from level ground. Taking g = 10 m/s², its horizontal range is:

  1. 433 m
  2. 250 m
  3. 108 m
  4. 217 m
Show answer and explanation

Answer: (D) 217 m

R = u² sin 2θ / g = 50² × sin 120° / 10 = 217 m.

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11. Rain drops are falling vertically with a speed of 20 m/s. A person runs horizontally towards the west with a speed of 20 m/s. To protect himself from the rain, he should hold his umbrella at an angle of:

  1. 45° with the vertical towards the east
  2. 45° with the vertical towards the west
  3. 30° with the vertical towards the west
  4. 60° with the vertical towards the east
Show answer and explanation

Answer: (B) 45° with the vertical towards the west

The velocity of rain relative to the person is v⃗_rm = v⃗_r - v⃗_m = -20 ĵ - (-20 î) = 20 î - 20 ĵ. Here +î is east and -î is west; since the man moves west (-20 î), the relative velocity has a horizontal component towards the east (+20 î) when viewed mathematically, meaning rain appears to come from the west. Thus, tan θ = |v_m| / |v_r| = 20 / 20 = 1, so θ = 45° with the vertical, tilted towards the direction of motion (west).

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12. A particle is executing uniform circular motion in a circle of radius R with constant speed v. What is the magnitude of the average acceleration over a time interval during which the particle completes a quarter of a revolution?

  1. 2√2 v² / (πR)
  2. 2v² / (πR)
  3. √2 v² / (πR)
  4. v² / (2πR)
Show answer and explanation

Answer: (A) 2√2 v² / (πR)

For a quarter revolution, the time taken is Δt = (πR/2) / v = πR / (2v). The initial and final velocity vectors are perpendicular to each other, so the magnitude of change in velocity is |Δv| = √(v² + v²) = √2 v. Thus, the average acceleration is |a_avg| = |Δv| / Δt = (√2 v) / [πR / (2v)] = 2√2 v² / (πR).

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