NEET Motion in a Plane PYQs & Practice Questions

10 free NEET Physics questions on Motion in a Plane, starting with previous year questions (2016–2024). Tap “Show answer” to check each one. No login needed.

Topics: Vector Algebra · Scalars and Vectors · Vector Components · Projectile Motion · Uniform Circular Motion - Velocity and Acceleration · Change in Velocity Direction · Projectile Motion - Maximum Height · Circular Motion - Tangential and Radial Acceleration · Relative Velocity - River Boat Problems · Circular Motion · Circular Motion - Projection · Projectile Motion - Speed at Highest Point

1. Q1NEET 2024Uniform Circular Motion - Velocity and Acceleration

   A particle moving with uniform speed in a circular path maintains:

   • A. constant velocity but varying acceleration.
   • B. varying velocity and varying acceleration.
   • C. constant velocity.
   • D. constant acceleration.
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   Answer: A — constant velocity but varying acceleration.

   In uniform circular motion, the direction of velocity changes continuously even though speed is constant, so velocity is not constant. Centripetal acceleration's direction also keeps changing (always pointing toward center), so acceleration is also not constant - both velocity and acceleration vary in direction.

2. Q2NEET 2023Change in Velocity Direction

   A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is:

   • A. along northward
   • B. along north-east
   • C. along south-west
   • D. along eastward
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   Answer: A — along northward

   Change in velocity ΔV = v(east) - v(south) = vî - (-vĵ) = vî + vĵ, which points in the north-east direction. Since force is in the direction of change in momentum/velocity, the force acts along north-east.

3. Q3NEET 2023Projectile Motion - Maximum Height

   A bullet is fired from a gun at the speed of 280 m/s in the direction 30° above the horizontal. The maximum height attained by the bullet is (g = 9.8 ms^-2, sin 30° = 0.5):

   • A. 2000 m
   • B. 1000 m
   • C. 3000 m
   • D. 2800 m
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   Answer: A — 2000 m

   Maximum height H = u²sin²θ/(2g) = (280×280×0.25)/(2×9.8) = 19600/19.6 = 1000 m.

4. Q4NEET 2022Projectile Motion - Speed at Highest Point

   A ball is projected with a velocity, 10 m/s, at an angle of 60 degrees with the vertical direction. Its speed at the highest point of its trajectory will be:

   • A. Zero
   • B. 5*sqrt(3) m/s
   • C. 5 m/s
   • D. 10 m/s
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   Answer: A — Zero

   Angle with vertical is 60 degrees, so angle with horizontal is 30 degrees. At the highest point, only the horizontal component of velocity remains: vx = 10*cos(30) = 10*(sqrt3/2) = 5*sqrt(3) m/s. (Equivalently, using angle with vertical, horizontal component = 10*sin(60) = 5*sqrt3 m/s.)

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5. Q5NEET 2022Projectile Motion

   A ball is projected with a velocity, 10 ms⁻¹, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be

   • A. Zero
   • B. 5√3 ms⁻¹
   • C. 5 ms⁻¹
   • D. 10 ms⁻¹
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   Answer: A — Zero

   Angle with vertical = 60°, so angle with horizontal = 30°. At highest point, speed = horizontal component = u·cos(30°) = 10×(√3/2) = 5√3 ms⁻¹.

6. Q6NEET 2019Relative Velocity - River Boat Problems

   The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is given by:

   • A. 60 degrees west
   • B. 45 degrees west
   • C. 30 degrees west
   • D. 0 degrees
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   Answer: B — 45 degrees west

   For shortest path crossing, the swimmer must angle upstream so the westward component of his velocity cancels the eastward river flow. sin(theta) = VR/VBR = 10/20 = 0.5, so theta = 30 degrees west of north.

7. Q7NEET 2019Circular Motion

   Two particles A and B are moving in uniform circular motion in concentric circles of radii rA and rB with speeds VA and VB respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be:

   • A. rB:rA
   • B. 1:1
   • C. rA:rB
   • D. VA:VB
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   Answer: A — rB:rA

   Angular speed omega = 2*pi/T. Since both particles have the same time period T, their angular speeds are equal, giving a ratio of 1:1.

8. Q8NEET 2019Circular Motion - Projection

   The radius of a circle, the period of revolution, initial position, and sense of revolution are indicated in the figure. A particle P is rotating on a circle of radius 3 m, starting at t=0 on the positive x-axis, moving counter-clockwise, with time period T = 4s. The y-projection of the radius vector of rotating particle P is:

   • A. y(t) = 3 cos(3*pi*t/2), where y in m
   • B. y(t) = 3 cos(pi*t/2), where y in m
   • C. y(t) = -3 cos(2*pi*t), where y in m
   • D. y(t) = 4 sin(pi*t/2), where y in m
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   Answer: A — y(t) = 3 cos(3*pi*t/2), where y in m

   Angular frequency omega = 2*pi/T = 2*pi/4 = pi/2 rad/s. The particle starts on the x-axis at t=0, so y-projection follows y = r*cos(omega*t) = 3*cos(pi*t/2).

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9. Q9NEET 2017Acceleration from position coordinates

   The x and y coordinates of the particle at any time are x = 5t − 2t² and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is:

   • A. 5 m/s²
   • B. −4 m/s²
   • C. −8 m/s²
   • D. 0
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   Answer: A — 5 m/s²

   vx = dx/dt = 5−4t, ax = dvx/dt = −4 m/s² (constant). vy = 10, ay = 0. Net acceleration = √(ax²+ay²) = √((−4)²+0²) = 4 m/s², directed as −4 m/s² along x.

10. Q10NEET 2016Circular Motion - Tangential and Radial Acceleration

    In the given figure, a = 15 m/s² represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 2.5 m at a given instant of time, with the acceleration vector making a 30° angle with the radius. The speed of the particle is:

    • A. 6.2 m/s
    • B. 4.5 m/s
    • C. 5.0 m/s
    • D. 5.7 m/s
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    Answer: C — 5.0 m/s

    Total acceleration a_r² + a_t² = 225 (15²). Since tan30° = a_t/a_r, a_r = a_t√3, giving (4/3)a_r² = 225, so a_r² = 168.75, a_r = 7.5√3/... solving: a_r = (15/2)√3 ≈ 12.99 m/s². Using v²/R = a_r: v² = 15×2.5×√3/2 ≈ 32.48, v ≈ 5.7 m/s.