Class XI · Chapter 03
Motion in a Plane — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering vector addition, projectile motion, and uniform circular motion.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. At the highest point of a projectile’s path, what is the direction of its velocity relative to its acceleration?
Answer
At the highest point, the vertical velocity component is momentarily zero, leaving only the horizontal component — velocity is horizontal. Acceleration (g) is always vertical, downward. So velocity is exactly perpendicular to acceleration at that instant.
At the highest point, the vertical velocity component is momentarily zero, leaving only the horizontal component — velocity is horizontal. Acceleration (g) is always vertical, downward. So velocity is exactly perpendicular to acceleration at that instant.
Q2. Derive an expression for the time of flight of a projectile launched with speed u at an angle θ above the horizontal.
Answer
Vertical motion: initial vertical velocity = u sinθ, acceleration = −g. The projectile returns to the launch height when net vertical displacement is zero:
0 = (u sinθ)t − ½gt² → t[u sinθ − ½gt] = 0.
The nonzero solution: t = 2u sinθ / g.
Vertical motion: initial vertical velocity = u sinθ, acceleration = −g. The projectile returns to the launch height when net vertical displacement is zero:
0 = (u sinθ)t − ½gt² → t[u sinθ − ½gt] = 0.
The nonzero solution: t = 2u sinθ / g.
Q3. (a) Derive the expression for the horizontal range of a projectile, and show that range is maximum at a launch angle of 45°. (b) A ball is projected at 30° with a speed of 20 m/s. Find its horizontal range. (g = 10 m/s²)
Answer
(a) R = (horizontal velocity) × (time of flight) = (u cosθ)(2u sinθ/g) = u²(2 sinθ cosθ)/g = u² sin(2θ)/g.
Since sin(2θ) has a maximum value of 1 at 2θ = 90° (θ = 45°), range is maximum at exactly 45°.
(a) R = (horizontal velocity) × (time of flight) = (u cosθ)(2u sinθ/g) = u²(2 sinθ cosθ)/g = u² sin(2θ)/g.
Since sin(2θ) has a maximum value of 1 at 2θ = 90° (θ = 45°), range is maximum at exactly 45°.
(b) R = u²sin(2θ)/g = (400 × sin60°)/10 = (400 × 0.866)/10 ≈ 34.6 m.
2NEET Style (Single-Correct MCQ)
Q4. Two vectors of equal magnitude A act at an angle of 120° to each other. The magnitude of their resultant is:
Solution
R = √(A²+A²+2A²cos120°) = √(2A² − A²) = √(A²) = A.
R = √(A²+A²+2A²cos120°) = √(2A² − A²) = √(A²) = A.
Q5. Centripetal acceleration of an object in uniform circular motion is given by:
Solution
ac = v²/r = ω²r, always directed toward the centre of the circular path.
ac = v²/r = ω²r, always directed toward the centre of the circular path.
Q6. The maximum height reached by a projectile depends on:
Solution
H = u²sin²θ/(2g) — depends only on u sinθ, the vertical (launch-angle-dependent) component of velocity.
H = u²sin²θ/(2g) — depends only on u sinθ, the vertical (launch-angle-dependent) component of velocity.
Q7. The velocity of rain, as observed by a person walking, is found using:
Solution
Relative velocity of rain with respect to the observer: vrain,observer = vrain − vobserver, computed as a full vector subtraction, not just scalar speeds.
Relative velocity of rain with respect to the observer: vrain,observer = vrain − vobserver, computed as a full vector subtraction, not just scalar speeds.
3JEE Main Style
Q8. Given vectors A = 3î + 4ĵ and B = î − 2ĵ, find the magnitude of A + B.
Solution
A + B = 4î + 2ĵ. Magnitude = √(4² + 2²) = √20 ≈ 4.47.
A + B = 4î + 2ĵ. Magnitude = √(4² + 2²) = √20 ≈ 4.47.
Q9. For a projectile launched at angle θ, the horizontal range is the same as for a launch angle of:
Solution
R = u²sin(2θ)/g. Since sin(2θ) = sin(180°−2θ) = sin(2(90°−θ)), angles θ and (90°−θ) give identical range, though different time of flight and maximum height.
R = u²sin(2θ)/g. Since sin(2θ) = sin(180°−2θ) = sin(2(90°−θ)), angles θ and (90°−θ) give identical range, though different time of flight and maximum height.
Q10. An object moves in a circle of radius 2 m at a constant speed of 4 m/s. Find its centripetal acceleration.
Solution
ac = v²/r = 16/2 = 8 m/s².
ac = v²/r = 16/2 = 8 m/s².
You must be logged in to post a comment.