Chapter test: Motion in a Plane
30 questions · Solve each question on paper first, then open "Show answer" to check your working.
Q1.
A particle moves from the point with position vector r₁ = 2î + 3ĵ (m) to r₂ = 5î + 7ĵ (m). Find the magnitude of its displacement in metres.Show answer
5 m — Δr = (5−2)î + (7−3)ĵ = 3î + 4ĵ → √(9+16) = 5
Q2.
Two forces of magnitudes 6 and 8 act perpendicularly. Find the magnitude of their resultant.Show answer
10 — θ = 90° → R = √(6² + 8²) = 10 (6-8-10)
Q3.
A 100 N force acts at 37° above the horizontal. Find its horizontal component in newtons.Show answer
80 N — F cos 37° = 100 × 0.8
Q4.
A ball is projected at 20 m/s at 30° above level ground (g = 10 m s⁻²). Find its time of flight in seconds.Show answer
2 s — T = 2u sin θ/g = 2 × 20 × 0.5/10
Q5.
For the same projection (u = 20 m/s, θ = 30°, g = 10), find the maximum height in metres.Show answer
5 m — H = (u sin θ)²/2g = 10²/20
Q6.
A car rounds a circle of radius 50 m at a constant 10 m/s. Find the magnitude of its centripetal acceleration in m s⁻².Show answer
2 m s⁻² — a = v²/r = 100/50, toward the centre
Q7.
Rain falls vertically at 35 m/s while a man walks horizontally at 12 m/s. Find the speed of rain relative to the man in m s⁻¹.Show answer
37 m s⁻¹ — √(12² + 35²) = √1369 (12-35-37 triple)
Q8.
A river 100 m wide flows at 3 m/s; a boat moves at 5 m/s in still water. Aiming for the shortest time, find the crossing time in seconds.Show answer
20 s — Shortest time: head straight across, t = 100/5
Q9.
The vector 4î − 3ĵ has magnitude and direction (from +x axis): (a) 5, 36.9° (b) 5, −36.9° (c) 7, −36.9° (d) 5, 53.1°Show answer
(b) — |v| = 5; components (+, −) → quadrant IV → −36.9° (323.1°)
Q10.
If |A + B| = |A − B| for nonzero A and B, the angle between them is: (a) 0° (b) 45° (c) 90° (d) 180°Show answer
(c) — Equating the ± diagonal formulas forces cos θ = 0
Q11.
At the highest point of a projectile's path (launched at θ ≠ 90°): (a) v = 0, a = 0 (b) v = 0, a = g (c) v = u cos θ horizontal, a = g downward (d) v = u sin θ verticalShow answer
(c) — vy = 0 at the top; vx = u cos θ survives; a = g down
Q12.
Two projectiles launched with equal speed at θ and 90° − θ (same level) always have the same range. They differ in: (a) nothing at all (b) range only (c) time of flight and maximum height (d) horizontal velocity componentShow answer
(c) — sin 2θ equal → R equal; T ∝ sin θ, H ∝ sin²θ differ
Q13.
In uniform circular motion, which quantity remains constant? (a) velocity vector (b) speed (c) acceleration vector (d) displacementShow answer
(b) — Speed constant; velocity and acceleration vectors rotate
Q14.
For A = 2î + 3ĵ and B = 4î − ĵ, which pair is correct? (a) A·B = 11, A×B = 14k̂ (b) A·B = 5, A×B = −14k̂ (c) A·B = 5, A×B = 14k̂ (d) A·B = −5, A×B = −14k̂Show answer
(b) — A·B = 8 − 3 = 5; A×B = (2×(−1) − 3×4)k̂ = −14k̂
Q15.
A ball is projected horizontally from a height. Which graph is a straight line? (a) y vs t (b) y vs x (c) x vs t (d) speed vs tShow answer
(c) — x = ut is linear in t; y vs t and y vs x are parabolas
Q16.
A boat (5 m/s in still water) crosses a 3 m/s current with zero drift. Its heading, from the perpendicular, is upstream at: (a) 53° (b) 37° (c) 30° (d) 60°Show answer
(b) — sin φ = 3/5 = 0.6 → φ = 37° upstream
Q17.
Assertion: In uniform circular motion the acceleration is directed toward the centre. Reason: The acceleration changes only the direction of the velocity, not its magnitude.Show answer
(a) — Central acceleration bends velocity without changing |v| — R explains A
Q18.
Assertion: Projectiles launched at θ and 90° − θ with equal speeds have the same horizontal range. Reason: They have the same maximum height.Show answer
(c) — R equal (sin 2θ equal); H ∝ sin²θ → heights differ, R is false
Q19.
Statement I: A boat whose still-water speed is less than the river's current cannot cross with zero drift. Statement II: Zero drift requires sin φ = vw/vbw, which must not exceed 1.Show answer
(a) — II supplies exactly the ≤ 1 condition that makes I true
Q20.
Statement I: At the highest point of a projectile's path its acceleration is zero. Statement II: At the highest point the vertical component of its velocity is zero.Show answer
(d) — I false (a = g at the apex); II true (vy = 0 there)
B1.
[Pattern mirror: NEET vector-angle style] The angle between A = î + ĵ and B = ĵ + k̂ is: (a) 30° (b) 45° (c) 60° (d) 90°Show answer
(c) — A·B = 1; |A| = |B| = √2 → cos θ = 1/2 → 60°
B2.
[Pattern mirror: JEE Main resultant style] Two equal vectors of magnitude A have a resultant also equal to A. The angle between them is: (a) 60° (b) 90° (c) 120° (d) 150°Show answer
(c) — R = 2A cos(θ/2) = A → cos(θ/2) = 1/2 → θ = 120°
B3.
[Pattern mirror: NEET projectile style] A projectile rises to a maximum height of 20 m (g = 10 m s⁻²). Its time of flight on level ground is: (a) 2 s (b) 2√2 s (c) 4 s (d) 4√2 sShow answer
(c) — T = 2√(2H/g) = 2√(4) = 4 s
B4.
[Pattern mirror: JEE Main river-boat style] A river is 200 m wide and flows at 2 m/s. A boat capable of 4 m/s in still water crosses in the shortest time. That time is: (a) 40 s (b) 50 s (c) 57.7 s (d) 66.7 sShow answer
(b) — Straight across: t = 200/4 = 50 s (drift = 2 × 50 = 100 m)
B5.
[Pattern mirror: NEET rain-man style] Rain falls vertically at 8 m/s; a cyclist moves at 6 m/s. The umbrella should make with the vertical an angle of about: (a) 37°, tilted forward (b) 37°, tilted backward (c) 53°, tilted forward (d) 0°Show answer
(a) — tan θ = 6/8 → θ ≈ 36.9° ≈ 37°, tilted forward (rain from ahead)
B6.
[Pattern mirror: JEE Main UCM scaling style] In UCM the speed doubles and the radius halves. The centripetal acceleration becomes: (a) 2a (b) 4a (c) 8a (d) a/2Show answer
(c) — a = v²/r → factor (2²) ÷ (1/2) = 4/0.5 = 8 → 8a
B7.
[Pattern mirror: NEET component style] A 10 N force makes 60° with the x-axis. Its y-component is: (a) 5 N (b) 5√3 N (c) 10√3 N (d) 10/√2 NShow answer
(b) — F sin 60° = 10 × (√3/2) = 5√3 ≈ 8.66 N
B8.
[Pattern mirror: JEE Main angular-quantity style] The angular velocity of the seconds hand of a watch is: (a) 2π rad/s (b) π/6 rad/s (c) π/30 rad/s (d) π/60 rad/sShow answer
(c) — 2π radians per 60 s → π/30 ≈ 0.105 rad/s
B9.
[Pattern mirror: NEET r(t) style] For r = 3tî + (4t − 2t²)ĵ (SI), the speed at t = 1 s is: (a) 5 m/s (b) 3 m/s (c) 4 m/s (d) 0Show answer
(b) — v = 3î + (4 − 4t)ĵ; at t = 1: v = 3î → speed 3 m/s
B10.
[Pattern mirror: JEE Main cross-product style] The area of the parallelogram whose adjacent sides are 3î and 4ĵ is: (a) 7 (b) 12 (c) 24 (d) 5Show answer
(b) — |A×B| = |3î × 4ĵ| = |12k̂| = 12 square units
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