Motion in a Plane

Correct Answer: B

Scalar quantities have only magnitude (e.g., mass, temperature, speed). They follow ordinary algebra rules. Vectors obey the triangle/parallelogram law of addition due to their directional nature.

Correct Answer: C — Impulse

Impulse = Force × time = change in momentum — it is a vector. Work, current (although it flows, it doesn't obey vector addition properly), and pressure are all scalar quantities.

Correct Answer: B — A + B = B + A

Vector addition is both commutative (A+B = B+A) and associative ((A+B)+C = A+(B+C)). Subtraction is NOT commutative: A−B ≠ B−A. This is why we use the head-to-tail method.

Correct Answer: B

A unit vector n̂ has |n̂| = 1. It has no dimensions or units. It is purely used to specify direction. Unit vectors î, ĵ, k̂ point along x, y, z axes. Any vector A = |A|n̂.

Correct Answer: C — Remains constant

Horizontal: ax = 0, so vx = v₀cosθ₀ = constant throughout. Vertical: ay = −g, so vy changes. At highest point, only vy = 0, not vx. This independence of horizontal and vertical motion is Galileo's great insight.

Correct Answer: C — A parabola

Eliminating t from x = (v₀cosθ₀)t and y = (v₀sinθ₀)t − ½gt² gives y = (tanθ₀)x − gx²/[2(v₀cosθ₀)²]. This is of the form y = ax + bx² — a parabola.

Correct Answer: B — 45°

R = v₀²sin2θ₀/g. R is maximum when sin2θ₀ is maximum = 1, i.e., 2θ₀ = 90° → θ₀ = 45°. At this angle, R_max = v₀²/g.

Correct Answer: A — Equal

Galileo's law: angles that exceed or fall short of 45° by equal amounts give equal ranges. 30° = 45° − 15° and 60° = 45° + 15°. R = v₀²sin2θ/g. sin(60°) = sin(120°) → equal ranges.

Correct Answer: C — Radially inward toward centre

Centripetal acceleration ac = v²/R, always directed toward the centre of the circle. The velocity is tangential. Although speed is constant, direction changes → acceleration exists (centripetal). ac is NOT constant (direction changes continuously).

Correct Answer: C — Zero (null vector)

The magnitude of centripetal acceleration is constant (v²/R), but its direction changes continuously — it always points inward. Over one full revolution, the vector averages to zero (null vector) by symmetry, since equal and opposite directions cancel.

Correct Answer: B — v = ωR

ω = Δθ/Δt (angular speed). Arc length Δs = RΔθ → v = Δs/Δt = R(Δθ/Δt) = Rω. Therefore v = ωR. Also: ac = v²/R = ω²R.

Correct Answer: B — Tangent to path

As Δt→0, the direction of average velocity (= Δr/Δt) becomes the instantaneous velocity, which is tangential to the path at that point. This is true for any 2D or 3D motion.

Correct Answer: C — 37 m s⁻¹

Since rain and wind velocities are perpendicular:
R = √(35² + 12²) = √(1225 + 144) = √1369 = 37 m s⁻¹
Angle with vertical: θ = tan⁻¹(12/35) = tan⁻¹(0.343) ≈ 19° (hold umbrella toward east at 19° from vertical).

Correct Answer: C — Actual path taken

Displacement = final position − initial position. It is the straight-line vector from start to finish, regardless of the path. Whether you go PABCQ or PDQ or PBEFQ, displacement PQ is the same.

Correct Answer: C

At highest point: vy = 0 (vertical stops momentarily), but vx = v₀cosθ₀ ≠ 0 (horizontal unchanged). Acceleration = g downward always (gravity never stops). Total velocity at top = v₀cosθ₀ (horizontal only).

Correct Answer: B — 2A cos(θ/2)

R = √(A² + A² + 2A²cosθ) = A√(2+2cosθ) = A√(2(1+cosθ)) = A√(4cos²(θ/2)) = 2Acos(θ/2). Check: θ=0°: R=2A ✓; θ=90°: R=A√2 ✓; θ=180°: R=0 ✓

Correct Answer: B

From r = r₀ + v₀t + ½at², each component can be treated independently: x = x₀ + v₀ₓt + ½aₓt² and y = y₀ + v₀ᵧt + ½aᵧt². This is Galileo's principle of independence of horizontal and vertical motions.

Correct Answer: C — (45° − α)

Galileo's statement: angles equidistant from 45° give equal ranges. For angle (45°+α): sin2θ = sin(90°+2α) = cos2α. For angle (45°−α): sin2θ = sin(90°−2α) = cos2α. Both equal → same range.

Correct Answer: C — 2v₀sinθ₀/g

Time to reach max height: tm = v₀sinθ₀/g. By symmetry of parabolic path, total time of flight Tf = 2tm = 2v₀sinθ₀/g. Note: Tf = 2tm confirms the path is symmetric about the highest point.

Correct Answer: C — Speed

In uniform circular motion: speed |v| = constant but direction changes → velocity vector changes. Acceleration magnitude |ac| = v²/R = constant but direction changes → acceleration vector changes. Only speed and |ac| are constant scalars.

Correct Answer: C — 5

|A| = √(Ax² + Ay²) = √(3² + 4²) = √(9 + 16) = √25 = 5
Direction: θ = tan⁻¹(4/3) = tan⁻¹(1.33) ≈ 53.1° from x-axis.

Correct Answer: A — 50 m

R_max = v₀²/g = 100 m → v₀² = 100g
For vertical throw (θ = 90°): h_max = v₀²sin²90°/2g = v₀²/2g = 100g/2g = 50 m

Correct Answer: B — ω = 2πν

One revolution = 2π radians. ν = 1/T revolutions per second. ω = 2π/T = 2πν rad/s. Also: v = 2πRν and ac = 4π²ν²R.

Correct Answer: B

Kinematic equations (v = v₀+at, etc.) require constant acceleration — meaning both magnitude AND direction must be constant. In uniform circular motion, |ac| is constant but its direction always points toward the centre (changes continuously) → equations don't apply.

Correct Answer: B — r = xî + yĵ + zk̂

In 3D, position vector r = xî + yĵ + zk̂, where x, y, z are components along x-, y-, z-axes. Magnitude: |r| = √(x²+y²+z²). For 2D (plane), z = 0, giving r = xî + yĵ.

For two perpendicular vectors (θ = 90°):
R = √(A² + B²) = √(6² + 8²) = √(36+64) = √100 = 10 N Direction with respect to A: tanα = B/A = 8/6 = 4/3 α = tan⁻¹(4/3) = 53.1° R = 10 N at 53.1° to the 6 N force

(a) Velocity and Acceleration: v(t) = dr/dt = 3.0î + 4.0tĵ m/s a(t) = dv/dt = 4.0ĵ m/s² Acceleration is 4.0 m/s² along y-direction (constant).

(b) At t = 1.0 s: v = 3.0î + 4.0ĵ m/s |v| = √(3² + 4²) = √25 = 5.0 m/s θ = tan⁻¹(4/3) = tan⁻¹(1.33) ≈ 53° with x-axis |v| = 5.0 m s⁻¹ at 53° to the x-axis

v₀ = 28 m/s, θ₀ = 30°, g = 9.8 m/s²

(a) Maximum height: hm = (v₀sinθ₀)²/2g = (28×sin30°)²/(2×9.8) = (28×0.5)²/19.6 = (14)²/19.6 = 196/19.6 = 10.0 m (a) hm = 10.0 m

(b) Time of flight: Tf = 2v₀sinθ₀/g = 2×28×0.5/9.8 = 28/9.8 ≈ 2.9 s (b) Tf = 2.9 s

(c) Horizontal range: R = v₀²sin2θ₀/g = 28²×sin60°/9.8 = 784×0.866/9.8 = 679/9.8 ≈ 69 m (c) R = 69 m

v₀y = 0 (horizontal throw), h = 490 m

(a) Time to reach ground: y = −½gt² → −490 = −½×9.8×t² t² = 490×2/9.8 = 100 → t = 10 s (a) t = 10 s

(b) Speed at impact: vx = 15 m/s (unchanged) vy = gt = 9.8×10 = 98 m/s (downward) v = √(vx² + vy²) = √(15² + 98²) = √(225 + 9604) = √9829 ≈ 99 m/s (b) Speed = 99 m s⁻¹

v₀x = 5.0 m/s, v₀y = 0; ax = 3.0 m/s², ay = 2.0 m/s²

Positions: x(t) = 5.0t + ½(3.0)t² = 5.0t + 1.5t² y(t) = 0 + ½(2.0)t² = 1.0t² (a) When x = 84 m: 5.0t + 1.5t² = 84 → 1.5t² + 5t − 84 = 0 t = [−5 + √(25 + 504)]/3 = [−5 + 23]/3 = 6 s y = 1.0 × 6² = 36 m (a) y = 36 m

(b) Speed at t = 6 s: vx = 5.0 + 3.0×6 = 23 m/s vy = 0 + 2.0×6 = 12 m/s v = √(23² + 12²) = √(529+144) = √673 ≈ 26 m/s (b) Speed ≈ 26 m s⁻¹

Boat velocity: vb = 25 km/h (North, i.e., +y direction)
Current: vc = 10 km/h at 60° east of south → 60° from south toward east
Angle between vb (North) and vc (60° E of S) = 180° − 60° = 120°

Using Law of Cosines: R² = vb² + vc² + 2vb·vc·cos120° R² = 25² + 10² + 2×25×10×(−0.5) R² = 625 + 100 − 250 = 475 R = √475 ≈ 21.8 km/h Direction (using Law of Sines): sinφ/vc = sinθ/R → sinφ = 10×sin120°/21.8 sinφ = 10×0.866/21.8 = 0.397 φ ≈ 23.4° Resultant ≈ 21.8 km/h at 23.4° from North

Resultant velocity: R = √(vr² + vw²) = √(35² + 12²) = √(1225+144) = √1369 = 37 m/s Angle with vertical: tanθ = vw/vr = 12/35 = 0.343 θ = tan⁻¹(0.343) ≈ 19° Rain appears to come at 19° from vertical toward east (since wind blows west, rain leans east).
Hold umbrella at 19° from vertical toward the east

R = 0.80 m, n = 14 rev, t = 25 s

Angular speed: ω = 2πn/t = 2π × 14/25 = 28π/25 ≈ 3.52 rad/s Centripetal acceleration: ac = ω²R = (3.52)² × 0.80 = 12.39 × 0.80 ≈ 9.91 m/s² Direction: always directed toward the centre of the circle (radially inward).
ac ≈ 9.9 m s⁻², directed toward centre of circle

R = 1.00 km = 1000 m, v = 900 km/h = 900/3.6 = 250 m/s

ac = v²/R = (250)²/1000 = 62500/1000 = 62.5 m/s² Ratio with g: ac/g = 62.5/9.8 ≈ 6.38 ac = 62.5 m s⁻² ≈ 6.4 g
This is why high-speed aircraft pilots experience about 6 times their body weight!

Constraint: max height h_m ≤ 25 m, v₀ = 40 m/s

Finding the critical angle: hm = (v₀sinθ)²/2g ≤ 25 (40sinθ)² ≤ 2×10×25 = 500 1600sin²θ ≤ 500 → sin²θ ≤ 0.3125 sinθ ≤ 0.559 → θ ≤ 33.99° ≈ 34° Range at θ = 34°: R = v₀²sin2θ/g = 40²×sin68°/10 = 1600×0.927/10 = 148.3 m Maximum horizontal distance ≈ 148.3 m

R = 12 cm = 0.12 m, n = 7 rev, t = 100 s

(a) ω = 2πn/t = 2π×7/100 = 14π/100 = 0.44 rad/s (b) v = ωR = 0.44 × 0.12 = 0.053 m/s = 5.3 cm/s (c) ac = ω²R = (0.44)²×0.12 = 0.1936×0.12 = 0.023 m/s² = 2.3 cm/s² (a) ω = 0.44 rad/s   (b) v = 5.3 cm/s   (c) ac = 2.3 cm/s²
Note: ac is NOT a constant vector (direction changes), but its magnitude is constant.

P and Q are diametrically opposite on a circle of R = 200 m.
Diameter PQ = 2R = 400 m.

Displacement for ALL three girls: Displacement = |PQ| = 2R = 2×200 = 400 m (displacement is the straight line from P to Q, regardless of path)

Path length = Displacement (400 m) for:
Only the girl who skates in a straight line along the diameter PQ — her path length equals displacement = 400 m.
The other girls have curved paths → path length > 400 m.
Displacement = 400 m for all; equal to path only for girl skating straight across diameter

v₀x = 0, v₀y = 10 m/s; ax = 8.0, ay = 2.0 m/s²

(a) Position equations: x = ½ × 8.0 × t² = 4.0t² y = 10.0t + ½ × 2.0 × t² = 10t + t² When x = 16 m: 4.0t² = 16 → t² = 4 → t = 2 s y = 10×2 + 2² = 20 + 4 = 24 m (a) t = 2 s; y = 24 m

(b) Speed at t = 2 s: vx = 0 + 8×2 = 16 m/s vy = 10 + 2×2 = 14 m/s v = √(16² + 14²) = √(256+196) = √452 ≈ 21.3 m/s (b) Speed ≈ 21.3 m s⁻¹

Step 1 — Find v₀ from R_max: R_max = v₀²/g = 100 m (at θ = 45°) v₀² = 100 × 10 = 1000 m²/s² v₀ = √1000 ≈ 31.6 m/s Step 2 — Max height (throw vertically, θ = 90°): hm = (v₀sin90°)²/2g = v₀²/2g = 1000/(2×10) = 50 m Maximum height = 50 m
Note: max height = R_max/2 always.

Height h = 3400 m, angle ∠AOB = 30° (O = observation point, A,B = aircraft positions), t = 10 s

The aircraft path is horizontal at height h. O is directly below the midpoint of AB.
Let M = midpoint of AB. Then OM = h = 3400 m, ∠AOM = 15°.
AM = h × tan(∠AOM) = 3400 × tan(15°) tan(15°) = 2 − √3 ≈ 0.2679 AM = 3400 × 0.2679 = 910.9 m Distance AB = 2 × AM: AB = 2 × 910.9 = 1821.8 m Speed of aircraft: v = AB/t = 1821.8/10 ≈ 182 m/s Speed of aircraft ≈ 182 m s⁻¹ ≈ 655 km/h