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Motion in 2D — IISER Physics MCQs with Solutions
Free IISER Physics Motion in 2D MCQs with step-by-step solutions covering Basics of 2D, Projectile Motion, Relative Motion in 2D, Rain Man Problem, River Boat Problem, Wind Problem. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Basics of 2D · easy · numerical
An insect crawls on a table from the point (2 m, 3 m) to the point (11 m, 15 m). What is the magnitude of its displacement?
A. 13 m
B. 21 m
C. 15 m ✓ Correct
D. 9 m
Solution: Δr = (11 − 2) î + (15 − 3) ĵ = 9 î + 12 ĵ (m), so |Δr| = √(9² + 12²) = √225 = 15 m.
Q2 — Basics of 2D · easy · numerical
A trolley moves from the point (4 m, −3 m) to the point (19 m, 17 m) in 5 s. What is the magnitude of its average velocity?
A. 7 m/s
B. 3 m/s
C. 5 m/s ✓ Correct
D. 4 m/s
Solution: Displacement = (15, 20) m with magnitude √(15² + 20²) = 25 m; average velocity = 25 m ÷ 5 s = 5 m/s.
Q3 — Basics of 2D · easy · numerical
A drone flies with velocity components 16 m/s towards the east and 30 m/s towards the north. What is its speed?
A. 23 m/s
B. 46 m/s
C. 30 m/s
D. 34 m/s ✓ Correct
Solution: The components are perpendicular, so speed = √(16² + 30²) = √1156 = 34 m/s.
Q4 — Basics of 2D · easy · numerical
A ship sails 21 km due east and then 28 km due north. What is the magnitude of its net displacement?
A. 28 km
B. 49 km
C. 35 km ✓ Correct
D. 25 km
Solution: The two legs are perpendicular, so |Δr| = √(21² + 28²) = √1225 = 35 km.
Q5 — Basics of 2D · easy · numerical
The position of a probe is given by r(t) = 10t î + 4t² ĵ (metres, seconds). What is its speed at t = 3 s?
A. 26 m/s ✓ Correct
B. 34 m/s
C. 14 m/s
D. 22 m/s
Solution: v = dr/dt = 10 î + 8t ĵ; at t = 3 s, v = (10, 24) m/s, so speed = √(100 + 576) = 26 m/s.
Q6 — Basics of 2D · easy · numerical
The position of a sphere is given by r(t) = 4.5t² î + 6t² ĵ (metres, seconds). What is the magnitude of its acceleration?
A. 10.5 m/s²
B. 21 m/s²
C. 7.5 m/s²
D. 15 m/s² ✓ Correct
Solution: a = d²r/dt² = 9 î + 12 ĵ (m/s²), so |a| = √(81 + 144) = √225 = 15 m/s².
Q7 — Basics of 2D · easy · numerical
A ball is rolled off a ledge horizontally at 7.5 m/s and falls with a constant downward acceleration of 10 m/s². What is the magnitude of its displacement 2 s after leaving the ledge?
A. 35 m
B. 25 m ✓ Correct
C. 20 m
D. 15 m
Solution: Horizontal: x = 7.5 × 2 = 15 m; vertical: y = ½ × 10 × 2² = 20 m. Displacement = √(15² + 20²) = 25 m.
Q8 — Basics of 2D · easy · numerical
A robot starts from rest and moves with constant acceleration a = 2.1 î + 2.8 ĵ (m/s²). What is its speed after 10 s?
A. 24.5 m/s
B. 35 m/s ✓ Correct
C. 49 m/s
D. 28 m/s
Solution: v = at = (21, 28) m/s after 10 s, so speed = √(441 + 784) = √1225 = 35 m/s.
Q9 — Projectile Motion · easy · numerical
A ball is projected from level ground at 25 m/s at 30° above the horizontal. What is its time of flight? (g = 10 m/s²)
A. 2 s
B. 5 s
C. 1.25 s
D. 2.5 s ✓ Correct
Solution: T = 2u sin θ / g = 2 × 25 × 0.5 / 10 = 2.5 s.
Q10 — Projectile Motion · easy · numerical
A sphere is launched at 20 m/s at 30° above the horizontal. What maximum height does it reach? (g = 10 m/s²)
A. 15 m
B. 5 m ✓ Correct
C. 10 m
D. 20 m
Solution: H = u² sin²θ / 2g = 400 × 0.25 / 20 = 5 m.
Q11 — Projectile Motion · easy · numerical
A robot launcher fires a projectile at 20 m/s at 45° above the horizontal. What is its range on level ground? (g = 10 m/s²)
A. 20 m
B. 40 m ✓ Correct
C. 80 m
D. 60 m
Solution: R = u² sin 2θ / g = 400 × sin 90° / 10 = 40 m.
Q12 — Projectile Motion · easy · numerical
A probe is projected at 30 m/s at 60° above the horizontal. What is its speed at the highest point of its path?
A. 15 m/s ✓ Correct
B. 30 m/s
C. 0 m/s
D. 15√3 ≈ 26 m/s
Solution: At the top the vertical component is zero, leaving only the horizontal component u cos θ = 30 × cos 60° = 15 m/s.
Q13 — Projectile Motion · easy · numerical
A sphere rolls off the edge of a 125 m high cliff, leaving it horizontally. How long does it take to reach the ground? (g = 10 m/s²)
A. 25 s
B. 2.5 s
C. 5 s ✓ Correct
D. 12.5 s
Solution: The fall is governed only by gravity: 125 = ½ × 10 × t², so t² = 25 and t = 5 s.
Q14 — Projectile Motion · easy · numerical
A ball is thrown horizontally at 15 m/s from the top of an 80 m tower. How far from the base of the tower does it land? (g = 10 m/s²)
A. 30 m
B. 80 m
C. 45 m
D. 60 m ✓ Correct
Solution: Fall time: 80 = 5t² gives t = 4 s; horizontal distance = 15 × 4 = 60 m.
Q15 — Projectile Motion · easy · numerical
A probe is launched horizontally at 16 m/s from a 45 m high platform. With what speed does it hit the ground? (g = 10 m/s²)
A. 46 m/s
B. 30 m/s
C. 16 m/s
D. 34 m/s ✓ Correct
Solution: Fall time: 45 = 5t² gives t = 3 s, so v_y = 10 × 3 = 30 m/s; impact speed = √(16² + 30²) = 34 m/s.
Q16 — Projectile Motion · easy · numerical
A trolley-mounted launcher fires a ball at 45° and it lands 62.5 m away on level ground. What was the launch speed? (g = 10 m/s²)
A. 12.5 m/s
B. 25 m/s ✓ Correct
C. 50 m/s
D. 35 m/s
Solution: At 45°, R = u²/g, so u = √(Rg) = √(62.5 × 10) = √625 = 25 m/s.
Q17 — Relative Motion in 2D · easy · numerical
Ship P sails due east at 9 km/h while ship Q sails due north at 12 km/h. What is the magnitude of the velocity of Q relative to P?
A. 3 km/h
B. 12 km/h
C. 21 km/h
D. 15 km/h ✓ Correct
Solution: v_QP = v_Q − v_P has perpendicular components 12 km/h (north) and 9 km/h (west), so |v_QP| = √(9² + 12²) = 15 km/h.
Q18 — Relative Motion in 2D · easy · numerical
Drone A flies due east at 54 km/h and drone B flies due north at 20 m/s. What is the speed of B relative to A?
A. 35 m/s
B. 25 m/s ✓ Correct
C. 20 m/s
D. 5 m/s
Solution: 54 km/h = 15 m/s; the relative velocity has perpendicular components 15 m/s and 20 m/s, so |v_BA| = √(15² + 20²) = 25 m/s.
Q19 — Relative Motion in 2D · easy · numerical
Trolley A moves due east at 12 m/s and trolley B moves due north at 12 m/s. In which direction does B appear to move as seen from A?
A. 45° west of north (towards the north-west) ✓ Correct
B. Towards the south-west
C. Due north
D. 45° east of north (towards the north-east)
Solution: v_BA = v_B − v_A = 12 ĵ − 12 î (m/s): equal components towards the north and the west, which points 45° west of north.
Q20 — Relative Motion in 2D · easy · numerical
Two insects leave the same crumb at the same moment, one crawling east at 0.9 cm/s and the other north at 1.2 cm/s. How far apart are they after 6 s?
A. 12.6 cm
B. 9 cm ✓ Correct
C. 1.8 cm
D. 4.5 cm
Solution: Their separation grows at the relative speed √(0.9² + 1.2²) = 1.5 cm/s, so after 6 s they are 1.5 × 6 = 9 cm apart.
Q21 — Relative Motion in 2D · easy · numerical
Probe A moves due east at 10 m/s and probe B moves due north at 24 m/s. What is the speed of A relative to B?
A. 26 m/s ✓ Correct
B. 34 m/s
C. 14 m/s
D. 17 m/s
Solution: The velocities are perpendicular, so |v_AB| = √(10² + 24²) = √676 = 26 m/s.
Q22 — Relative Motion in 2D · easy · numerical
Robot A has velocity 7 î + 2 ĵ (m/s) and robot B has velocity −2 î − 10 ĵ (m/s). What is the speed of A relative to B?
A. 12 m/s
B. 9 m/s
C. 21 m/s
D. 15 m/s ✓ Correct
Solution: v_AB = v_A − v_B = (7 − (−2), 2 − (−10)) = (9, 12) m/s, so |v_AB| = √(81 + 144) = 15 m/s.
Q23 — Relative Motion in 2D · easy · numerical
Two spheres are rolled from the same corner of a hall at the same moment, one east at 2.1 m/s and the other north at 2.8 m/s. How far apart are they after 10 s?
A. 17.5 m
B. 49 m
C. 35 m ✓ Correct
D. 25 m
Solution: Relative speed = √(2.1² + 2.8²) = 3.5 m/s, so separation after 10 s = 3.5 × 10 = 35 m.
Q24 — Relative Motion in 2D · easy · numerical
Two ships leave the same port together, one heading due east at 16 km/h and the other due north at 30 km/h. How far apart are they after 30 minutes?
A. 34 km
B. 23 km
C. 17 km ✓ Correct
D. 8.5 km
Solution: Relative speed = √(16² + 30²) = 34 km/h; in 0.5 h the separation is 34 × 0.5 = 17 km.
Q25 — Rain Man Problem · easy · numerical
Rain is falling vertically at 12 m/s. A man walks along a straight road at 9 m/s. At what angle from the vertical must he hold his umbrella to keep the rain off?
A. tan⁻¹(4/3) ≈ 53° from the vertical
B. 45° from the vertical
C. tan⁻¹(3/4) ≈ 37° from the vertical ✓ Correct
D. 30° from the vertical
Solution: The rain appears tilted by tanθ = v(man)/v(rain) = 9/12 = 3/4, so θ = tan⁻¹(0.75) ≈ 37° from the vertical, tilted forward.
Q26 — Rain Man Problem · easy · numerical
Rain falls vertically downward at 20 m/s. A cyclist rides on a level road at 15 m/s. What is the speed of the rain relative to the cyclist?
A. 5 m/s
B. 25 m/s ✓ Correct
C. 35 m/s
D. 18 m/s
Solution: Relative speed = √(v(rain)² + v(cyclist)²) = √(20² + 15²) = √625 = 25 m/s.
Q27 — Rain Man Problem · easy · numerical
Rain is coming straight down at 24 km/h. A woman runs at 10 km/h. What is the apparent speed of the rain with respect to her?
A. 14 km/h
B. 26 km/h ✓ Correct
C. 34 km/h
D. 22 km/h
Solution: Apparent speed = √(24² + 10²) = √(576 + 100) = √676 = 26 km/h.
Q28 — Rain Man Problem · easy · numerical
A cyclist moving at 15 km/h finds that rain, which is actually falling vertically, appears to strike him at 37° to the vertical. What is the true speed of the rain? (tan 37° = 3/4)
A. 25 km/h
B. 12 km/h
C. 20 km/h ✓ Correct
D. 11.25 km/h
Solution: tan 37° = v(cyclist)/v(rain), so v(rain) = 15/(3/4) = 20 km/h.
Q29 — Rain Man Problem · easy · numerical
To a cyclist, rain appears to fall at 28 m/s making an angle of 30° with the vertical. The rain is actually falling vertically. What is the cyclist’s speed?
A. 28 m/s
B. 14√3 ≈ 24.2 m/s
C. 7 m/s
D. 14 m/s ✓ Correct
Solution: The horizontal component of the apparent velocity equals the cyclist’s speed: v = 28 sin 30° = 28 × 0.5 = 14 m/s.
Q30 — Rain Man Problem · easy · numerical
Rain falls vertically at 21 m/s. A passenger sits in a bus moving at 28 m/s on a straight road. What is the speed of the rain as seen through the bus window?
A. 7 m/s
B. 49 m/s
C. 35 m/s ✓ Correct
D. 24.5 m/s
Solution: Apparent speed = √(21² + 28²) = √(441 + 784) = √1225 = 35 m/s.