Kinematics and Laws of Motion — Practice

Question 1

A ball is thrown vertically upward at 20 m/s. The maximum height it reaches is (g = 10 m/s²):
(a) 20 m (b) 40 m (c) 10 m (d) 2 m

Accepted Answer

(a) 20 m.

H = u²/2g = 20²/(2×10) = 400/20 = 20 m. It reaches the top in u/g = 2 s.

Question 2

A car moving at 20 m/s decelerates uniformly at 4 m/s². The distance it covers before stopping is:
(a) 50 m (b) 40 m (c) 100 m (d) 25 m

Accepted Answer

(a) 50 m.

v² = u² + 2as with v = 0: 0 = 400 − 2(4)s, so s = 400/8 = 50 m.

Question 3

A projectile is launched at 20 m/s at 45° to the horizontal. Its range is (g = 10 m/s²):
(a) 40 m (b) 20 m (c) 80 m (d) 45 m

Accepted Answer

(a) 40 m.

R = u²sin2θ/g = 20² × sin90° / 10 = 400 × 1 / 10 = 40 m.

Question 4

A body is projected at 20 m/s at 30° to the horizontal. Its maximum height is (g = 10 m/s²):
(a) 5 m (b) 10 m (c) 2.5 m (d) 20 m

Accepted Answer

(a) 5 m.

H = u²sin²θ/2g = 20² × (sin30°)² / (2×10) = 400 × 0.25 / 20 = 100/20 = 5 m.

Question 5

Two projectiles are launched at the same speed at 30° and 60°. Compared with each other, their horizontal ranges are:
(a) equal (b) in ratio 1:2 (c) in ratio 2:1 (d) in ratio 1:√3

Accepted Answer

(a) equal.

R = u²sin2θ/g. For 30°, sin60°; for 60°, sin120° = sin60°. Complementary angles give the same sin2θ, hence the same range.

Question 6

A 50 kg person stands in a lift accelerating upward at 2 m/s². The normal force on them is (g = 10 m/s²):
(a) 600 N (b) 500 N (c) 400 N (d) 100 N

Accepted Answer

(a) 600 N.

N = m(g + a) = 50 × (10 + 2) = 600 N — greater than the true weight of 500 N because the lift accelerates upward.

Question 7

The apparent weight of a person in a lift whose cable snaps (free fall) is:
(a) zero (b) mg (c) 2mg (d) mg/2

Accepted Answer

(a) zero.

In free fall the lift and person accelerate together at g, so the normal force N = m(g − g) = 0 — they are weightless, exactly like an orbiting astronaut.

Question 8

A block slides down a smooth (frictionless) incline of 30°. Its acceleration is (g = 10 m/s²):
(a) 5 m/s² (b) 10 m/s² (c) 8.7 m/s² (d) 2.5 m/s²

Accepted Answer

(a) 5 m/s².

On a frictionless incline a = g sinθ = 10 × sin30° = 10 × 0.5 = 5 m/s².

Question 9

The coefficient of static friction between a block and an incline is 1/√3. The angle of repose (at which the block just begins to slide) is:
(a) 30° (b) 45° (c) 60° (d) 15°

Accepted Answer

(a) 30°.

At the angle of repose tanθ = μ_s = 1/√3, so θ = tan⁻¹(1/√3) = 30°.

Question 10

A car moves at 10 m/s around a circular track of radius 20 m. Its centripetal acceleration is:
(a) 5 m/s² (b) 2 m/s² (c) 200 m/s² (d) 0.5 m/s²

Accepted Answer

(a) 5 m/s².

a_c = v²/r = 10²/20 = 100/20 = 5 m/s², directed toward the centre.

Question 11

For a car to round a curve of radius r at speed v with no reliance on friction, the road should be banked at an angle θ where:
(a) tanθ = v²/rg (b) tanθ = rg/v² (c) sinθ = v²/rg (d) tanθ = v/rg

Accepted Answer

(a) tanθ = v²/rg.

Resolving the normal force, the horizontal component N sinθ supplies the centripetal force and the vertical component N cosθ balances gravity; dividing gives tanθ = v²/rg.

Question 12

A flat (unbanked) curve has radius 20 m and coefficient of friction 0.5. The maximum speed without skidding is (g = 10 m/s²):
(a) 10 m/s (b) 5 m/s (c) 20 m/s (d) 100 m/s

Accepted Answer

(a) 10 m/s.

Friction supplies the centripetal force: μmg = mv²/r, so v_max = √(μrg) = √(0.5 × 20 × 10) = √100 = 10 m/s.

Question 13

A 4 kg gun fires a 0.02 kg bullet at 200 m/s. The recoil speed of the gun is:
(a) 1 m/s (b) 2 m/s (c) 0.5 m/s (d) 4 m/s

Accepted Answer

(a) 1 m/s.

Momentum is conserved and initially zero: m_bullet v_bullet = M_gun V. So V = (0.02 × 200)/4 = 4/4 = 1 m/s, opposite to the bullet.

Question 14

A 0.2 kg ball hits a wall horizontally at 10 m/s and rebounds at 10 m/s. The magnitude of the impulse on the ball is:
(a) 4 kg·m/s (b) 2 kg·m/s (c) 0 (d) 40 kg·m/s

Accepted Answer

(a) 4 kg·m/s.

Impulse = change in momentum = m(v_final − v_initial) = 0.2 × (10 − (−10)) = 0.2 × 20 = 4 kg·m/s. The reversal of direction doubles it compared with mv.

Question 15

A body starts from rest and accelerates uniformly to 20 m/s in 10 s. The displacement (area under the v–t graph) is:
(a) 100 m (b) 200 m (c) 50 m (d) 20 m

Accepted Answer

(a) 100 m.

The v–t graph is a triangle of base 10 s and height 20 m/s; area = ½ × 10 × 20 = 100 m. This equals ½at² with a = 2 m/s².

Practice — Kinematics and Laws of Motion

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