Coefficient of Restitution & Rebound — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Coefficient of Restitution & Rebound MCQs with step-by-step solutions (8 questions). Part of Centre of Mass, Momentum & Collisions. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Coefficient of Restitution & Rebound · easy · theory
The coefficient of restitution e is defined as:
A. (Relative speed of separation)/(relative speed of approach) ✓ Correct
B. The impulse ratio
C. The ratio of final to initial kinetic energy
D. The ratio of the masses
Solution: e = v_sep/v_app along the line of impact; e = 1 elastic, e = 0 perfectly inelastic, 0 < e < 1 real collisions.
Q2 — Coefficient of Restitution & Rebound · medium · theory
A ball dropped from height h rebounds to height h₁. The coefficient of restitution with the floor is:
A. √(h₁/h) ✓ Correct
B. h₁/h
C. (h₁/h)²
D. 1 − h₁/h
Solution: Speeds scale as √h: e = v_up/v_down = √(2gh₁)/√(2gh) = √(h₁/h).
Q3 — Coefficient of Restitution & Rebound · easy · numerical
A ball dropped from 5 m rebounds to 3.2 m. The coefficient of restitution is:
A. 0.8 ✓ Correct
B. 0.64
C. 0.36
D. 0.5
Solution: e = √(3.2/5) = √0.64 = 0.8.
Q4 — Coefficient of Restitution & Rebound · medium · numerical
A ball with e = 0.5 is dropped from 8 m. The height after the SECOND bounce is:
A. 0.5 m ✓ Correct
B. 1 m
C. 4 m
D. 2 m
Solution: h_n = e^(2n)h = (0.5)⁴×8 = 8/16 = 0.5 m. Trap: each bounce multiplies the height by e², not e.
Q5 — Coefficient of Restitution & Rebound · hard · numerical
A 2 kg ball at 6 m/s hits a stationary 2 kg ball head-on with e = 0.5. Their final velocities are:
A. 3 m/s and 3 m/s
B. 0 and 6 m/s
C. 2 m/s and 4 m/s
D. 1.5 m/s and 4.5 m/s ✓ Correct
Solution: Momentum: v₁ + v₂ = 6; restitution: v₂ − v₁ = 0.5×6 = 3 ⇒ v₂ = 4.5, v₁ = 1.5 m/s.
Q6 — Coefficient of Restitution & Rebound · medium · numerical
A ball strikes a floor at 10 m/s (normal incidence) and rebounds at 8 m/s. The coefficient of restitution is:
A. 0.64
B. 0.8 ✓ Correct
C. 1.25
D. 0.2
Solution: e = 8/10 = 0.8.
Q7 — Coefficient of Restitution & Rebound · hard · numerical
A ball falls from 20 m onto a floor (e = 0.5). The total TIME until it stops bouncing is (g = 10):
A. 8 s
B. 4 s
C. 2 s
D. 6 s ✓ Correct
Solution: t = t₀(1 + e)/(1 − e) with t₀ = √(2h/g) = 2 s: total = 2×(1.5/0.5) = 6 s.
Q8 — Coefficient of Restitution & Rebound · medium · numerical
A ball hits a smooth floor at 45° to the vertical with speed v, and e = 1/√2 (only the normal component is reduced). The rebound direction makes an angle with the vertical of:
A. tan⁻¹(√2) ≈ 54.7° ✓ Correct
B. 30°
C. 45°
D. 60°
Solution: Horizontal component v/√2 is unchanged (smooth floor); vertical becomes e·v/√2 = v/2. tanθ = (v/√2)/(v/2) = √2 ⇒ θ ≈ 54.7° — the bounce flattens.