Coefficient of Restitution & Rebound — NEET Physics MCQs with Solutions
Free NEET 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 ratio of final to initial kinetic energy
C. The impulse ratio
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)²
C. h₁/h
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. 0.36
B. 0.5
C. 0.8 ✓ Correct
D. 0.64
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. 4 m
B. 2 m
C. 1 m
D. 0.5 m ✓ Correct
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. 1.5 m/s and 4.5 m/s ✓ Correct
D. 2 m/s and 4 m/s
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. 0.8 ✓ Correct
B. 0.64
C. 1.25
D. 0.2
Solution: e = 8/10 = 0.8.
Q7 — Coefficient of Restitution & Rebound · easy · numerical
A ball dropped from 1.8 m rebounds with e = 2/3. The rebound height is:
A. 1.2 m
B. 0.8 m ✓ Correct
C. 0.6 m
D. 0.9 m
Solution: h₁ = e²h = (4/9)×1.8 = 0.8 m.
Q8 — Coefficient of Restitution & Rebound · medium · numerical
A ball dropped from height h keeps bouncing with e = 1/√2. The TOTAL distance travelled before it stops is:
A. 2h
B. 3h ✓ Correct
C. 4h
D. h
Solution: D = h(1 + e²)/(1 − e²) = h(1.5/0.5) = 3h — the classic geometric series.