Q: A ball dropped from height h rebounds to height h₁. The coefficient of restitution with the floor is:

√(h₁/h). Speeds scale as √h: e = v_up/v_down = √(2gh₁)/√(2gh) = √(h₁/h).

Q: A ball dropped from 5 m rebounds to 3.2 m. The coefficient of restitution is:

0.8. e = √(3.2/5) = √0.64 = 0.8.

Q: A ball with e = 0.5 is dropped from 8 m. The height after the SECOND bounce is:

0.5 m. h_n = e^(2n)h = (0.5)⁴×8 = 8/16 = 0.5 m. Trap: each bounce multiplies the height by e², not e.

Q: A 2 kg ball at 6 m/s hits a stationary 2 kg ball head-on with e = 0.5. Their final velocities are:

1.5 m/s and 4.5 m/s. Momentum: v₁ + v₂ = 6; restitution: v₂ − v₁ = 0.5×6 = 3 ⇒ v₂ = 4.5, v₁ = 1.5 m/s.

Q: For a collision with 0 < e < 1, the kinetic energy after the collision:

Decreases, by ½μv_rel²(1 − e²). ΔKE = ½μv_app²(1 − e²): e parametrises how much of the relative-motion energy survives — zero loss at e = 1, maximal at e = 0.

Q: A bouncing ball loses 36% of its kinetic energy in a single bounce off the floor. Its coefficient of restitution with the floor is:

0.8. Retained KE fraction = e² = 1 − 0.36 = 0.64 ⇒ e = 0.8. (Rebound height would likewise be 64% of the drop height.)

Question 1

The coefficient of restitution e is defined as:

Accepted Answer

(Relative speed of separation)/(relative speed of approach). e = v_sep/v_app along the line of impact; e = 1 elastic, e = 0 perfectly inelastic, 0 < e < 1 real collisions.

Question 2

A ball dropped from height h rebounds to height h₁. The coefficient of restitution with the floor is:

Accepted Answer

√(h₁/h). Speeds scale as √h: e = v_up/v_down = √(2gh₁)/√(2gh) = √(h₁/h).

Question 3

A ball dropped from 5 m rebounds to 3.2 m. The coefficient of restitution is:

Accepted Answer

0.8. e = √(3.2/5) = √0.64 = 0.8.

Question 4

A ball with e = 0.5 is dropped from 8 m. The height after the SECOND bounce is:

Accepted Answer

0.5 m. h_n = e^(2n)h = (0.5)⁴×8 = 8/16 = 0.5 m. Trap: each bounce multiplies the height by e², not e.

Question 5

A 2 kg ball at 6 m/s hits a stationary 2 kg ball head-on with e = 0.5. Their final velocities are:

Accepted Answer

1.5 m/s and 4.5 m/s. Momentum: v₁ + v₂ = 6; restitution: v₂ − v₁ = 0.5×6 = 3 ⇒ v₂ = 4.5, v₁ = 1.5 m/s.

Question 6

A ball strikes a floor at 10 m/s (normal incidence) and rebounds at 8 m/s. The coefficient of restitution is:

Accepted Answer

0.8. e = 8/10 = 0.8.

Question 7

For a collision with 0 < e < 1, the kinetic energy after the collision:

Accepted Answer

Decreases, by ½μv_rel²(1 − e²). ΔKE = ½μv_app²(1 − e²): e parametrises how much of the relative-motion energy survives — zero loss at e = 1, maximal at e = 0.

Question 8

A bouncing ball loses 36% of its kinetic energy in a single bounce off the floor. Its coefficient of restitution with the floor is:

Accepted Answer

0.8. Retained KE fraction = e² = 1 − 0.36 = 0.64 ⇒ e = 0.8. (Rebound height would likewise be 64% of the drop height.)

Coefficient of Restitution & Rebound — IISER Physics MCQs with Solutions

Questions with solutions