Skip to content

Coefficient Of Restitution

Hardphysics

A rubber ball is dropped from a height of 2 m onto a hard floor and rebounds to a height of 1.28 m after the first bounce. What is the coefficient of restitution between the ball and the floor?

Select the correct option:

🔒 Solution Hidden from View

Submit your answer to unlock the detailed step-by-step solution.

About This Question

Subject
physics
Chapter
work, energy and power
Topic
coefficient of restitution
Difficulty
Hard
Year
2025
Tags
coefficient of restitutionrebound heightimpact speed ratiobouncing ballenergy loss on impact

Solution

Correct Answer:

As discussed in NCERT Class 11, Chapter 6 (Work, Energy and Power), the coefficient of restitution between a falling object and a fixed surface equals the ratio of the rebound speed to the impact speed. Using energy conservation, speed is proportional to the square root of the drop or rebound height, so e = sqrt(h_rebound / h_drop) = sqrt(1.28 / 2) = sqrt(0.64) = 0.8. The coefficient lies between 0 and 1 for a real bounce, with 1 being perfectly elastic. The option 0.64 mistakenly takes the height ratio itself without the square root. The option 0.9 overestimates the bounce. The option 0.5 underestimates it. A plausibility check: the ball loses energy on impact, so the rebound height of 1.28 m is less than the 2 m drop, giving e below 1, and the value 0.8 fits a fairly bouncy but non-ideal rubber ball.

This hard difficulty physics question is from the chapter work, energy and power, covering the topic of coefficient of restitution. It appeared in the 2025 exam.

Looking for more practice? Explore all physics questions or browse work, energy and power questions on RankGuru.