
A ball is bouncing down a flight of stairs. The coefficient of restitution is $e$ . The height of each step is $d$ and the ball descends one step each bounce. After each bounce it rebounds to a height $h$ above the next lower step. The height is large compared with the width of the step so that the impacts are effectively head-on. Find the relationship between $h$ and $d$.
(A) $h = \dfrac{d}{{1 - {e^2}}}$
(B) $h = \dfrac{d}{{1 + {e^2}}}$
(C) $h = \dfrac{d}{{1 + {e^{}}}}$
(D) $h = \sqrt {\dfrac{d}{{1 - {e^2}}}} $
Answer
161.4k+ views
Hint We are here asked to find the relationship between $h$ and $d$. Also we are given with the coefficient of restitution. Thus, it would be easier to go through the path of restitution analysis.
Formulae Used:
${v^2} - {u^2} = 2ah$
Where, $v$ is the final velocity of the particle, $u$ is the initial velocity of the particle, $a$ is the acceleration on it and $h$ is the height of the particle.
$e = \dfrac{{{v_2}}}{{{v_1}}}$
Where, $e$ is the coefficient of restitution, ${v_2}$ is the velocity of the particle after collision and ${v_1}$ is the velocity of the particle before collision.
Complete Step By Step Solution
Here,
For the first bounce of the ball on the top most step,
${v_1}^2 - {0^2} = 2ah$
We took ${u_1} = 0$ as the initial velocity of the ball was zero.
Thus, we get
$2ah = {v_1}^2 \cdot \cdot \cdot \cdot (1)$
Now,
After the first bounce, the new height of the ball is $(h - d)$,
Thus, we get
$2a(h - d) = {v_2}^2 \cdot \cdot \cdot \cdot (2)$
Now,
Applying$\dfrac{{Equation(1)}}{{Equation(2)}}$, we get
$\dfrac{h}{{h - d}} = \dfrac{{{v_1}^2}}{{{v_2}^2}}$
After further evaluation, we get
$\dfrac{{{v_2}}}{{{v_1}}} = \sqrt {\dfrac{{h - d}}{h}} $
But,
$e = \dfrac{{{v_2}}}{{{v_1}}}$
Thus, equating both, we get
$e = \sqrt {\dfrac{{h - d}}{h}} $
Further, we get
${e^2} = \dfrac{{h - d}}{h} \Rightarrow h{e^2} = h - d \Rightarrow h = \dfrac{d}{{1 - {e^2}}}$
Hence, the correct option is (A).
Note We have used the route of using the fundamental formula for coefficient of restitution as that allows us to directly relate all the given parameters. Moreover, in the given case it is given that the collision is head on which means completely elastic, but if the collision was inelastic, then the evaluation will be somewhat different.
Formulae Used:
${v^2} - {u^2} = 2ah$
Where, $v$ is the final velocity of the particle, $u$ is the initial velocity of the particle, $a$ is the acceleration on it and $h$ is the height of the particle.
$e = \dfrac{{{v_2}}}{{{v_1}}}$
Where, $e$ is the coefficient of restitution, ${v_2}$ is the velocity of the particle after collision and ${v_1}$ is the velocity of the particle before collision.
Complete Step By Step Solution
Here,
For the first bounce of the ball on the top most step,
${v_1}^2 - {0^2} = 2ah$
We took ${u_1} = 0$ as the initial velocity of the ball was zero.
Thus, we get
$2ah = {v_1}^2 \cdot \cdot \cdot \cdot (1)$
Now,
After the first bounce, the new height of the ball is $(h - d)$,
Thus, we get
$2a(h - d) = {v_2}^2 \cdot \cdot \cdot \cdot (2)$
Now,
Applying$\dfrac{{Equation(1)}}{{Equation(2)}}$, we get
$\dfrac{h}{{h - d}} = \dfrac{{{v_1}^2}}{{{v_2}^2}}$
After further evaluation, we get
$\dfrac{{{v_2}}}{{{v_1}}} = \sqrt {\dfrac{{h - d}}{h}} $
But,
$e = \dfrac{{{v_2}}}{{{v_1}}}$
Thus, equating both, we get
$e = \sqrt {\dfrac{{h - d}}{h}} $
Further, we get
${e^2} = \dfrac{{h - d}}{h} \Rightarrow h{e^2} = h - d \Rightarrow h = \dfrac{d}{{1 - {e^2}}}$
Hence, the correct option is (A).
Note We have used the route of using the fundamental formula for coefficient of restitution as that allows us to directly relate all the given parameters. Moreover, in the given case it is given that the collision is head on which means completely elastic, but if the collision was inelastic, then the evaluation will be somewhat different.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Class 11 JEE Main Physics Mock Test 2025

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
