Answer
Verified
105.3k+ 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
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main