Question

Hail Storms are observed to strike the surface of the frozen lake at $30^\circ $ with the vertical and rebound at $60^\circ $ with vertical. Assume contact to be smooth, the coefficient of restitution is:
A. $e = \dfrac{1}{{\sqrt 3 }}$
B. $e = \dfrac{1}{3}$
C. $e = \sqrt 3 $
D. $e = 3$

Answer 1

Hint: In order to solve this question we need to understand collision.Collision is physical contact between two bodies after moving with some initial speed towards each other and interacting under mutual force when they are close to each other. Net force in collision is always zero so from Newton's second law of motion, momentum would always be constant. So momentum is always conserved during collision which is momentum before collision must equals to momentum after collision.

Complete step by step answer:
Collisions are of two types, one is elastic collision and other is non-elastic collision. In elastic collision both momentum and energy is conserved while in in-elastic collision only momentum is conserved. Coefficient of restitution is defined as ratio of velocity of separation to velocity of approachSo if hail storms collide with earth at an angle of $30^\circ $ with vertical. Let the initial velocity be “$u$” and the final velocity is “$v$”.
Since, acceleration in $x$ direction is, ${a_x} = 0$.
So, Force in $x$ direction, ${F_x} = m{a_x}$.
Putting values we get, ${F_x} = 0$.
So from newton’s second law of motion, ${F_x} = \dfrac{{d{p_x}}}{{dt}}$ we get, $\dfrac{{d{p_x}}}{{dt}} = 0$
So, Momentum in the “x” direction is, ${p_x} = $ constant.
So before collision momentum in x direction, ${p_{bx}} = m{u_x}$
Since, ${u_x} = u\sin \;30^\circ $

Putting values we get, ${p_{bx}} = mu\sin 30^\circ $
${p_{bx}} = \dfrac{{mu}}{2}$
Similarly after collision momentum in x direction, ${p_{ax}} = m{v_x}$
Since, ${v_x} = v\sin \;60^\circ $
Putting values we get, ${p_{ax}} = mv\sin 60^\circ $
${p_{ax}} = \dfrac{{mv\sqrt 3 }}{2}$
So from conservation of momentum in x direction, ${p_{bx}} = {p_{ax}}$
Putting values we get, $\dfrac{{mu}}{2} = \dfrac{{mv\sqrt 3 }}{2}$
$\dfrac{v}{u} = \dfrac{1}{{\sqrt 3 }}$
Since coefficient of restitution is defined as,
$e = \dfrac{v}{u}$
Putting values we get,
$\therefore e = \dfrac{1}{{\sqrt 3 }}$

So the correct option is A.

Note: It should be remembered that momentum would be conserved only in that direction in which force is zero, here there is no acceleration or force in $x$ direction so momentum in this direction would be conserved whereas acceleration in $y$ direction is non zero and it is equal to acceleration due to gravity so in this direction momentum is not conserved.