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# Ball A is falling vertically downwards with velocity ${v_1}$ . it strikes elastically with a wedge moving horizontally with velocity ${v_2}$ as shown in figure. What is the ratio of $\dfrac{{{v_1}}}{{{v_2}}}$, when that ball bounces back in vertically upward direction relative to the wedge:A. $\sqrt 3$B. $\dfrac{1}{{\sqrt 3 }}$C. $\dfrac{1}{{\sqrt 2 }}$D.$\dfrac{m}{2}$ Verified
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Hint: In the given data a ball is falling towards the downwards with some velocity and it strikes to the wedge which was moving horizontally with some velocity if the direction of two velocities changes then the ratio of those velocities is at some angle given in the data. Now by using momentum of inertia with a given angle we are finding the ratio of velocities.

Complete step-by-step solution:
Given data, velocity ${v_1}$ and another velocity ${v_2}$
Here the forces applied perpendicular to the wedge, so momentum of the charge is initial momentum of inertia and final momentum inertia and the momentum changes only in the direction of the wedge,
${P_i} = {P_f}$

Thus, $m{v_1}\cos {30^ \circ } = m{v_2}\sin {30^ \circ }$
Here mass is same and velocities are different, as we discussed earlier at some angle is applied
Then we get the ratio of two velocities is,
$\dfrac{{{v_1}}}{{{v_2}}} = \tan {30^ \circ }$
From trigonometric equations tan value is,
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{1}{{\sqrt 3 }}$
Hence we have proved the ratio of velocities is $\dfrac{1}{{\sqrt 3 }}$

Note: Momentum is defined as the product of mass in motion and velocity. Thus from the given velocities we have proved the ratio of velocities, hence the correct option is b. in the given data the mass is constant and velocities are different and given some angle at the wedge of the objects. Hence we have proved from the momentum of the inertia formula.