The velocity of a sphere after it hits a vertical wall which is parallel to y-axis is $(-^i + 3^j) m / s$ on a smooth horizontal surface. The coefficient of restitution between ball and wall is $e = 0.5$ . Find the velocity of sphere just before collision.

^i + 3^j

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- 2^i + 3^j

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-^i + 3^j

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2^i + 3^j

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Solution

The correct option is A $^i + 3^j$
R.E.F image
let the figure be:-
Note :- In this figure
the velocities are
resolved along x
& y axes.
let's assure the velocity of ball before
collision as $_{i} = v_{x}^i + v_{y}^j$
we know that :- $e = \frac{R e l a t i v e v e l o c i t y a f t e r c o l l i s i o n (x - a x i s)}{R e l a t i v e v e l o c i t y b e f o r e c o l l i s i o n (x - a x i s)}$
$e = \frac{1}{V_{x}}$
but $e = 0.5$
$\Rightarrow 0.5 = \frac{1}{v_{x}}$
$\Rightarrow v_{x} = 2 m / s$
And since velocity along y-axis will
remain unchanged.
$∴ v_{y} = 3 m / s$
So $_{i} = 2^i + 3^j$

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The velocity of a sphere after it hits a vertical wall which is parallel to y-axis is (−^i+3^j)m/s on a smooth horizontal surface. The coefficient of restitution between ball and wall is e=0.5. Find the velocity of sphere just before collision.

The velocity of a sphere after it hits a vertical wall which is parallel to y-axis is $(-^i + 3^j) m / s$ on a smooth horizontal surface. The coefficient of restitution between ball and wall is $e = 0.5$ . Find the velocity of sphere just before collision.