The velocity vector of a sphere after it hits a vertical wall which is parallel to $^j$ is $(-^i + 3^j)$ on a smooth horizontal surface. The coefficient of restitution between ball and wall is (1/2). Find the velocity vector of sphere immediately 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 D $2^i + 3^j$
So,
$v_{x} = 2 m / s ˆ i$
$v_{y}$ remain same before $\in$ collision
$\begin{matrix} v_{y} = 3 ˆ j ¯ ¯ ¯ v = 2 ˆ i + 3 ˆ j \end{matrix}$
So,
Option $D$ is correct answer.

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The velocity vector of a sphere after it hits a vertical wall which is parallel to ^j is (−^i+3^j) on a smooth horizontal surface. The coefficient of restitution between ball and wall is (1/2). Find the velocity vector of sphere immediately before collision :

The correct option is D 2^i+3^jSo,vx=2m/sˆivy remain same before ∈ collisionvy=3ˆj¯¯¯v=2ˆi+3ˆj So,Option D is correct answer.

The velocity vector of a sphere after it hits a vertical wall which is parallel to $^j$ is $(-^i + 3^j)$ on a smooth horizontal surface. The coefficient of restitution between ball and wall is (1/2). Find the velocity vector of sphere immediately before collision :

The correct option is D $2^i + 3^j$
So,
$v_{x} = 2 m / s ˆ i$
$v_{y}$ remain same before $\in$ collision
$\begin{matrix} v_{y} = 3 ˆ j ¯ ¯ ¯ v = 2 ˆ i + 3 ˆ j \end{matrix}$
So,
Option $D$ is correct answer.