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A sphere is moving with velocity vector ...

A sphere is moving with velocity vector `2hati+2hatj` immediately before it hits a vertical wall. The wall is parallel to `hatj` and the coefficient of restitution of the sphere and the wall is `e=1/2`. Find the velocity of the sphere after it hits the wall?

A

`hati-hatj`

B

`-hati+2hatj`

C

`-hati-hatj`

D

`2hati-hatj`

Text Solution

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The correct Answer is:
B
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A smooth sphere is moving on a horizontal surface with velocity vector 2hati+2hatj immediately before it hits a vertical wall. The wall is parallel to hatj and the coefficient of restitution of the sphere and the wall is e=1/2 . Find the velocity of the sphere after it hits the wall?

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Knowledge Check

  • A smooth sphere is moving on a horizontal surface with velocity vector 2hati+2hatj immediately before it hits a vertical wall. The wall is parallel to hatj vector and the coefficient of restitution between the sphere and the wall is e = 1//2 . The velocity vector of the sphere after it hits the wall is

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    B
    `-hati+2hatj`
    C
    `-hati-hatj`
    D
    `2hati-hatj`

  • A smooth sphere is moving on a horizontal surface with velocity vector 3hati+hatj immediately before it hits a vertical wall. The wall is parallel to the vector hatj and the coefficient of restitution between the wall and the sphere is 1//3 . The velocity vector of the sphere after it hits the wall is :

    A
    `3hati-1/3 hatj`
    B
    `-hati+hatJ`
    C
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    D
    `-hati-1/3hatj`

  • A smooth sphere is moving on a horizontal surface with velocity vector 3hat(i) + hat(j) immediately before it hits a vertical wall. The wall is parallel to the vector hat(j) and the coefficient of restitution between the wall and sphere is (1)/(3) . The velocity vector of the sphere after it hits the wall is :-

    A
    `hat(i) + hat(j)`
    B
    `3hat(i) - (1)/(3) hat(j)`
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