The reflecting surface of a plane mirror is vertical. A particle is projected in a vertical plane which is also perpendicular to the mirror. The initial speed of the particle is `10m//s` and the angle of projection is `60^@` from the normal of the mirror. The point of projection is at a distance 5m from the mirror. The particle moves towards the mirror. Just before the particle touches the mirror, the velocity of approach of the particle and the image is:

Options

A. 10 m/s
B. 5 m/s
C. `10sqrt(3) m//s`
D. `5sqrt(3) m//s`

Answer

A

Explanation

`(V_(I,m))x=-(V_(o,m))x`
`(V_(1)-V_(m))x=-(V_(0)-V_(m))x`
`V_(ix)-O=V_(ox)`
`V_(ix)=-5 m//s`
`V_(app)=5-(-5)=10 m//s`