Velocity of a particle of mass `2 kg` change from `vecv_(1) =-2hati-2hatjm/s` to `vecv_(2)=(hati-hatj)m//s` after colliding with as plane surface.

Options

A. The angle made by the plane surface with the positive `x`-axis is `90^(@)+tan^(-1)(1/3)`
B. The angle made by the plane surface with the positive `x`- axis is `tan^(-1)(1/3)`
C. The direction of change in momentum make an angle `tan^(-1)(1/3)` with the positive `x`-axis
D. The direction of charge in momentum makes an angle `90^(@)+tan^(-1)(1/3)` with the plane surface.

Answer

A, C

Explanation

`DeltavecP=2(hati-hatj)-(-2hati-2hatj)`
`=6hati+2hatj` kg-m/s
So `DeltavecP` makes angle `tan^(-1)(1/3)` with positive `x`- axis which is also the direction of normal force exerted on the ball. So the plane surface makes an angle `(pi)/2+tan^(-1)(1/3)` with positive `x`- axis.