A projectile of mass `m` is fired with a velocity `v` from point P at an angle `45^@`. Neglecting air resistance, the magnitude of the change in momentum leaving the point P and arriving at Q is

Options

A. ` mv`
B. `2 mv`
C. `sqrt 2mv`
D. `zero`

Answer

C

Explanation

Initial momentum of nody along horizontal direction, ` P_x = mv cos 45^@ = ( mv)/(sqrt 2)` Initial momentum along vertical upward direction
` P_y = mv sin 45^@ = (mv)/(sqrt2 )`
:. Total momentum ` = vec p_x hat I + vec P_y hat j = (mv)/( sqrt 2) hat i + (mv)/(sqrt 2) jat j`
The body will strike the horizontal surface with velocity (v) hence its momentum along vertical downward direction ` P'_y = mv sin 45^2 = (mv)/(sqrt 2)`
Total momentume ltBrgt 1 vec P' = vec P'_y aht i+ vec P'_y hat j = (mv)/(sqrt 2) hat i- (mv)/(sqrt 2)`
Chang in moventum ` = Delta vec P = vec P - vec p'`
` = sqrt 2 mv`.