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

Explanation

Initial moventum of body along horizontal direction, ` P-x = mv cis 45^@ = (mv)/ (sqrt 2)` ,
Initial momentum along vertical upward direction , ` P_y = mv sin 45^@ = (mv)/( sqrt)`
:. Total momentum,
` vec P P_x hat I +P_y hat j = ( mv)/(sqrt2) hat i+ (mv)/(sqrt 2) hat j`
The body will strike the horizontal sutface with velocity (v), hence its momentum along vertical downward direction,
` P_y' = mv sin 45^@= (mv)/(sqrt 2)`
momentume along horizontal direction, ltbr. ` P_x ' = mv cos 45^@ = (mv)/(sqrt 2)` ltbRgt :. total momentum,
` vec P = P_x ' hat i+ p_y' (-hat j) = (mv)/(sqrt 2) hat i- (mv)/(sqrt 2) hat j`
:. Change in momentum, ` Delta vec P = vec P- vec P'`
`= ((mv)/(sqrt0 hat i + (mv)/(sqrt 2) hat j) - (( mv)/(sqrt 2) hat i - (mv)/( sqrt 2) hat j)`
` = ( 2mv)/(sqrt 2) hat j`
:. ` | Delta vec P | = ( 2 mv)/ (sqrt 2) = sqrt 2 mv)`.