A projectile of mass m is thrown with a velocity v making an angle `60^(@)` with the horizontal. Neglecting air resistance , the change in momentum from the departure A to its arrival at B is

Answer

B

Explanation

(b): change in momentum will act only in Y-direction, while change in momentum in X-direction will be zero
`DeltaP_(x)=0, DeltaP_(y) = - "mu" sin theta-"mu" sin theta`
`= -2 "mv" sin theta`
` = - 2 "mu" (sin 60^(@))`
`= - sqrt3 "mu"`
`DeltaP=sqrt((DeltaP_(x))^(2)+(DeltaP_(y))^(2))`
`=sqrt3 "mu"`