A particle of mass m is projected with velocity making an angle of `45^(@)` with the horizontal When the particle lands on the level ground the magnitude of the change in its momentum will be .

Options

A. `mvsqrt2`
B. zero
C. 2 mv
D. `mv//sqrt2`

Answer

A

Explanation

Only vertical component of velocity changes during projectile motion so, there is change in momentum in vertical direction only.


Change in momentum `=DeltavetP`
`=DeltaP_(x) hati+DeltaP_(y)hatj". since "DeltaP_(x)=0`
`=DeltaP_(x)hatj=vecP_(fy)-vecP_(iy)`,
`P_(iy)=mv sin 45^(@), P_(fy)=-mv sin 45^(@)`
`DeltavecP=-mv sin 45^(@),-mv sin 45^(@)`
`=-2mv sin 45^(@)=-sqrt2mv_(j)|Delta vecP|`
`sqrt2mv`
`-ve` sign shows the direction of change in momentum in `-ve` y-direction.