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

Options

A. `2m v`
B. `m v//sqrt2`
C. `m v sqrt2`
D. zero

Answer

c

Explanation

As is clear from when the particle lands on the level ground at A there is no change in linear momentum along the horizontal
However , the vertical component of velocity is reversed
`:.` change in momentum in vertical direction
` = m upsilon sin 45^(@) - ( - m upsilon sin 45^(@))`
` = 2 m upsilon sin 45^(@) = 2 mu upsilon xx (1)/(sqrt2) = sqrt(2) m upsilon`
.