A ball of mass `0.2kg ` is thrown from a height of 1m and with an initial velocity of `sqrt10 m //s` at an angle of `45^(@)` with the horizontal. Assuming, acceleration due to gravity `g =10m//s ^(2),` ten modulus of momentum increment during the total time of motion in kg m/s is

Options

A. `(2 + sqrt10)/(sqrt10)`
B. `(1+ sqrt10)/(sqrt5)`
C. `(1+ sqrt5)/(sqrt5)`
D. `(sqrt5-1)/(sqrt5)`

Answer

C

Explanation

There is no change in velocity in x-direction so momentum change occurs onhly in y-direction.

Now initial momentum in y direction is
`p _(y) `(initial) = `mv_(y) =0.2 (sqrt10sin 45^(@)hatj)= (1)/(sqrt5). hatjkg ^(-) ms ^(-) `
Final mmentum in y-direction is
`py _("final") = mv _(y) =m (sqrt( u ^(2) + 2gh))`
`=(0.2 sqrt((5^(2)) + 2 ( -10) (-1)hatj)=- 1 hatj kg -ms ^(-1)` (downwards)
So, change in momentum
`Delta p = py _(("final")) - px _(("initial")) =- 1 hatj - (1)/(sqrt5) hatj `
`=-(1+ (1)/(sqrt5))hatj kg ^(-) ms ^(-1)`
`implies = |Delta p | = 1 + (1)/(sqrt5) = (1+ sqrt5)/(sqrt5)`