A particle is projected at time t=0 from a point P on the ground with a speed `v_0,` at an angle of `45^@` to the horizontal. Find the magnitude and direction of the angular momentum of the particle about P at tiem `t= v_0//g`

Options

A. `(mv_(0)^(3))/(2sqrt(2)g)`
B. `(mv_(0)^(3))/(sqrt(2)g)`
C. `(3mv_(0)^(3))/(sqrt(2)g)`
D. `(sqrt(2)mv_(0)^(3))/(g)`

Answer

a

Explanation

Let L = angular momentum of the particle about P at time t = `v_(0)//g`
At time `t=v_(0)/g`:
Let `p_(x)`= x-component of momentum of particle
`v_(x)` =x-componentof velocity of particle
x= displacement along x-axis.
Let `p_(y), v_(y)` and y denote the quantities along y-axis
Angular momentum , `L=xp_(y)-yp_(x)`..............(i)
or `L=x(mv_(y))-y(mv_(x))`or `v_(x) = v_(0) cos45^(@) = v_(0)/sqrt(2)`. .................(ii)
It remains constant all along.
`v_(y)` = vertical velocity of particle
or `v_(y) = v_(0)sin45^(@) - g xx (v_(0)/g)` `[therefore v=u+at]`
or `v_(y) = (v_(0)/sqrt(2) -v^(0))` ..................(iii)
To find x and y:
or `x=v_(0)/sqrt(2) xx v_(0)/g= v_(0)^(2)/sqrt(2)g therefore x=v_(0)^(2)/sqrt(2g)`...........(iv)
`y=v_(0)sinthetat-("gt"^(2))/2`
`y=(v_(0)/sqrt(2)) xx v_(0)/g-g/2(v_(0)/g)^(2)` .........(v)
`therefore` Substituting (ii), (iii), (iv) and (v) in (i)
`L=m[(v_(0)^(2)/sqrt(2g) xx (v_(0)/sqrt(2) - v_(0)) - (v_(0)^(2)/sqrt(2g) - v_(0)^(2)/(2g))v_(0)/sqrt(2)]`
or `L=(mv_(0)^(3))/(g) xx (-1/(2sqrt(2))) = (-mv_(0)^(3))/(2sqrt(2)g)`
L is `bot` to plane of motion and is directed away from the reader, i.e., along -ve z direction.