On a frictionless horizontal surface , assumed to be the ` x-y` plane , a small trolley `A` is moving along a straight line parallel to the `y-axis `( see figure) with a constant velocity of `(sqrt(3)-1) m//s ` . At a particular instant , when the line `OA` makes an angle of `45(@)` with the `x - axis ` , a ball is thrown along the surface from the origin `O`. Its velocity makes an angle `phi` with the `x -axis and it hits the trolley .
(a) The motion of the ball is observed from the frame of the trolley . Calculate the angle `theta` made by the velocity vector of the ball with the ` x-axis in this frame .
(b) Find the speed of the ball with respect to the surface , if ` phi = (4 theta )//(4)`.

Explanation

Method 1 : (a) Since the ball hits the trolley, relative to trolley, the velocity of ball should be directed towards the trolley. Hence, in the frame of trolley, the ball will appear to be moving towards `OA`, or in the frame of trolley, ball's velocity will make an angle of `45^@`.
(b) `phi = (4 theta)/(3) = (4 xx 45^@)/(3) = 60^@`
Using sine rule `(V_B)/(sin 135^@) = (V_A)/(sin 15^@)`
`rArr V_B = 2 m s^-1`
Method 2 : (a) Let `A` stands for trolley and `B` for ball.
Relative velocity of `B` will respect to `A (vec v_(B A))` should be along `OA` for the ball to hit the trolley.
Hence, `vec v_(B A)` will make an angle of `45^@` with positive x - axis.
(b) `tan theta = v_(B Ay)/v_(B A x) = tan 45^@ or v_(B A y) = v_(B A x)`...(i)
Further `v_(B Ay) = v_(B y) - v_(A y) or v_(B A x) = v_(B x) - 0` ...(ii)
`v_(B y) - (sqrt(3 - 1))`
`tan theta = v_(B y)/v_(B x) or v_(B y) = v_(B x) tan phi`
From (i),(ii),(iii), and (iv), we get
`v_(B x)= ((sqrt(3 - 1)))/(tan phi - 1) and v_(B y) = ((sqrt(3 - 1)))/(tan theta -1) tan phi`
`phi = (4 theta)/(3) = (4)/(3) (45^@)`
Speed of ball w.r.t. surface,
`v_B = sqrt(v_(B x)^2 + v_(B y)^2) = (sqrt( 3- 1))/(tan phi - 1) sec phi`
Substituting `phi = 60^@`, we get `v_B = 2 ms^-1`.
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