Find the point on the curve `y^2=a x` the tangent at which makes an angle of 45^0 with the x-axis.

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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