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 φ 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 θ 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 φ = 4θ /3  .

Solution:  (a) If the trolley A is at rest, anything that comes to hit it must be along the line OA, so the angle should be \( \theta=45^{0}\) with the x- axis.

(b) Let the ball is given a velocity v at an angle φ = 4θ /3, where \( \theta=45^{0}\), the angle we get is \( \phi=60^{0}\). The observer at trolley will see it coming at angle \( \theta=45^{0}\) so the relative velocity . \(\vec{v}_{rel} = \vec{v}_{ball} - \vec{v}_{Trolley}\) will have same x and y components.

Here \(\vec{v}_{ball} =v cos 60^{0}\vec{i} + v sin 60^{0}\vec{j}\)

and \(\vec{v}_{Trolley} = (\sqrt{3} -1)\vec{j}\).

Now the x- component of \(v_{rel}\) = \(\frac{v}{2}\)

and the y -component of \(v_{rel}\) = \(\frac{\sqrt{3}}{2}v - \sqrt{3}+1\). Since both the components are equal, we get v= 2m/s after a little bit of manipulation.