Question

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 $x-axis$ (see figure) with a constant velocity of $(\sqrt 3 - 1)m/s.$. At a particular instant when the line OA makes an angle of $45^{\circ}$ with the $x$-axis, a ball is thrown along the surface from the origin $O$. Its velocity makes an angle $\varnothing$ 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 $\varnothing = 4\theta / 3.$

Answer 1

Let A stands for trolley and $B$ for ball.
Relative velocity of $B$ with respect to $A (v_{BA})$ should be along $OA$ for the ball to hit the trolley.
Hence, $v_{BA}$ will make an angle of $45^{\circ}$ with positive x-axis.
$\theta = 45^{\circ}$
(b) Let $v$= absolute velocity of ball.
$\varnothing = \frac{4 \theta}{3}=\frac{4}{3} (45^{\circ})=60^{\circ} \rightarrow$ with x-axis
$ \therefore v_B = (v\, \cos\, \theta)\hat{i}+(v\, \sin\, \theta)\hat{j}=\frac{v}{2}\hat{i}+\frac{\sqrt 3 v}{2}\hat{j}$
$v_A = (\sqrt 3 -1)\hat{j}$
$ \therefore v_{BA} = \frac{v}{2}\hat{i}+(\frac{\sqrt3 v}{2}-\sqrt 3 + 1)\hat{j}$
Since, $v_{BA}$ is at $45^{\circ}$
$ \therefore \frac{v}{2}+\frac{\sqrt3 v}{2}-\sqrt 3 + 1$ or $v = 2\,m/s$