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 $(\sqrt{3}-1)\,\text{m/s}$. 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 $\phi$ 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 $\phi = 40/3$.

Step 1: Define the coordinates and initial positions.

Let the origin $O(0,0)$ be the projection point of the ball, and let the trolley $A$ move along the line $x = a$.

$$\vec{r}_A(0) = (a, a)$$

Step 2: Write the position vectors as functions of time.

The trolley moves parallel to the $y$-axis with constant velocity $v_A = \sqrt{3}-1$.

$$\vec{r}_A(t) = (a, a + v_A t)$$

The ball is projected from the origin with speed $u$ at an angle $\phi$ with the $x$-axis.

$$\vec{r}_{\text{ball}}(t) = (u\cos\phi\,t, u\sin\phi\,t)$$

Step 3: Apply the condition for the ball to hit the trolley.

At the instant of collision $t$, the position vectors of the ball and the trolley must be equal.

$$u\cos\phi\,t = a$$ $$u\sin\phi\,t = a + v_A t$$

Substitute the first equation into the second equation.

$$u\sin\phi\,t = u\cos\phi\,t + v_A t$$

Divide both sides by $t$.

$$u\sin\phi - u\cos\phi = v_A$$

Step 4: Determine the relative velocity and the angle $\theta$ in the trolley's frame.

The velocity of the ball relative to the trolley is the difference of their velocities.

$$\vec{v}_{\text{rel}} = \vec{v}_{\text{ball}} - \vec{v}_A$$ $$\vec{v}_{\text{rel}} = (u\cos\phi, u\sin\phi - v_A)$$

Substitute $u\sin\phi - v_A = u\cos\phi$ into the relative velocity vector.

$$\vec{v}_{\text{rel}} = (u\cos\phi, u\cos\phi)$$

Calculate the angle $\theta$ of this relative velocity vector with the $x$-axis.

$$\tan\theta = \frac{u\cos\phi}{u\cos\phi} = 1$$ $$\theta = 45^\circ$$

Step 5: Calculate the speed of the ball with respect to the surface.

The given angle satisfies $\tan\phi = 4/3$.

$$\sin\phi = \frac{4}{5}$$ $$\cos\phi = \frac{3}{5}$$

Substitute these values into the collision relation.

$$u\left(\frac{4}{5} - \frac{3}{5}\right) = v_A$$ $$\frac{u}{5} = \sqrt{3}-1$$ $$u = 5(\sqrt{3}-1)\,\text{m/s}$$ $$u \approx 3.66\,\text{m/s}$$

Answer: (a) $\theta = 45^\circ$ (b) $u = 5(\sqrt{3}-1)\,\text{m/s} \approx 3.66\,\text{m/s}$