A bullet of mass 10 g travelling horizontally with a speed of 50 7...

By conservation of linear momentum during the collision:
$m u = (m + M) v$
$0.01 \times 50\sqrt{7} = (0.01 + 0.49) v$
$0.5\sqrt{7} = 0.5 v \Rightarrow v = \sqrt{7} \text{ m/s}$

Let the block leave the semicircular track at an angle $\theta$ above the horizontal diameter. At this point, the normal reaction $N = 0$. The component of weight along the radius provides the necessary centripetal force:
$(m + M)g \sin\theta = \dfrac{(m + M)v'^2}{R}$
$v'^2 = gR \sin\theta = 10 \times 0.2 \times \sin\theta = 2\sin\theta$

By conservation of mechanical energy from the bottom of the track to this point:
$\dfrac{1}{2}(m + M)v^2 = \dfrac{1}{2}(m + M)v'^2 + (m + M)gR(1 + \sin\theta)$
$v^2 = v'^2 + 2gR(1 + \sin\theta)$
$7 = 2\sin\theta + 2(10)(0.2)(1 + \sin\theta)$
$7 = 2\sin\theta + 4(1 + \sin\theta)$
$7 = 4 + 6\sin\theta \Rightarrow 6\sin\theta = 3 \Rightarrow \sin\theta = \dfrac{1}{2}$
\theta = 30^{}^{{}^{\circ}}

The velocity of the block when it leaves the track is:
v' = \sqrt{2 \sin 30^{}^{{}^{\circ}}} = 1 \text{ m/s}

Taking the junction of the straight and curved parts as the origin $(0,0)$, the position of the block when it leaves the track is:
x_0 = -R \cos 30^{}^{{}^{\circ}} = -0.2 \times \dfrac{\sqrt{3}}{2} = -0.1\sqrt{3} \text{ m}
y_0 = R + R \sin 30^{}^{{}^{\circ}} = 0.2 + 0.2 \times 0.5 = 0.3 \text{ m}

The velocity vector is tangent to the path, making an angle of 90^{}^{{}^{\circ}} - \theta = 60^{}^{{}^{\circ}} with the horizontal.
v_x = v' \cos 60^{}^{{}^{\circ}} = 1 \times 0.5 = 0.5 \text{ m/s}
v_y = v' \sin 60^{}^{{}^{\circ}} = 1 \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2} \text{ m/s}

For the subsequent projectile motion, the time $t$ to hit the horizontal part $(y = 0)$ is given by:
$y = y_0 + v_y t - \dfrac{1}{2}gt^2$
$0 = 0.3 + \dfrac{\sqrt{3}}{2} t - 5t^2$
$50t^2 - 5\sqrt{3}t - 3 = 0$

Solving for $t$ using the quadratic formula:
$t = \dfrac{5\sqrt{3} \pm \sqrt{75 - 4(50)(-3)}}{100} = \dfrac{5\sqrt{3} \pm \sqrt{675}}{100} = \dfrac{5\sqrt{3} \pm 15\sqrt{3}}{100}$
Taking the positive root, $t = \dfrac{20\sqrt{3}}{100} = \dfrac{\sqrt{3}}{5} \text{ s}$

The horizontal position at this time is:
$x = x_0 + v_x t = -0.1\sqrt{3} + 0.5 \times \dfrac{\sqrt{3}}{5} = -0.1\sqrt{3} + 0.1\sqrt{3} = 0 \text{ m}$

Thus, the block hits exactly at the junction of the straight and curved parts.

Answer: Exactly at the junction of the straight and curved parts