Solve the following :


A bullet of mass $10 \mathrm{~g}$ moving horizontally at a speed of $50 \sqrt{7} \mathrm{~m} / \mathrm{s}$ strikes a block of mass $490 \mathrm{~g}$ kept on a frictionless track as shown in figure. The bullet remains inside the block and the system proceeds towards the semicircular track of radius $0.2 \mathrm{~m}$. Where will the block strike the horizontal part after leaving the semicircular track?


Use C.O.L.M,

$\left(10 \times 10^{-3} \times 50 \sqrt{7}\right)+0=(490+10) \times 10^{-3} \mathrm{~V}$

$V=\sqrt{7} \mathrm{~m} / \mathrm{s}$

$F_{c}=\frac{m v_{B}^{2}}{r}=m g \sin \theta$

$\left(V_{B}=\right.$ velocity at position $\left.B\right)$

Use C.O.E.L

$\frac{1}{2} m\left(V_{B}^{2}-V^{2}\right)=-m g 0.2(1+\sin \theta)$

$\frac{1}{2}(g r \sin \theta-7)=g(0.2)(1+\sin \theta) \quad\{r=0.2\}$

$\sin \theta=\frac{1}{2} \theta=30^{\circ}$

$\alpha=$ Angle of projection $=\frac{\pi}{2}-30^{\circ}=60^{\circ}$

$t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 0.2(1+\sin \theta)}{g}}$

$=\sqrt{\frac{0.2 \times 3}{10}}=0.247$

Distance in horizontal direction

$x=v \cos \alpha \times t=\sqrt{9.8 \times 2 \times \frac{1}{2}} \times 0.247$

$=0.196 \mathrm{~m}$

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