Question

A bullet of mass 10 g moving horizontally at a speed of $50\sqrt{7}ms/$ strikes a block of mass 490 g kept on a frictionless track as shown in figure (9-E17). the bullet remains inside the block and the system proceeds towards the semicircular track of radius 0.2 m. Where will the block strike the horizontal part after leaving the semicircular track ?

Answer 1

Mass of block = 490 gm

Mass of bullet = 10 gm

Since the bullet embedded inside the block,

it is a plastic collision

Initial velocity of bullet

$v_1=50\sqrt{7}m/s$

Velocity of the block is $v_2=0$

Let final velocity of both = v

Hence $10\times {10}^{-3}\times 50\sqrt{7}+490\times {10}^{-3}\times 0=(490+10)$

$={10}^{-3}\times V_A$

$\therefore$ $V_A=\sqrt{7}m/s$

When the block loses the contact at 'D' the component mg will act on it,

$\frac{m(V_B{)}^2}{r}=mgsin\theta$

$\Rightarrow (V_B{)}^2=grsin\theta ...(i)$

Putting work energy principle

$\left(\frac{1}{2}\right)m\times (V_B{)}^2-\left(\frac{1}{2}\right)\times m\times {\left(V_A\right)}^2=-mg(0.2+0.2sin\theta )$

$\Rightarrow \left(\frac{1}{2}\right)\times grsin\theta -\left(\frac{1}{2}\right)\times {\left(\sqrt{7}\right)}^2=g(0.2+0.2sin\theta )$

$\Rightarrow 3.5-\left(\frac{1}{2}\right)+9.8+0.2+sin\theta =9.8+0.2(1+sin\theta )$

$\Rightarrow 3.5-0.98sin\theta =1.96+1.96sin\theta$

$\Rightarrow sin\theta =\left(\frac{1}{2}\right)$

$\Rightarrow \theta ={30}^{\circ }$

$\therefore$ Angle of projection = ${90}^{\circ }-{30}^{\circ }={60}^{\circ }$

$\therefore$ time of reaching the ground = $\sqrt{\frac{2h}{g}}$

$=\sqrt{\frac{2\times (0.2+0.2\times sin{30}^{\circ })}{9.8}}=0.247sec$

$\Rightarrow$ Distance travelled in horizontal direction

$S=vcosq\times t$

$=\sqrt{grsin\theta }cos\theta \times t$

$=\sqrt{9.8\times 2\times \left(\frac{1}{2}\right)}\frac{\sqrt{3}}{2}\times 0.247$

$=0.196m$

Total distance = $(0.2-0.2cos{30}^{\circ }+0.196)$

$=0.22m$