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  • A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The

A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 30 m and the bullet at a distance of 120 m from the foot of the post. The value of initial velocity of the bullet will be (if g=10\; m/s^{2} ) :

Option: 1

360 m/s


Option: 2

400 m/s


Option: 3

60 m/s 


Option: 4

120 m/s


Answers (1)

best_answer

Time to reach ground will be same for both

\mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 20}{10}}=2 \mathrm{sec}

Range of bullet = 120

120=\mathrm{v}_2(2) \Rightarrow \mathrm{v}_2=60 \mathrm{~m} / \mathrm{sec}

Range of ball = 30

30=\mathrm{V}_1(2) \Rightarrow \mathrm{v}_1=15 \mathrm{~m} / \mathrm{sec}

Now apply momentum conservation

\begin{aligned} & P_i=P_f \\ & P_{\text {ball }}+P_{\text {bullet }}=P_{\text {ball }}+P_{\text {balles }} \\ & 0+\left(\frac{10}{1000}\right) v_0=\left(\frac{200}{1000}\right)(15)+\left(\frac{10}{1000} \times 60\right) \\ & 10 \mathrm{v}_0=3000+600 \\ & \mathrm{v}_0=\frac{3600}{10} \Rightarrow \mathrm{v}_0=360 \mathrm{~m} / \mathrm{sec} \end{aligned}

 

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