In a spring gun having spring constant \(100\, {N} / {m}\) a small ball \('B'\) of mass \(100\, {g}\) is put in its barrel (as shown in figure) by compressing the spring through \(0.05\, {m}\). There should be a box placed at a distance \('d'\) on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of \(2\, {m}\) above the ground. The value of \(d\) is \(....{m} .\) \(\left(g=10\, {m} / {s}^{2}\right)\)

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Answer & Solution

Correct Answer

(C) \(1\)

Detailed explanation

\(\frac{1}{2} k x^{2}=\frac{1}{2} m v^{2}\)

\(k x^{2}=m v^{2}\)

\(V=x \sqrt{\frac{k}{m}}=0.05 \sqrt{\frac{100}{1}}=0.05 \times 10 \sqrt{10}\)

\(v=0.5 \sqrt{10}\)

From \(h=\frac{1}{2} g t^{2}\)

\(t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 2}{10}}=\frac{2}{\sqrt{10}}\)

\(\therefore d=v t=0.5 \sqrt{10} \times \frac{2}{\sqrt{10}}=1\, {m}\)

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