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. (Take, g =10 m / s²)

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answer is 0.003.

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Detailed Solution

Given, k = 100N/m

m = 100g = 0.1 kg

x = 0.05 m and H = 2m

By energy conservation,

$\begin{matrix} \frac{1}{2} {kx}^{2} = \frac{1}{2} {mv}^{2} \Rightarrow v = x \sqrt{\frac{k}{m}} \\ = 0 .05 \times \sqrt{\frac{100}{0 .1}} = 0 .5 \sqrt{10} {ms}^{- 1} . . . (i) \end{matrix}$

Time of flight of ball, $t = \sqrt{\frac{2 H}{g}}$

$\Rightarrow t = \sqrt{\frac{2 \times 2}{10}} = \frac{2}{\sqrt{10}} s$ ...(ii)

$\begin{array}{rcl} ∴ Range of ball, d & = & vt \\ = & 0 .5 \sqrt{10} \times (\frac{2}{\sqrt{10}}) [From Eqs . (i) and (ii)] \\ = & 0 .5 \times 2 = 30 \times 10^{- 3} m = 0 .003 m \end{array}$

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