Question:
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. (g = 10 m/s\(^2\)).
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. (g = 10 m/s\(^2\)).
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Spring energy \(\frac12kx^2\) becomes the ball's kinetic energy; then treat it as a horizontal projectile falling 2 m and use \(d = vt\).
Updated On: Nov 4, 2026
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Correct Answer: 1
Solution and Explanation
Energy stored in the spring: \(\frac12 kx^2 = \frac12\times100\times(0.05)^2 = 0.125\) J.
This becomes kinetic energy of the ball (m = 0.1 kg): \(\frac12 mv^2 = 0.125 \Rightarrow v^2 = 2.5\ \text{m}^2/\text{s}^2\), \(v = \sqrt{2.5}\) m/s (horizontal).
Time of fall from 2 m: \(h=\frac12 gt^2 \Rightarrow t=\sqrt{\frac{2\times2}{10}}=\sqrt{0.4}\) s.
Horizontal distance \(d = vt = \sqrt{2.5\times0.4} = \sqrt{1} = 1\) m.
Answer: 1
This becomes kinetic energy of the ball (m = 0.1 kg): \(\frac12 mv^2 = 0.125 \Rightarrow v^2 = 2.5\ \text{m}^2/\text{s}^2\), \(v = \sqrt{2.5}\) m/s (horizontal).
Time of fall from 2 m: \(h=\frac12 gt^2 \Rightarrow t=\sqrt{\frac{2\times2}{10}}=\sqrt{0.4}\) s.
Horizontal distance \(d = vt = \sqrt{2.5\times0.4} = \sqrt{1} = 1\) m.
Answer: 1
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