JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 21)
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/s2).
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Answer
1
Explanation
Given, k = 100 N/m
m = 100 g = 0.1 kg
x = 0.05 m and H = 2 m
By energy conservation,
$${1 \over 2}k{x^2} = {1 \over 2}m{v^2} \Rightarrow v = x\sqrt {{k \over m}} $$
$$ = 0.05 \times \sqrt {{{100} \over {0.1}}} = 0.5\sqrt {10} $$ ms$$-$$1 .... (i)
Time of flight of ball, $$t = \sqrt {{{2H} \over g}} $$
$$ \Rightarrow t = \sqrt {{{2 \times 2} \over {10}}} = {2 \over {\sqrt {10} }}$$s .... (ii)
$$\therefore$$ Range of ball, d = vt
$$ = 0.5\sqrt {10} \times \left( {{2 \over {\sqrt {10} }}} \right)$$ [From Eqs. (i) and (ii)]
$$ = 0.5 \times 2 = 1 m$$
m = 100 g = 0.1 kg
x = 0.05 m and H = 2 m
By energy conservation,
$${1 \over 2}k{x^2} = {1 \over 2}m{v^2} \Rightarrow v = x\sqrt {{k \over m}} $$
$$ = 0.05 \times \sqrt {{{100} \over {0.1}}} = 0.5\sqrt {10} $$ ms$$-$$1 .... (i)
Time of flight of ball, $$t = \sqrt {{{2H} \over g}} $$
$$ \Rightarrow t = \sqrt {{{2 \times 2} \over {10}}} = {2 \over {\sqrt {10} }}$$s .... (ii)
$$\therefore$$ Range of ball, d = vt
$$ = 0.5\sqrt {10} \times \left( {{2 \over {\sqrt {10} }}} \right)$$ [From Eqs. (i) and (ii)]
$$ = 0.5 \times 2 = 1 m$$
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