JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Kinematics

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.

Answer: 1

Solution

The compressed spring hands all its energy to the ball, so

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

and

v = x \frac{k}{m} = 0.05 \frac{100}{0.1} = 0.05 \times 31.6 = 1.58 m/s

The ball leaves horizontally from a height of

2

m, so its vertical motion starts from rest and

t = \frac{2 h}{g} = \frac{2 \times 2}{10} = 0.632 s

The horizontal velocity stays constant through the flight, so the box must be placed at

d = v t = 1.58 \times 0.632 = 1

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Kinematics

Solution

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