JEE MAIN - Physics (2021 - 27th August Evening Shift - No. 22)
A bullet of 10 g, moving with velocity v, collides head-on with the stationary bob of a pendulum and recoils with velocity 100 m/s. The length of the pendulum is 0.5 m and mass of the bob is 1 kg. The minimum value of v = ____________ m/s so that the pendulum describes a circle. (Assume the string to be inextensible and g = 10 m/s2)
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Answer
400
Explanation
$$V' = \sqrt {5gR} = \sqrt {5 \times 10 \times 0.5} $$
V' = 5 m/s
m1 V = m2 $$\times$$ 5 $$-$$ m1 $$\times$$ 100
$${{10} \over {1000}} \times V = 5 - {{10} \over {1000}} \times 100$$
V = 400 m/s
V' = 5 m/s
m1 V = m2 $$\times$$ 5 $$-$$ m1 $$\times$$ 100
$${{10} \over {1000}} \times V = 5 - {{10} \over {1000}} \times 100$$
V = 400 m/s
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