Q86 / 90 · JEE Main 2021 (27 Aug Shift 2) · Physics

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-1. The length of the pendulum is 0.5 m and mass of the bob is 1 kg. The minimum value of v in m s-1, so that the pendulum describes a circle. (Assume the string to be inextensible and g=10 m s-2 )

Figure

Correct answer: 400

Solution

Final velocity of system (bullet + mass) should be =5gr

Now M5gr=m(v+100)

thus v=Mm5gr-100=500-100=400 m s-1

v=400 m s-1

Hence, the correct answer is 400.

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