Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

A rod of mass m and length L, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets embedded in it. The combined system now rotates with the angular speed ω about the pivot. The maximum angular speed ωM is achieved for x = xM. Then

solution : according to conservation of angular momentum about the suspension point,

L = Iω

here L = mvx and I = moment of inertia about suspension point

= mL²/3 + mx² [ using parallel axis theorem ]

now, mvx = m(L²/3 + x²)ω

⇒ω = 3vx/(L² + 3x²)

therefore option (A) is correct choice.

for maximum value of ω ⇒dω/dx = 0

⇒3v [ (L² + 3x²) - 6x(x)]/(L² + 3x²) = 0

⇒L² - 3x² = 0

⇒x = L/√3 , option (C) is correct choice.

so the angular speed ω = 3v(L/√3)/(L² + L²)

= √3v/2L

so the option (D) is also correct choice.

Therefore options (A), (C) and (D) are correct choices.