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uniform thin rod of length 0.500 m and mass 4.00 kg can rotate in a horizontal plane about a vertical axis through its centre. The rod is at rest when a 3.00 g bullet travelling in the rotation plane is fired into one end of the rod. In the view from above, the bullet's path makes angle θ=60.0o with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullet's speed just before impact?

Respuesta :

The speed of bullet just before the impact would be 1.28 x 10^3 meter per second.

Let the speed of the bullet before impact be 'V₀.'

According to the question, we are given the mass of the rod 'M' = 4 kg. Length of the rod 'l' = 0.5 meter. The angle of the bullet's path 'Θ' = 60 degree and the angular velocity 'ω' = 10 rad/s.

   For bullet, the mass 'M' = 3 g = 3 x 10^3 kg. and the velocity is V₀.

  According to the law of conservation of momentum, we get

L(bullet) = l (bullet + rod)

mV₀ (l/2 sinΘ) = Iω

mV₀ x (l/2 sinΘ) = [ ml²/ 12 + m (l/2) ²] x ω

By cancelling the like terms, we get the equation as,

V₀ = (m +3m) x l/6m sin Θ

  By calculating the above equation, we get,

V₀ = 1.28 x 10^3 meter per second.

  Therefore, the speed of bullet before the impact is 1.28 x 10^3 meter per second.

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