Treat a planet as a uniform density sphere of mass M, radius R, and rotation per

Question

Treat a planet as a uniform density sphere of mass M, radius R, and rotation
period T. A large asteroid of mass aM is moving at speed ßR/T. The
asteroid collides perfectly inelastically with the planet at its equator while
traveling tangentially to the surface of the planet in the same direction as
the velocity of the contact point on the equator. What is the rotation period
of the planet after the asteroid collides with the planet?

Extra info: moment of inertia of a solid sphere of mass M and radius R is 2/5 MR^2

Explanation / Answer

Given: Mass of the planet = M Radious of the planet = R Intial time period of the planet Moment inertia of the planet I 1= 2/5 M R2 Let ,intial avgualr speed = W2 = ? (Before hit by the Asteriod) Mass of the Asteriod = M when it hits the planet at its equator The distance from the centre of the Planet to asteriod is = R Thus, Moment of inertia of the Asteriod + planet I2 = 2/5 MR2 + 2/5M R2 = 4/5 MR2 Intial velocity of the Asteriod is V = R / T thus, Angular velocity = W 2= V / R = / T It is known by the formula, we have , I 1 W1 = I2 W2 thus, W1 = 2 W2                      = 2 (/ T )                but , W1 = 2 / T'     where T' is the time period of the      Asteriod after hits the planet thus, 2 / T ' = 2 (/ T )                  T ' = / T                      = 2 (/ T )                but , W1 = 2 / T'     where T' is the time period of the      Asteriod after hits the planet thus, 2 / T ' = 2 (/ T )                  T ' = / T

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