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A thin circular wooden hoop of mass m and radius Rrests on a horizontal frictionless plane. A bullet, also of mass m, moving with horizontal velocity v, strikes the hoop and becomes embedded in it as shown in the figure. (a) (2) Calculate the center of mass velocity (b) (2) Calculate the angular momentum of the system about the CM. (c) (32) Calculate the angular velocity w of the hoop. (d) (4) Calculate the kinetic energy of the system, before and after collision (e) (48) Find a point of the hoop which remains at rest after collision

Answer :

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The center of mass velocity of the system after the collision is: vcm = (mv + 0) / (2m) = v/2. the angular momentum of the system about the CM is: Lcm = Icm * wcm = (1/2)mR^2 * wcm. the angular velocity of the hoop is: w = Lcm / Icm = (Lcm / (1/2)mR^2) = 2Lcm / mR^2.

The center of mass velocity of the system after the collision can be calculated using the conservation of linear momentum. The linear momentum of the bullet before the collision is mv, and the linear momentum of the hoop before the collision is 0, since it is at rest. The linear momentum of the system after the collision is the sum of the linear momenta of the bullet and the hoop, which is mv. Therefore, the center of mass velocity of the system after the collision is: vcm = (mv + 0) / (2m) = v/2. The angular momentum of the system about the center of mass (CM) can be calculated using the equation: Lcm = Icm * wcm where Lcm is the angular momentum, Icm is the moment of inertia of the system about the CM, and wcm is the angular velocity of the system about the CM.

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