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Hak

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- Homework Statement
- An asteroid has a spherical shape and uniform mass distribution. Its radius is ##R## and its mass ##M##. The asteroid is stationary in interstellar space when it is struck by another much smaller asteroid of mass ##m## and negligible radius relative to ##R##. The minor asteroid has velocity ##v## before the impact, and its direction is tangent to the surface of the major asteroid. The impact is completely inelastic, so that the minor asteroid remains embedded inside the major one, but it is not destroyed; it is just embedded inside the rock. The gravity is negligible.

After the collision has occurred, what is the value of the force acting on the minor asteroid?

- Relevant Equations
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Before the collision, the total linear momentum of the system is ##p = mv##, where ##m## is the mass of the minor asteroid and ##v## is its velocity. The total angular momentum of the system is ##L = mRv##, where ##R## is the radius of the major asteroid. Since there are no external forces or torques acting on the system, both ##p## and ##L## are conserved during and after the collision.

After the collision, the minor asteroid becomes embedded inside the major one, so they move together as a single rigid body. The linear momentum of this body is still ##p = mv##, so its velocity is ##v' = \frac{p}{(M + m)}##, where ##M## is the mass of the major asteroid. The angular momentum of this body is also still ##L = mRv##, so its angular velocity is ##\omega' = \frac{L}{I}##, where ##I## is the moment of inertia of the body.

The moment of inertia of a sphere with uniform mass distribution and radius ##R## is ##I = (2/5)MR^2##. However, since there is a small mass ##m## embedded inside the sphere at a distance ##R## from its center, we need to use the parallel axis theorem to find the moment of inertia of the body. The parallel axis theorem states that ##I = I_{cm} + md^2##, where ##I_{cm}## is the moment of inertia about the center of mass and ##d## is the distance from the center of mass to the axis of rotation. In this case, ##I_cm = (2/5)MR^2## and ##d = R##, so we get ##I = (\frac{2}{5})MR^2 + mR^2 = (\frac{2}{5} + \frac{m}{M})MR^2##.

Now we can find the angular velocity ##\omega'## by plugging in ##L## and ##I## into ##\omega' = \frac{L}{I}##. We get ##\omega' = \frac{(mRv)}{[(\frac{2}{5} + \frac{m}{M})MR^2]} = \frac{(5v)}{[2R(\frac{5}{2} + \frac{m}{M})]}##. This means that after the collision, both asteroids rotate together with an angular velocity ##\omega'## around their common center of mass.

The force acting on the minor asteroid after the collision is due to its centripetal acceleration toward the center of rotation. The centripetal acceleration is given by ##a_c = \omega'^2 R##, where ##\omega'## is the angular velocity and ##R## is the distance from the center. The force is then given by ##F = ma_c##, where ##m## is the mass of the minor asteroid. Plugging in ##\omega'## and ##R##, we get $$F = m(\omega '^2 R) = m \left[\frac{(5v)}{\left[2R \left(\frac{5}{2} + \frac{m}{M}\right)\right]}\right]^2 R = \frac{(25mv^2)}{\left[4R \left(\frac{5}{2} + \frac{m}{M}\right)^2 \right]}$$.

Where am I going wrong?

After the collision, the minor asteroid becomes embedded inside the major one, so they move together as a single rigid body. The linear momentum of this body is still ##p = mv##, so its velocity is ##v' = \frac{p}{(M + m)}##, where ##M## is the mass of the major asteroid. The angular momentum of this body is also still ##L = mRv##, so its angular velocity is ##\omega' = \frac{L}{I}##, where ##I## is the moment of inertia of the body.

The moment of inertia of a sphere with uniform mass distribution and radius ##R## is ##I = (2/5)MR^2##. However, since there is a small mass ##m## embedded inside the sphere at a distance ##R## from its center, we need to use the parallel axis theorem to find the moment of inertia of the body. The parallel axis theorem states that ##I = I_{cm} + md^2##, where ##I_{cm}## is the moment of inertia about the center of mass and ##d## is the distance from the center of mass to the axis of rotation. In this case, ##I_cm = (2/5)MR^2## and ##d = R##, so we get ##I = (\frac{2}{5})MR^2 + mR^2 = (\frac{2}{5} + \frac{m}{M})MR^2##.

Now we can find the angular velocity ##\omega'## by plugging in ##L## and ##I## into ##\omega' = \frac{L}{I}##. We get ##\omega' = \frac{(mRv)}{[(\frac{2}{5} + \frac{m}{M})MR^2]} = \frac{(5v)}{[2R(\frac{5}{2} + \frac{m}{M})]}##. This means that after the collision, both asteroids rotate together with an angular velocity ##\omega'## around their common center of mass.

The force acting on the minor asteroid after the collision is due to its centripetal acceleration toward the center of rotation. The centripetal acceleration is given by ##a_c = \omega'^2 R##, where ##\omega'## is the angular velocity and ##R## is the distance from the center. The force is then given by ##F = ma_c##, where ##m## is the mass of the minor asteroid. Plugging in ##\omega'## and ##R##, we get $$F = m(\omega '^2 R) = m \left[\frac{(5v)}{\left[2R \left(\frac{5}{2} + \frac{m}{M}\right)\right]}\right]^2 R = \frac{(25mv^2)}{\left[4R \left(\frac{5}{2} + \frac{m}{M}\right)^2 \right]}$$.

Where am I going wrong?

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