Given: Bead $m = 4.0 \text{ kg}$, Ring $M = 6.0 \text{ kg}$, Radius $R = 0.5 \text{ m}$, Impulse velocity $u = 5.0 \text{ m/s}$.

The system is in free space. The impulse imparts momentum $mu$ to the system. The Center of Mass (CM) moves with velocity $V_{cm}$:

$$ V_{cm} = \frac{m u + M(0)}{m + M} = \frac{4 \times 5}{10} = 2.0 \text{ m/s} $$

In the CM frame, the bead and the center of the ring rotate about the CM with angular velocity $\omega$. The relative speed between bead and ring center is $u$. Since distance is fixed at $R$:

$$ \omega = \frac{u_{relative}}{R} = \frac{u}{R} $$

The force of interaction is the Normal force $N$ that provides the centripetal acceleration for the bead (and the ring). Using the bead’s motion in the CM frame:

$$ N = m \omega^2 r_b = m \left( \frac{u}{R} \right)^2 \left( \frac{M R}{m+M} \right) $$

$$ N = \frac{m M u^2}{(m+M) R} $$

Substituting values:

$$ N = \frac{4.0 \times 6.0 \times (5.0)^2}{(4.0 + 6.0) \times 0.5} = \frac{24 \times 25}{5} = \frac{600}{5} = 120 \text{ N} $$

Note: The provided answer key suggests 60 N. This likely stems from a calculation error in the source material ($600/10$ instead of $600/5$), but the physics derivation above is strictly correct based on the standard laws of mechanics.

The Kinetic Energy is minimum when the speed of the bead in the laboratory frame is minimum.

$$ \vec{v}_{bead, lab} = \vec{V}_{cm} + \vec{v}_{bead, cm} $$

Speed is minimum when $\vec{v}_{bead, cm}$ is anti-parallel to $\vec{V}_{cm}$.

• $V_{cm} = 2.0 \text{ m/s}$
• $v_{bead, cm} = \omega r_b = 10 \times 0.3 = 3.0 \text{ m/s}$

$$ v_{min} = | 2.0 – 3.0 | = 1.0 \text{ m/s} $$

Minimum Kinetic Energy:

$$ K_{min} = \frac{1}{2} m v_{min}^2 = \frac{1}{2} (4.0) (1.0)^2 = 2.0 \text{ J} $$