A 3.50-g bullet, traveling at 250 m/s, hits a hinged, light board at right angles, at a distance of 55.0 cm from its hinges, and emerges on the other side of the board with a speed of 150 m/s. The board is initially stationary but is free to rotate about its hinges and has a moment of inertia of 0.275 kg∙m². What is the angular speed of the board after the bullet emerges? A) 0.700 rad/s B) 3.05 rad/s C) 6.10 rad/s D) 0.0212 rad/s E) 22.6 rad/s

Question

Quizplus · Accepted Answer

Explanation:Momentum is conserved in the x direction:$$m_bv_{bi}=m_bv_{bf}+I\omega$$$$(3.50	imes10^{-3}	ext{ kg})(250	ext{ m/s})=(3.50	imes10^{-3}	ext{ kg})(150	ext{ m/s})+(0.275	ext{ kg$\cdot$m}^2)\omega$$$$\omega=\boxed{0.700	ext{ rad/s}}$$

A 350-G Bullet, Traveling at 250 M/s, Hits a Hinged, Light