A block of mass m = 1 , kg slides with veloci | vedclass

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(B) Step $1$: Conservation of angular momentum about the pivot $O$ during the collision.$L_i = L_f$$mvl = I_{total} \omega$$mvl = (\frac{Ml^2}{3} + ml

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(B) Step $1$: Conservation of angular momentum about the pivot $O$ during the collision.$L_i = L_f$$mvl = I_{total} \omega$$mvl = (\frac{Ml^2}{3} + ml^2) \omega$Substituting the values: $1 	imes 6 	imes 1 = (\frac{2 	imes 1^2}{3} + 1 	imes 1^2) \omega$$6 = (\frac{2}{3} + 1) \omega = \frac{5}{3} \omega$$\omega = \frac{18}{5} \, rad/s = 3.6 \, rad/s$Step $2$: Conservation of mechanical energy after the collision.The kinetic energy of the system is converted into potential energy as the rod swings by an angle $	heta$.$K_i = U_f$$\frac{1}{2} I_{total} \omega^2 = (M + m) g h_{cm}(1 - \cos 	heta)$Where $h_{cm}$ is the height of the center of mass from the pivot.$h_{cm} = \frac{M(l/2) + m(l)}{M + m} = \frac{2(0.5) + 1(1)}{2 + 1} = \frac{2}{3} \, m$$\frac{1}{2} (\frac{5}{3}) (\frac{18}{5})^2 = (2 + 1) 	imes 10 	imes \frac{2}{3} (1 - \cos 	heta)$$\frac{1}{2} 	imes \frac{5}{3} 	imes \frac{324}{25} = 20 (1 - \cos 	heta)$$\frac{54}{5} = 20 (1 - \cos 	heta)$$1 - \cos 	heta = \frac{54}{100} = 0.54$$\cos 	heta = 1 - 0.54 = 0.46$$	heta = \cos^{-1}(0.46) \approx 62.6^{\circ} \approx 63^{\circ}$

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