A large block of wood of mass M = 5.99 , kg i | vedclass

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(C) Given:Mass of block $M = 5.99 \, kg$Mass of bullet $m = 10 \, g = 0.01 \, kg$Total mass $M + m = 5.99 + 0.01 = 6.00 \, kg$Vertical height $h = 9.8

Vedclass · Accepted Answer

$831.5$

Vedclass · Answer

(C) Given:Mass of block $M = 5.99 \, kg$Mass of bullet $m = 10 \, g = 0.01 \, kg$Total mass $M + m = 5.99 + 0.01 = 6.00 \, kg$Vertical height $h = 9.8 \, cm = 0.098 \, m$Acceleration due to gravity $g = 9.8 \, ms^{-2}$Step $1$: Apply conservation of mechanical energy for the (block $+$ bullet) system after the collision.$(M + m)gh = \frac{1}{2}(M + m)v_1^2$$v_1 = \sqrt{2gh} = \sqrt{2 	imes 9.8 	imes 0.098} = \sqrt{19.6 	imes 0.098} = \sqrt{1.9208} \approx 1.3856 \, m/s$Step $2$: Apply conservation of linear momentum during the collision.$mv = (M + m)v_1$$v = \frac{(M + m)v_1}{m} = \frac{6.00 	imes 1.3856}{0.01} = 600 	imes 1.3856 = 831.36 \, m/s$Rounding to the nearest option,the speed is $831.5 \, m/s$.

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