A bullet of mass 10, g moving horizontally wi | vedclass

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(B) Given:Mass of bullet,$m = 10\, g = 0.01\, kg$Initial speed of bullet,$u = 400\, m s^{-1}$Mass of block,$M = 2\, kg$Vertical rise of block,$h = 10\

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) Given:Mass of bullet,$m = 10\, g = 0.01\, kg$Initial speed of bullet,$u = 400\, m s^{-1}$Mass of block,$M = 2\, kg$Vertical rise of block,$h = 10\, cm = 0.1\, m$Let $v_1$ be the speed of the block immediately after the collision and $v$ be the speed of the bullet after emerging from the block.Applying the principle of conservation of mechanical energy to the block after the collision:$\frac{1}{2} M v_1^2 = Mgh$$v_1 = \sqrt{2gh} = \sqrt{2 	imes 9.8 	imes 0.1} = \sqrt{1.96} = 1.4\, m s^{-1}$(Using $g = 9.8\, m s^{-2}$ for precision).Applying the principle of conservation of linear momentum for the system (bullet + block) during the collision:$mu = Mv_1 + mv$$0.01 	imes 400 = 2 	imes 1.4 + 0.01 	imes v$$4 = 2.8 + 0.01v$$0.01v = 1.2$$v = 120\, m s^{-1}$

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