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

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(A) Let $m = 10\,g = 0.01\,kg$ be the mass of the bullet,$u = 400\,m/s$ be its initial velocity,$M = 2\,kg$ be the mass of the block,and $h = 10\,cm =

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(A) Let $m = 10\,g = 0.01\,kg$ be the mass of the bullet,$u = 400\,m/s$ be its initial velocity,$M = 2\,kg$ be the mass of the block,and $h = 10\,cm = 0.1\,m$ be the vertical rise.By the law of conservation of linear momentum during the collision: $mu = Mv + mV'$,where $v$ is the velocity of the block immediately after impact and $V'$ is the final velocity of the bullet.When the block rises to height $h$,its kinetic energy is converted into potential energy: $\frac{1}{2}Mv^2 = Mgh \Rightarrow v = \sqrt{2gh}$.Using $g = 10\,m/s^2$,$v = \sqrt{2 	imes 10 	imes 0.1} = \sqrt{2} \approx 1.414\,m/s$.Substituting values into the momentum equation: $(0.01)(400) = (2)(\sqrt{2}) + (0.01)V'$.$4 = 2.828 + 0.01V' \Rightarrow 0.01V' = 1.172$.$V' = 117.2\,m/s$. Rounding to the nearest provided option,$V' \approx 120\,m/s$.

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