A lead bullet moving with velocity V strikes | vedclass

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Question

(A) The initial kinetic energy of the bullet is $K.E. = \frac{1}{2} MV^2$.Given that $75 \%$ of this energy is converted into heat,the heat energy pro

Vedclass · Accepted Answer

$\frac{3 V^2}{8 Js}$

Vedclass · Answer

(A) The initial kinetic energy of the bullet is $K.E. = \frac{1}{2} MV^2$.Given that $75 \%$ of this energy is converted into heat,the heat energy produced is $Q = 0.75 	imes K.E. = \frac{3}{4} 	imes \frac{1}{2} MV^2 = \frac{3}{8} MV^2$.The heat energy required to raise the temperature of the bullet by $\Delta T$ is given by $H = M s \Delta T$.Using the mechanical equivalent of heat $J$,the relationship between heat energy in Joules $(W)$ and calories $(Q)$ is $W = JQ$.Here,the work done (energy converted) is $W = \frac{3}{8} MV^2$ and the heat absorbed is $Q = M s \Delta T$.Thus,$\frac{3}{8} MV^2 = J (M s \Delta T)$.Solving for the increase in temperature $\Delta T$:$\Delta T = \frac{3 MV^2}{8 J M s} = \frac{3 V^2}{8 Js}$.

$A$ lead bullet moving with velocity $V$ strikes a wall and stops. If $75 \%$ of its energy is converted into heat,then the increase in temperature is ($s=$ specific heat of lead,$J=$ mechanical equivalent of heat).

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