A lead bullet moving with velocity 'v' strike | vedclass

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(B) The kinetic energy of the bullet is $K.E. = \frac{1}{2} mv^2$.Given that $50\%$ of this energy is converted into heat energy $(Q)$.Therefore,the h

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$\frac{v^2}{4 sJ}$

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(B) The kinetic energy of the bullet is $K.E. = \frac{1}{2} mv^2$.Given that $50\%$ of this energy is converted into heat energy $(Q)$.Therefore,the heat energy produced is $Q = \frac{1}{2} 	imes (\frac{1}{2} mv^2) = \frac{1}{4} mv^2$.The heat required to raise the temperature of the bullet by $\Delta T$ is given by $Q = ms \Delta T$.Since $J$ is the mechanical equivalent of heat,we use the relation $W = JQ$,where $W$ is the work done (or energy converted).Equating the energy converted to heat: $\frac{1}{4} mv^2 = J(ms \Delta T)$.Solving for $\Delta T$: $\Delta T = \frac{mv^2}{4 Jms} = \frac{v^2}{4 Js}$.

$A$ lead bullet moving with velocity '$v$' strikes a wall and stops. If $50\%$ 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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