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(B) The kinetic energy of the lead ball is $K = \frac{1}{2}mV^2$.Given that $50\%$ of this energy is converted into heat,the heat energy produced is $

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$\frac{V^2}{4JS}$

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(B) The kinetic energy of the lead ball is $K = \frac{1}{2}mV^2$.Given that $50\%$ of this energy is converted into heat,the heat energy produced is $Q = 0.5 	imes K = \frac{1}{2} 	imes \frac{1}{2}mV^2 = \frac{1}{4}mV^2$.The heat produced is also given by the formula $Q = mS\Delta	heta$,where $m$ is the mass,$S$ is the specific heat,and $\Delta	heta$ is the change in temperature.Using the mechanical equivalent of heat $J$,we relate work and heat as $W = JQ$.Here,the work done (or energy converted) is $\frac{1}{4}mV^2 = J(mS\Delta	heta)$.Canceling $m$ from both sides,we get $\frac{V^2}{4} = JS\Delta	heta$.Therefore,the increase in temperature is $\Delta	heta = \frac{V^2}{4JS}$.

$A$ lead ball moving with a velocity $V$ strikes a wall and stops. If $50\%$ of its energy is converted into heat,then what will be the increase in temperature? (Specific heat of lead is $S$)

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