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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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$\frac{v^{2}}{J}$

$\frac{v^{2} S}{2 J}$

$\frac{2 v^{2}}{J S}$

$\frac{v^{2}}{4 J S}$

answer is B.

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Detailed Solution

$KE = \frac{{MV}^{2}}{2} N o w, h a l f o f t h e K E i s c o n v e r t e d i n t o h e a t, t h e r e f o r e w e h a v e J \times (ΔT) \times M \times S = \frac{{MV}^{2}}{4} \Rightarrow ΔT = \frac{V^{2}}{4 J S}$

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