A lead bullet of mass 10 g travelling at 300 | vedclass

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Question

(C) Mass of the bullet,$m = 10 \ g = 10 	imes 10^{-3} \ kg$.Velocity of the bullet,$v = 300 \ m/s$.Kinetic energy of the bullet,$KE = \frac{1}{2}mv^2

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(C) Mass of the bullet,$m = 10 \ g = 10 	imes 10^{-3} \ kg$.Velocity of the bullet,$v = 300 \ m/s$.Kinetic energy of the bullet,$KE = \frac{1}{2}mv^2 = \frac{1}{2} 	imes (10 	imes 10^{-3}) 	imes (300)^2 = 0.005 	imes 90000 = 450 \ J$.Heat produced,$Q = 50\% 	ext{ of } KE = 0.5 	imes 450 = 225 \ J$.The heat absorbed by the bullet raises its temperature,given by $Q = mc\Delta T$.Here,$c = 150 \ J/kg \cdot ^{\circ}C$.$225 = (10 	imes 10^{-3}) 	imes 150 	imes \Delta T$.$225 = 1.5 	imes \Delta T$.$\Delta T = \frac{225}{1.5} = 150^{\circ}C$.

$A$ lead bullet of mass $10 \ g$ travelling at $300 \ m/s$ strikes against a block of wood and comes to rest. Assuming $50\%$ of heat is absorbed by the bullet,the increase in its temperature is (Specific heat of lead $= 150 \ J/kg \cdot ^{\circ}C$) (in $^{\circ}C$)

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