Arun
Last Activity: 5 Years ago
Explanation: K.E. of bullet = (1/2) mV2. After striking KE is converted to heat. Hence Q = (1/2) mV2. Given is that 25% heat is absorbed by obstacle, hence Q' = 0.75 × (1/2) mV2 (1) This Q' will melt the bullet. ΔT = 327 – 27 = 300°C. If C is specific heat & L is latent heat then Q' = (mCΔT) + (m × L) ∴ form (1) [{0.75} / 2] mV2 = mCΔT + mL ∴ 0.375 V2 = C ∙ ΔT + L V2 = [{C ∙ ΔT + L} / {0.375}] = [{(0.03 × 300) + 6} / {0.375}] = 40kcal/kg = 40 × 4.2 × 103 J/kg ∴ V2 = 168000 ∴ V = 409.90 m/s