An ice bullet of temperature – 10°C is fired upon an object. The bullet spends 20% of its kinetic energy in penetrating into the object and remaining kinetic energy is spent into melting the bullet at 0°C. The latent heat of fusion of ice is 3·4 × 10⁵ J kg⁻¹ and its specific heat capacity is 2000 J kg⁻¹ K⁻¹. Which one among the following is the correct initial speed of the bullet ?
Initial kinetic energy (KE) = (1/2)mv².
20% of KE is spent on penetration = 0.2 * (1/2)mv² = 0.1mv².
Remaining KE is spent on heating the ice from -10°C to 0°C and melting it at 0°C.
Remaining KE = KE – 0.1mv² = 0.9mv².
Energy required to heat mass m of ice from -10°C to 0°C:
Q_heat = m * c_ice * ΔT = m * (2000 J kg⁻¹ K⁻¹) * (0 – (-10)) K = m * 2000 * 10 = 20000m J.
Energy required to melt mass m of ice at 0°C:
Q_melt = m * L = m * (3.4 × 10⁵ J kg⁻¹) = 340000m J.
Total energy required for heating and melting = Q_heat + Q_melt = 20000m + 340000m = 360000m J.
This energy comes from the remaining KE:
0.9mv² = 360000m
0.9v² = 360000
v² = 360000 / 0.9 = 3600000 / 9 = 400000.
v = √400000 = √((4 × 10⁵) * 10/10) = √(40 × 10⁴) = 100 * √40 = 100 * √(4 * 10) = 100 * 2√10 = 200√10 m/s.
Wait, let’s re-read the problem carefully: “remaining kinetic energy is spent into melting the bullet at 0°C”. This phrasing usually means only the phase change energy is considered from the remaining KE. If this interpretation is correct, the 80% KE is used *only* for melting at 0°C, implying the energy for heating from -10°C to 0°C is either negligible or sourced otherwise (e.g., from the 20% penetration energy, which is unlikely). Let’s retry with this interpretation:
Remaining KE (80% of total KE) = 0.8 * (1/2)mv² = 0.4mv².
This is spent on melting the bullet at 0°C:
0.4mv² = Q_melt = m * L = m * 3.4 × 10⁵
0.4v² = 3.4 × 10⁵
v² = (3.4 × 10⁵) / 0.4 = (3.4 × 10⁵) / (4/10) = (3.4 × 10⁵) * (10/4) = (34/4) × 10⁵ = 8.5 × 10⁵
v = √(8.5 × 10⁵) = √(85 × 10⁴) = 100 * √85.
√85 is approximately 9.22. v ≈ 922 m/s. This doesn’t match any option well.
Let’s go back to the first interpretation, where 80% KE covers both heating and melting.
0.8 * (1/2)mv² = m * c_ice * ΔT + m * L
0.4v² = m(c_ice * ΔT + L)
0.4v² = 2000 * 10 + 3.4 × 10⁵ = 20000 + 340000 = 360000
v² = 360000 / 0.4 = 900000
v = √900000 = √(90 × 10⁴) = 100 * √90 = 100 * √(9 × 10) = 100 * 3√10 = 300√10 m/s.
300√10 ≈ 300 * 3.162 = 948.6 m/s. This matches option A exactly.
The phrase “melting the bullet at 0°C” in the problem statement, combined with providing the specific heat capacity of ice and the initial temperature of -10°C, strongly suggests that the energy required to raise the temperature to 0°C must also be accounted for by the available KE. Therefore, the 80% KE is used for the entire process from -10°C to liquid water at 0°C (heating ice + melting ice).