A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is :
(Given : initial temperature of the bullet = 127°C, Melting point of the bullet = 327°C, Latent heat of fusion of lead = 2.5 × 10⁴ J kg⁻¹, Specific heat capacity of lead = 125 J/kg K)

Solution

Physics sir · 4 steps

Only 40% of the kinetic energy becomes heat; equate it to warming plus melting, and mass cancels.

1. Heat needed. The bullet first warms from 127°C to its melting point 327°C, a rise of 200 K. Then it melts. Per kilogram this needs c Δ T + L = 125 × 200 + 2.5 × 10⁴ = 2.5 × 10⁴ + 2.5 × 10⁴ = 5 × 10⁴ J kg⁻¹.
2. Energy balance. Only 40% of the kinetic energy becomes heat in the bullet. Mass m appears on both sides, so it cancels.

   0.4 × 1/2 m v² = m(c Δ T + L)

3. Solve for v. 0.2 v² = 5 × 10⁴, so v² = 2.5 × 10⁵ m² s⁻² and v = 500 m s⁻¹.
4. So. The initial speed of the bullet is 500 m s⁻¹, option (b).

The idea to keep. To melt a solid that starts below its melting point, add both heats: c Δ T to reach the melting point, then L to melt it. Apply the efficiency only to the energy supplied.

Answer from the NTA final answer key. Solution by PrepWiser.