A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030 cal/(g – ᵒC) (1 cal = 4.2 × 10⁷ ergs) close to :

Solution

Physics sir · 4 steps

Give the bullet only half its kinetic energy as heat, then turn joules into calories before using c.

1. Kinetic energy. With m = 5 g = 5 × 10⁻³ kg, the bullet's energy is 1/2 × 5 × 10⁻³ × 210² = 110.25 J.
2. Half to bullet. The bullet keeps only half of this as heat, 55.125 J. Since 1 cal = 4.2 × 10⁷ erg = 4.2 J, this is 55.125/4.2 = 13.125 cal.

   Δ T = Q/mc = (13.125 cal)/(5 g × 0.030 cal g⁻¹ °C⁻¹)

3. Divide. The denominator is 0.15 cal °C⁻¹, so Δ T = 13.125/0.15 = 87.5 °C. Mass cancels in the end: Δ T = v²/4c with c = 126 J kg⁻¹ K⁻¹ gives 44100/504 = 87.5 too.
4. So. The bullet warms up by about 87.5 °C, which is option (a).

The idea to keep. Mechanical energy becomes heat through Q = mcΔ T, but first take only the share that goes into the body, and keep units matched: joules with J/kg K, or calories with cal/g °C.

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