Question

When a lead bullet hits a large solid target at high speed, its entire kinetic energy is converted into heat. For simplicity, assume that all that heat is concentrated in the bullet rather than the target. If the bullet’s initial temperature is 29°C, how fast would it have to move before the collision in order to have 29 % of its mass melted after the collision? Lead has a melting temperature 327.3 °C, a specific heat of 128 J/kg · K and latent heat of fusion 24500 J/kg. Answer in units of m/s.

Answer 1

Answer:

$v = 300.96 m/s$

Explanation:

As we know that 29% of the mass will melt due to heat developed in the bullet

so here we will have

$Q = Mc\Delta T + 0.29M L$

$Q = M(128\times (327.3 - 29) + 0.29(24500))$

$Q = M(38182.4 + 7105)$

$Q = 45287.4 M$

now we know this is all due to kinetic energy of the bullet

so we will have

$\frac{1}{2}Mv^2 = Q$

$\frac{1}{2}Mv^2 = 45287.4 M$

$v = 300.96 m/s$