A lead bullet penetrates into a solid object and melts Assuming that `50%` of its K.E. was used to heat it , calculate the initial speed of the bullet , The initial temp , of bullet is `27^(@)c` and its melting point is `327^(@)C` Latent heat of fasion of lead ` = 2.5 xx 10^(4) J kg^(-1)` and sp heat capacity of lead `= 125 J kg^(-1) K^(-1)`

Explanation

Let the mass of the bullet = m.
Heat required to take the bullet from `27^(@)`C to `327^(@)`C
Q = ms(`T_(2)-T_(1))` = `mxx125 xx(327-27))`
`=mxx(125)(300)=m xx(3.75 xx 10^(4)Jkgi^(-1)`
Heat required to melt the bullet = `mL=mxx(2.5 xx10^(4)Jkg^(-1)`
If the initial speed be v, the kinetic energy is `1/2mv^(2)` and hence, the heat developed is `1/4(1/2mv^(2))` = `1/8mv^(2)`.
Thus, `1/8mv^(2)`=Q=`(m(3.75+2.5)xx10^(4)Jkg^(-1)`
or `v=707.11ms^(-1)`