Question

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 `= 1.25 J kg^(-1) K^(-1)`

Answer 1

Here, let m be the masss of the lead bulled
Heat required to raise its temp from `27^(@)C` to `372^(2)C`
`Delta Q_(1) = I m Delta T = 1.25 xxx m xx (327 - 27) = 3.75 xx 10^(6)m` joule
Heat required to melt the bullet `Delta Q_(2) = ml = m xx 2.5 xx 10^(4)J`
If upsilon is initial velocity of the bullet , then K.E. of bullet `= (1)/(2) m upsilon^(2)`
As Heat developed `= (1)/(2) K.E. = (1)/(2) xx (1)/(2) m upsilon^(2)`
`:. 3.75 xx 10^(4) m + 2.5 xx 10^(4) m = (1)/(4) m upsilon^(2)`
`6.25 xx 10^(4)m = (1)/(4) m upsilon^(2)`
` upsilon= sqrt(4 xx 6.25 xx 10^(4)) = 5 xx 10^(2) m//s`