Two identical pieces of ice fly towards each other with equal and opposite velocities and are converted into water upon impact. If their initial temperature is -12^(@)C, then the minimum possible velocity of either piece before impact is (S_("ice") = 0.50 "cal" gm^(-1) ""^(@)C^(-1),1"cal" = 4.2J)

Two identical pieces of ice fly towards each other with equal and oppo

1.	Two identical pieces of ice fly towards each other with equal and opposite velocities and are converted into water upon impact. If their initial temperature is -12^(@)C, then the minimum possible velocity of either piece before impact is (S_("ice") = 0.50 "cal" gm^(-1) ""^(@)C^(-1),1"cal" = 4.2J)
Answer» 850 m/sec 600m/sec 1000m/sec 500m/sec Solution :Having equal and opposite momenta, the two pieces of ice comes into rest and the loss of their kinetic ENERGIES gets converted into heat to melt it into water. i.e., Loss in K.E. = Latent heat + specific heat `rArr 2 xx (1)/(2) mV^(2) = 2ML + 2m S_("ice") Delta theta rArr V= sqrt(2(L + S_("ice") Delta theta))` `therefore V= sqrt(2(80 xx 1000 + 0.5 xx 1000 xx 12) 4.2)m//s= 850 m//s`

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Two identical pieces of ice fly towards each other with equal and opposite velocities and are converted into water upon impact. If their initial temperature is -12^(@)C, then the minimum possible velocity of either piece before impact is (S_("ice") = 0.50 "cal" gm^(-1) ""^(@)C^(-1),1"cal" = 4.2J)

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