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  4. A bullet of mass 10× 10-3kg moving with a speed of 20 text ms-1 hits an ice block (0?C) of 990 g kept at rest on a frictionless floor and gets embedded in it. If ice takes 50% of KE lost by the system, the amount of ice melted (in grams) approximately is: (J = 4.2 J/cal) (Latent heat of ice = 80 cal/g).

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Q. A bullet of mass $ 10\times {{10}^{-3}}kg $ moving with a speed of $ 20\text{ }m{{s}^{-}}1 $ hits an ice block (0?C) of 990 g kept at rest on a frictionless floor and gets embedded in it. If ice takes 50% of KE lost by the system, the amount of ice melted (in grams) approximately is: (J = 4.2 J/cal) (Latent heat of ice = 80 cal/g).

EAMCETEAMCET 2006Thermal Properties of Matter

Solution:

Kinetic energy of bullet $ =\frac{1}{2}m{{v}^{2}} $ $ =\frac{1}{2}\times 10\times {{10}^{-3}}\times 20\times 20 $ $ =2J=\frac{2}{4.2}\,cal $ Heat gained by ice to melt $ =\frac{2}{4.2}\times \frac{50}{100} $ or $ mL=\frac{1}{4.2} $ or $ m=\frac{1}{4.2\times 80}g $ $ =0.002976\,g $ $ =0.002976\,g $