Question

a bullet of mass 10g moving with speed of 20 m/s hits an ice block of mass 990g kept on a frictionless floor and gets struck in it. how much ice will melt if 50% of lost KE goes to ice ? ( initial tempertauteof ice block and bullet=0 celcius)

Answer 1

"Answer: The bullet has melted 0.003 g of ice.

Given,

Mass of the bullet $= 10 g = 10\quad \times \quad { 10 }^{ -3 }$

Speed of the bullet = 20 m/s

Mass of the ice = 990 g

Energy required to melt ice = 50% of kinetic energy

Total kinetic energy required $=\quad \frac { 1 }{ 2 } m{ v }^{ 2 }$

$=\quad \frac { 1 }{ 2 } \quad \times \quad 10\quad \times \quad { 10 }^{ -3 }\quad \times \quad { 20 }^{ 2 }$

Total kinetic energy = 2 J

50% of total kinetic energy $=\quad \frac { 20\quad \times \quad 50 }{ 100 }$

50% of total kinetic energy = 1 J

In this case, the latent heat is amount of heat required to solid (ice) into water.

Latent heat of ice = 334 J/g

Amount of energy required to convert x g of ice,

$Kinetic\quad energy\quad =\quad mass\quad \times \quad latent\quad heat$

$1\quad J\quad =\quad x\quad g\quad \times \quad 334\quad J/g$

$x\quad =\quad \frac { 1 }{ 334 }$

$x\quad =\quad 2.9\quad \times \quad { 10 }^{ -3 }\quad =\quad 0.003\quad g$

The bullet has melted 0.003 g of ice."

Answer 2

Answer:

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