Question

A bullet of mass $10\times {10}^{-3}kg$ moving with a speed of $20m{s}^{-1}$ hits an ice block $\left(0^{o}c\right)$ of $990g$ kept at rest on a frictionless floor and gets embedded in it. If ice takes $50\%$ of $K.E$, for melting, then the mass of the ice block that has melted is($J=4.2J/Cal$) (Latent heat of ice $=80cal/g$)

• A. $6$
• B. $3$
• C. $6\times {10}^{-3}$
• D. $3\times {10}^{-3}$

Answer 1

The correct option is D $3\times {10}^{-3}$

Kinetic energy of the bullet is $=\frac{1}{2}\times m\times v^2=2J$

Now half of it is used by the ice to melt it, i.e. $1J=1/4.2$ cal

Latent heat of the ice is $L=80cal/gram$.

Let M is the amount of ice melts:

$M\times L=1/4.2$

$M=0.003$gm

And ice will only melt. The temperature will be same.