Question

A bullet of mass $10\text{g}$ moving with a speed of $20\text{m/s}$ hits an ice block of mass $990\text{g}$ kept on a floor and gets stuck in it. How much ice will melt if $50\%$ of the lost KE goes to ice? (initial temperature of the ice block and bullet = $0^{\circ }\text{C}$)

• A. $0.001\text{g}$
• B. $0.002\text{g}$
• C. $0.003\text{g}$
• D. $0.004\text{g}$

Answer 1

The correct option is C $0.003\text{g}$

$\begin{array}{cc}KE& =\frac{1}{2}{m}_B{v}^2\\ & =\frac{1}{2}\times \frac{10}{1000}\times (20{)}^2\\ & =\frac{1}{2}\times \frac{1}{2}\times 400=2J\end{array}$

So 50% of this is lost which is $1J$ and this is adsorbed by ice to melt into water.


$80cal\to 1g$
$80\times 4\cdot 2J\to 1g$
$1J\to \frac{1}{80\times 4.2}=0.003g$

So 0.003 g of ice melts into water.