Question

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

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

Answer 1

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


As there is no external force,
$\therefore$ Applying P.C.L.M
$(m_b+m_{icc})v=m_b{v}_b+m_{ice}{v}_{ice}$
$\Rightarrow (0.01+0.99)v=0.01\times 20+m_{ice}\times 0$
$\therefore v=\frac{0.01\times 20}{0.01+0.99}=0.2\text{m/s}$
Loss in kinetic energy
$=|K.E_{final}-K.E_{initial}|$
$=|\frac{1}{2}\times (m_b+m_{ice})v^2-\frac{1}{2}{m}_b{v}_b^2|$
$=|\frac{1}{2}\times 1\times {0.2}^2-\frac{1}{2}\times 0.01\times {20}^2|$
$=1.98\text{J}$
Let $m^{\prime }\text{g}$ of ice melted.
According to question,
$0.5\times$ loss of kinetic energy = Heat gained by ice for melting
$0.5\times 1.98=m^{\prime }\times L_f$
where
$L_f$ = Latent heat of fusion of ice $=80\text{cal/g}$
$=80\times 4.2\text{J/g}$

$\therefore 0.5\times 1.98=m^{\prime }\times 80\times 4.2$
$\therefore m^{\prime }\approx 3\times {10}^{-3}\text{kg}$