Question

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

Answer 1

Velocity of bullet $+$ ice block, $V=\frac{\left(10gm\right)\times \left(20m/s\right)}{1000gm}=0.2m/s$
Loss of $K.E.=\frac{1}{2}m{v}^2-\frac{1}{2}(m+M)V^2$
$=\frac{1}{2}[0.01\times (20{)}^2-1\times \left(0.2{)}^2\right]=1.98J$
$\therefore$ Heat generated $=\frac{1.98}{4.2}cal$.
$\therefore$ Heat received by ice block $=\frac{1.98}{4.2\times 2}cal=0.24cal$.
$\therefore$ Mass of ice melted $=\frac{\left(0.24cal\right)}{\left(80cal/gm\right)}=0.003gm$.