Question

A bullet of mass $10g$ moving with a speed of $20m/s$ hits an ice block of mass $990g$ kept on a frictionless 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 $=0^{\circ }C$).

• A. $0.001g$
• B. $0.002g$
• C. $0.003g$
• D. $0.004g$

Answer 1

Answer: (C)

$0.003g$

Here, velocity of bullet + ice block,

$V=\frac{10g\ast 20m/s}{1000}g=0.2m/s$

Loss of KE$=\frac{1}{2}m{v}^2-\frac{1}{2}(m+M)V^2$

=$\frac{1}{2}[0.01\ast (20{)}^2-1\ast (0.2{)}^2$]

=$1.98J$

Therefore, Heat received by the ice block=$\frac{1.98}{4.2\ast 2}cal$=$0.24cal$

Mass of ice melted=$\frac{0.24cal}{80cal/g}$=$0.003g$