Question

A block moving horizontally on a smooth surface with a speed of $40\text{m/s}$ splits into two parts with masses in the ratio of $1:2$ If the smaller part moves at $60\text{m/s}$ in the same direction, then the fractional change in kinetic energy is

• A. $\frac{1}{8}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{1}{8}$

Let initial mass of of block be $3m$ which breaks into two parts of masses $m,2m$
Given that intial mass of 3m is $4\text{m/s}$,mass of $m$ after explosion is $4\text{m/s}$
let mass of $2m$ after collision is $v$
From conservation of momentum
$3m\left(40\right)=m\left(60\right)+2m\left(v\right)$
$v=30\text{m/s}$
Initial Kinetic energy is $K_i=\frac{1}{2}3m(40{)}^2$
Final kinetic energy is $K_f=\frac{1}{2}m(60{)}^2+\frac{1}{2}2m(30{)}^2$
Fractional change in kinetic energy $\frac{K_f-K_i}{K_i}=\frac{{60}^2+2(30{)}^2-3(40{)}^2}{3(40{)}^2}=\frac{3600+1800-4800}{4800}=\frac{600}{4800}=\frac{1}{8}$