Question

A block moving horizontally on a smooth surface with a speed of $20{\text{ms}}^{-1}$ bursts into two equal pieces continuing in the same direction. If one of the points moves at $30{\text{ms}}^{-1}.$ Find the fractional change in kinetic energy.

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{16}$
• D. $\frac{1}{8}$

Answer 1

The correct option is A $\frac{1}{4}$



In collision, there is no external force involved. Thus, linear momentum will be conserved.

$P_i=P_f$

$m\times 20=\frac{m}{2}\times 30+\frac{m}{2}\times v$

$\Rightarrow 5m=\frac{m}{2}\times v$

$\therefore v=10{\text{ms}}^{-1}$

Change in kinetic energy is,

$\Delta KE=K{E}_f-K{E}_i$

$\Delta KE=(\frac{1}{2}\times \frac{m}{2}\times (30{)}^2+\frac{1}{2}\times \frac{m}{2}\times (10{)}^2)-(\frac{1}{2}\times m\times (20{)}^2)$

$\Delta KE=\frac{m}{2}(450+50-400)=50m$

Thus, the fractional change in kinetic energy is,

$\frac{\Delta KE}{K{E}_i}=\frac{50m}{\frac{1}{2}\times m\times (20{)}^2}=\frac{1}{4}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(\text{A}\right)$ is the correct answer.

Why this question ? Concept: Fraction change in KE is $\frac{\Delta KE}{K{E}_i}$