Particle A moving with a velocity $v=10 \mathrm{~m} / \mathrm{s}$ experienced a head on collision with a stationary particle B of the same mass.Asaresultof collision, thekinetic energyofthesystem decreased by $1 \%$. The speed of particle A after collision in

• Correct answer
  A

  10 m/s
• B

  0.05 m/s
• C

  5 m/s
• D

  $10 \sqrt{2} \mathrm{~m} / \mathrm{s}$

Solution

By law of conservation of momentum $\mathrm{m} \times 10+0=\mathrm{mV}_1+\mathrm{mV}_2$
Now, $$ \begin{aligned} & \frac{\Delta \mathrm{K}}{\mathrm{K}}=0.01 \Rightarrow \frac{\frac{1}{2} \mathrm{~m} \times 10^2-\frac{1}{2} \mathrm{mV}_1^2-\frac{1}{2} \mathrm{mV}_2^2}{\frac{1}{2} \mathrm{~m} \times 10^2}=0.01 \\ & \Rightarrow \quad \frac{100-\mathrm{V}_1^2-\mathrm{V}_2^2}{100}=0.01 \\ & \Rightarrow \quad \mathrm{V}_1^2+\mathrm{V}_2^2=99 \end{aligned} $$ $\begin{aligned} & \text { Now, }\left(\mathrm{V}_1+\mathrm{V}_2\right)^2=\mathrm{V}_1^2+\mathrm{V}_2^2+2 \mathrm{~V}_1 \mathrm{~V}_2 \\ & \Rightarrow \quad 100=99+2 \mathrm{~V}_1 \mathrm{~V}_2 \Rightarrow \quad \mathrm{V}_1 \mathrm{~V}_2=0.5 \\ & \text { So, }\left(\mathrm{V}_1-\mathrm{V}_2\right)^2=\left(\mathrm{V}_1+\mathrm{V}_2\right)^2-4 \mathrm{~V}_1 \mathrm{~V}_2 \\ & =100-4 \times 0.5=98\end{aligned}$
Adding (i) \& (ii), we get $$ 2 \mathrm{~V}_1=10+7 \sqrt{2} \Rightarrow \mathrm{V}_1=9.94 \mathrm{~m} / \mathrm{s}=10 \mathrm{~m} / \mathrm{s} $$

Hence, the correct answer is option: 'A'.