A man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by \(1…
A man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by \(1 \mathrm{~ms}^{-1}\) and then has the same kinetic energy as the boy. The initial speed of the boy is
\(\sqrt{2}+1 \mathrm{~ms}^{-1}\)
\(2(\sqrt{2}+1) \mathrm{ms}^{-1}\)
\(\sqrt{2} \mathrm{~ms}^{-1}\)
\(2 \mathrm{~ms}^{-1}\)
Solution
According to the question, let the mass of man be \(m\).
So, mass of boy is \(\frac{m}{2}\).
Let speed of man be \(v_1\) and that of boy be \(v_2\).
\(\therefore\) Kinetic energy of man \(=\frac{1}{2}\) kinetic energy of boy
or \(\mathrm{KE}_{\operatorname{man}}=\frac{1}{2} \mathrm{KE}_{\text {boy }}\)
\(\frac{1}{2} m v_1^2=\frac{1}{2} \times \frac{1}{2} \times \frac{m}{2} v_2^2 \text { or } \frac{v_1^2}{v_2^2}=\frac{1}{4}\)
\(\Rightarrow \quad \frac{v_1}{v_2}=\frac{1}{2}\)
\(\therefore \quad v_1=\frac{1}{2} v_2\)...(i)
If the velocity of man increased by \(1 \mathrm{~m} / \mathrm{s}\) then, new velocity of man, \(v_1^{\prime}=\left(\frac{v_2}{2}+1\right)[\therefore\) From Eq. (i)]
Now, kinetic energy of man (KE) = kinetic energy of Boy \(\left(\mathrm{KE}_{\mathrm{B}}\right)\)
\(\begin{aligned}
\therefore \quad \frac{1}{2} \times m \times\left(\frac{v_2}{2}+1\right)^2 & =\frac{1}{2} \times \frac{m}{2} \times v_2^2 \\
\left(\frac{v_2}{2}+1\right)^2 & =\left(\frac{v_2}{\sqrt{2}}\right)^2 \\
\frac{v_2}{2}+1 & =\frac{v_2}{\sqrt{2}} \\
\sqrt{2} v_2+2 \sqrt{2} & =2 v_2 \\
2 \sqrt{2} & =2 v_2-\sqrt{2} v_2 \\
\sqrt{2} v_2(\sqrt{2}-1) & =2 \sqrt{2} \\
v_2= & \frac{2}{(\sqrt{2}-1)} \\
v_2= & \frac{2}{(\sqrt{2}-1)} \times \frac{(\sqrt{2}+1)}{(\sqrt{2}+1)}=\frac{2(\sqrt{2}+1)}{(2-1)} \\
v_2= & 2(\sqrt{2}+1)
\end{aligned}\)
So, the initial speed of the boy, \(v_2=2(\sqrt{2}+1)\).