Question

A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by \(1\ \text{m/s}\) so as to have the same kinetic energy as that of the boy. The original speed of the man is

• A. \((\sqrt{2} - 1)\ \text{m/s}\)
• B. \(\sqrt{2}\ \text{m/s}\)
• C. \(\frac{1}{\sqrt{2} - 1}\ \text{m/s}\)
• D. \(\frac{1}{\sqrt{2}}\ \text{m/s}\)

Hint:

Rationalize: \(\frac{1}{\sqrt{2}-1} = \sqrt{2}+1 \approx 2.414\ \text{m/s}\).

Answer 1

The Correct Option is C

Step 1: Concept:
KE = \(\frac{1}{2}mv^2\). Let mass of man = \(M\), boy = \(m = M/2\).
Step 2: Explanation:
Initial: \(\frac{1}{2}Mv^2 = \frac{1}{2}\left(\frac{1}{2}m v_b^2\right) = \frac{1}{4}m v_b^2\). But \(m = M/2 \Rightarrow \frac{1}{2}Mv^2 = \frac{1}{4} \times \frac{M}{2} \times v_b^2 = \frac{M}{8}v_b^2\).
\(\Rightarrow \frac{1}{2}v^2 = \frac{1}{8}v_b^2 \Rightarrow v_b^2 = 4v^2 \Rightarrow v_b = 2v\).
After speed up: \(\frac{1}{2}M(v+1)^2 = \frac{1}{2}m v_b^2 = \frac{1}{2} \times \frac{M}{2} \times 4v^2 = Mv^2\).
\(\Rightarrow \frac{1}{2}(v+1)^2 = v^2 \Rightarrow (v+1)^2 = 2v^2 \Rightarrow v+1 = \sqrt{2}v \Rightarrow 1 = v(\sqrt{2} - 1) \Rightarrow v = \frac{1}{\sqrt{2} - 1}\).
Step 3: Conclusion:
Thus, original speed = \(\frac{1}{\sqrt{2} - 1}\ \text{m/s}\).