Question

A running man has half the kinetic energy of a boy of half his mass. The man speeds up by \(1 \, {m/s}^{-1}\) to have the same kinetic energy as the boy. The initial speed of the man is:

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

Hint:

When dealing with kinetic energy, remember that it is proportional to the square of the velocity. Small changes in speed can result in significant changes in kinetic energy.

Answer 1

The Correct Option is C

Step 1: Establish the kinetic energy relationship. Let \( m \) be the mass of the man and \( v \) his initial speed. 
Then, his kinetic energy is \(\frac{1}{2}mv^2\). 
The boy's mass is \( \frac{m}{2} \) and his kinetic energy is \( K \). Given that the man's kinetic energy is half of the boy's, we have: \[ \frac{1}{2}mv^2 = \frac{1}{2}K \] Step 2: Calculate the man's new speed. When the man speeds up by \(1 \, {m/s}^{-1}\), his new speed is \( v + 1 \) and his new kinetic energy equals the boy's: \[ \frac{1}{2}m(v+1)^2 = K \] Step 3: Solve for \( v \). Equating the kinetic energies before and after the speed increase, and solving for \( v \), we find: \[ v = \frac{1}{(\sqrt{2} - 1)} \, {m/s}^{-1} \]