Solution:
Let \(m_A = 2m_R\). Given \(K_A = \frac{1}{3}K_R ⇒ \frac{1}{2}m_A v'^2 = \frac{1}{3}\left(\frac{1}{2}m_R v^2\right)\). Substituting \(m_A = 2m_R\), we get \(2v'^2 = \frac{1}{3}v^2 ⇒ v' = \frac{v}{\sqrt{6}}\).
Arun has \(\left(\frac{1}{3}\right)^{\text{rd}}\) of the kinetic energy as that of Raunak when they both run. If Raunak has half the mass of Arun, then the relationship between the speed of Arun \(v'\) and speed of Raunak \(v\) is
Leave a Reply