CUET Preparation Today
The blocks A and B shown in figure have masses MA = 5 kg and MB = 4 kg. The system is released from rest. The speed of B after A has travelled a distance 1 m along the incline is :
|
\(\frac{\sqrt{3}}{2}\sqrt{g}\) \(\frac{\sqrt{3}}{4}\sqrt{g}\) \(\frac{1}{2}\sqrt{g}\) \(\frac{1}{2\sqrt{3}}\sqrt{g}\) |
\(\frac{1}{2\sqrt{3}}\sqrt{g}\) |
If a moves down the incline by 1 m, B shall move up by \(\frac{1}{2}\) m. If the speed of B is v, then the speed of A will be 2 v. By Conservation of Energy : Gain in KE = Loss in PE \(\frac{1}{2}m_A (2v)^2 + \frac{1}{2}m_B v^2 = m_Ag \frac{3}{5} - m_Bg \frac{1}{2}\) Solving it, we get : \(v = \frac{1}{2} \sqrt{\frac{g}{3}}\) |
