JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A wedge of mass \(M = 4\,m\) lies on a frictionless plane. A particle of mass \(m\) approaches the wedge with speed \(v\). There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by
- A \(\frac{{2{v^2}}}{{7g}}\)
- B \(\frac{{{v^2}}}{g}\)
- C \(\frac{{2{v^2}}}{{5g}}\)
- D \(\frac{{{v^2}}}{2g}\)
Answer & Solution
Correct Answer
(C) \(\frac{{2{v^2}}}{{5g}}\)
Step-by-step Solution
Detailed explanation
Let mass attains height \('h'\) on wedge and at that time. both attain velocity \(v_f\) Conservation of momentum: \(mv = (m + 4m)\)\(V_f\) \(...(A)\) \(COE:\) \(\frac{1}{2}m{v^2} = \frac{1}{2}\left( {m + 4m} \right)V_f^2 + mgh\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( B \right)\)…
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