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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

A piece of wood of mass \(0.03\, kg\) is dropped from the top of a \(100\, m\) height building. At the same time, a bullet of mass \(0.02\, kg\) is fired vertically upward, with a velocity \(100\, ms^{- 1}\), from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is ........ \(m\). \((g = 10\, ms^{-2})\)

  1. A \(20\)
  2. B \(30\)
  3. C \(40\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(40\)

Step-by-step Solution

Detailed explanation

\(\begin{array}{l} {Y_{cm}}from\,ground\, = \frac{{0.03 \times 100}}{{0.05}} = 60\,m\\ {V_{cm}} = \frac{{0.02 \times 100}}{{0.05}} = 40\,m/s\\ H = \frac{{V_{cm}^2}}{{2g}} = \frac{{40 \times 40}}{{20}} = 80\,m \end{array}\) Height above building \(= 80 - 40 = 40\, m\)
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