JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2019
- Hard
- Numerical
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 ........ \(m\). \((g = 10\, ms^{-2})\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(40\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
{Y_{cm}}from\,ground\, = \frac{{0.03 \times 100}}{{0.05}} = 60\,m\\
{V_{cm}} = \frac{{0.02 \times 100}}{{0.05}} = 40\,m/s\\
H = \frac{{V_{cm}^2}}{{2g}} = \frac{{40 \times 40}}{{20}} = 80\,m
\end{array}\)
{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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