JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2021
- Hard
- MCQ
A large block of wood of mass \(M =5.99\, kg\) is hanging from two long massless cords. A bullet of mass \(m =10\, g\) is fired into the block and gets embedded in it. The (block \(+\) bullet) then swing upwards, their centre of mass rising a vertical distance \(h =9.8\,cm\) before the (block \(+\) bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is: (Take \(g =9.8\, ms ^{-2}\) ) (in \(m/s\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(831.5\)
Step-by-step Solution
Detailed explanation
From energy conservation,
\([\) after bullet gets embedded till the system comes momentarily at rest \(]\)
\(( M + m ) g h =\frac{1}{2}( M + m ) v _{1}^{2}\)
\(\left[ v _{1}\right.\) is velocity after collision]
\(\therefore v _{1}=\sqrt{2 gh }\)
Applying momentum conservation, (just before and just after collision)
\(mv =( M + m ) v _{1}\)
\(v =\left(\frac{ M + m }{ m }\right) v _{1}=\frac{6}{10 \times 10^{-3}} \times \sqrt{2 \times 9.8 \times 9.8 \times 10^{-2}}\)
\(\approx 831.55\, m / s\)
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