JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(m/s\))

- A \(846.5\)
- B \(821.5\)
- C \(831.5\)
- D \(886.4\)
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,…
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