Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

17 of 130

Question 17 of 130

A large block of wood of mass M=5.99kgM=5.99kg is hanging from two long massless cords. A bullet of mass m=10gm=10g is fired into the block and gets embedded in it. The (block ++ bullet) then swing upwards, their center of mass rising a vertical distance h=9.8cmh=9.8cm before the (block ++ bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before the collision is: (Take g=9.8ms−2g=9.8{ms}^{-2})

Show answer and solution

Answer: (C) 831.4ms−1831.4m{s}^{-1}

From energy conservation,
[after bullet gets embedded till the system comes momentarily at rest]
(M+m)gh=12(M+m)v12\left(M+m\right)gh=\dfrac{1}{2}\left(M+m\right){v}_{1}^{2}
[v1{v}_{1} is velocity after collision]
∴v1=2σh∴{v}_{1}=\sqrt{2\sigma h}
Applying momentum conservation, (just before and just after collision)
mv=(M+m)v1mv=\left(M+m\right){v}_{1}
v=(M+mm)v1=610×10−3×2×9.8×9.8×10−2v=\left(\dfrac{M+m}{m}\right){v}_{1}=\dfrac{6}{10\times {10}^{-3}}\times \sqrt{2\times 9.8\times 9.8\times {10}^{-2}}
≈831.55ms−1\approx 831.55m{s}^{-1}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
Revise Rotational Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Center of Mass, Momentum & Collision (momentum), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.