A large block of wood of mass M=5.99 kg is hanging from two long massless cords. A bullet of mass m=10g is fired into the block and gets embedded in…
JEE Main 2021 — Physics Mechanics
2021mcqmedium
A large block of wood of mass M=5.99kg is hanging from two long massless cords. A bullet of mass m=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=9.8cm 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−2)
Official previous-year question
Held on 16 Mar 2021 · Verified 6 Jul 2026.
Options
A
841.4ms−1
B
811.4ms−1
C
831.4ms−1
D
821.4ms−1
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Solution
From energy conservation,
[after bullet gets embedded till the system comes momentarily at rest]
(M+m)gh=21(M+m)v12
[v1 is velocity after collision]
∴v1=2σh
Applying momentum conservation, (just before and just after collision)
mv=(M+m)v1
v=(mM+m)v1=10×10−36×2×9.8×9.8×10−2
≈831.55ms−1
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