Q62 / 90 · JEE Main 2021 (16 Mar Shift 2) · Physics

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 center 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 the collision is: (Take g=9.8 m s-2)

Figure

  • A
    841.4 m s-1
  • B
    811.4 m s-1
  • Correct answer
    C
    831.4 m s-1
  • D
    821.4 m s-1

Solution

From energy conservation,

[after bullet gets embedded till the system comes momentarily at rest]

M+mgh=12M+mv12
[v1 is velocity after collision] 

∴ v1=2σh

Applying momentum conservation, (just before and just after collision)

mv=M+mv1

v=M+mmv1=610×10-3×2×9.8×9.8×10-2

≈831.55 m s-1

Hence, the correct answer is option: 'C'.

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