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…

Question

A large block of wood of mass M = 5 . 99 &#160; kg is hanging from two long massless cords. A bullet of mass m = 10 &#160; 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 &#160; 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 &#160; m &#160; s - 2 )

Aicharya · Accepted Answer

From energy conservation,

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

$(M + m) g h = \frac{1}{2} (M + m) v_{1}^{2}$
[ $v_{1}$ is velocity after collision]

$∴ v_{1} = \sqrt{2 σ h}$

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

$m v = (M + m) v_{1}$

$v = (\frac{M + m}{m}) v_{1} = \frac{6}{10 \times 10^{- 3}} \times \sqrt{2 \times 9.8 \times 9.8 \times 10^{- 2}}$

$\approx 831.55 m s^{- 1}$

A large block of wood of mass M = 5 . 99   kg is hanging from two long massless cords. A bullet of mass…

Solution

A large block of wood of mass M = 5 . 99 &#160; kg is hanging from two long massless cords. A bullet of mass…

Solution

A large block of wood of mass M = 5 . 99 kg is hanging from two long massless cords. A bullet of mass…