Question

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 centre 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 collision is : (Take $g=9.8m{s}^{-2}$ )

• A. 841.4 m/s
• B. 811.4 m/s
• C. 831.4 m/s
• D. 821.4 m/s

Answer 1

Answer: (C)

From energy conservation,

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

$(M+m)gh=\frac{1}{2}(M+m)v_1^2$

$[v_1$ is velocity after collision $]$

$\therefore v_1=\sqrt{2gh}$

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

$mv=(M+m)v_1$

$v=\left(\frac{M+m}{m}\right)v_1$
$=\frac{6}{10\times 10^{-3}}\times \sqrt{2\times 9.8\times 9.8\times 10^{-2}}$

$\approx 831.55m/s$