Question

A spring-block pendulum is shown in the figure. The system is hanging in equilibrium. A bullet of mass $m/2$ moving with a speed $u$ hits the block from down as shown in the figure. Find the amplitude of oscillation now.

• A. $\sqrt{\frac{m{u}^2}{8k}+{\left(\frac{mg}{2k}\right)}^2}$
• B. $\sqrt{\frac{m{u}^2}{k}+{\left(\frac{mg}{k}\right)}^2}$
• C. $\sqrt{\frac{m{u}^2}{6k}+{\left(\frac{mg}{2k}\right)}^2}$
• D. $\sqrt{\frac{m{u}^2}{4k}-{\left(\frac{mg}{2k}\right)}^2}$

Answer 1

The correct option is C $\sqrt{\frac{m{u}^2}{6k}+{\left(\frac{mg}{2k}\right)}^2}$

Let h be the elongation in spring

$\therefore mg=kh$

If a bullet of mass $m/2$ gets embedded in block

New mass, $m^{\prime }=m+\frac{m}{2}=\frac{3m}{2}$

New mean position of block will be let say at a depth of $h_1$ from old mean position ; then

$\frac{3}{2}mg=k(h+h_1)$

$\Rightarrow h_1=\frac{mg}{2k}$ $(\because mg=kh)$

Just after impact due to inelastic collision if the velocity of block becomes $v_1$,

According to momentum conservation

$\frac{m}{2}u=\frac{3m}{2}v$

$v=\frac{u}{3}$

Now, the block executes SHM and at $t=0$ block is at a distance $h_1=\frac{mg}{2k}$ above its mean position and having a velocity $\frac{u}{3}$

If A is amplitude $\Rightarrow \frac{u}{3}=\omega \sqrt{A^3-{\left(\frac{mg}{2k}\right)}^2}$

or $\frac{u^2}{9}=\frac{2k}{3m}[A^2-{\left(\frac{mg}{2k}\right)}^2[$ $[\because \omega =\sqrt{\frac{k}{\left(3m/2\right)}}]$

$\Rightarrow A=\sqrt{\frac{m{v}^2}{6k}+{\left(\frac{mg}{2k}\right)}^2}$ (Option C)