Question

A block of mass $m$ is suspended from one end of a light spring as shown. The origin $O$ is considered at distance equal to natural length of spring from ceiling and vertical downward direction as positive $Y-axis.$ When the system is in equilibrium a bullet of mass $m/3$ moving in vertical upward direction with velocity $v_0$ strikes the block and embeds into it. As a result, the block (with bullet embedded into it) moves up and starts oscillating.
The amplitude of oscillation would be:

• A. $\sqrt{{\left(\frac{4mg}{3k}\right)}^2+\frac{m{v}_0^2}{12k}}$
• B. $\sqrt{\frac{m{v}_0^2}{12k}+{\left(\frac{mg}{3k}\right)}^2}$
• C. $\sqrt{\frac{m{v}_0^2}{6k}+{\left(\frac{mg}{k}\right)}^2}$
• D. $\sqrt{\frac{m{v}_0^2}{6k}+{\left(\frac{4mg}{3k}\right)}^2}$

Answer 1

Answer: (D)

$\sqrt{\frac{m{v}_0^2}{6k}+{\left(\frac{4mg}{3k}\right)}^2}$

Initially in equilibrium let the elongation is spring be $y_0$ then $mg=k{y}_0$
$y_0=\frac{mg}{k}$
As the bullet strikes the block with velocity $v_0$ and gets embedded into it, the velocity of the combined mass can be computed by using the principle of moment conservation.
$\frac{m}{3}{v}_0=\frac{4m}{3}v\Rightarrow v=\frac{v_0}{4}$
Let new mean position is at distance $y$ from the origin, then
$ky=\frac{4m}{3}g\Rightarrow y=\frac{4mg}{3k}$
Now , the block executes $SHM$ about mean position defined by $y=4mg/3k$ with time period $T=2\pi \sqrt{4m/3k}$. At $t=0$, the combined mass is at a displacement of $(y-y_0)$ from mean position and is moving with velocity $v$, then by using $v=\omega \sqrt{A^2-x^2}$, we can find the amplitude of motion.

${\left(\frac{v_0}{4}\right)}^2=\frac{3k}{4m}[A^2-(y-y_0{)}^2]=\frac{3k}{4m}[A^2-{\left(\frac{mg}{3k}\right)}^2]$

$\Rightarrow A=\sqrt{\frac{m{v}_0^2}{12k}+{\left(\frac{mg}{3k}\right)}^2}$