Question

The time taken by block-bullet system to move from $y=\frac{mg}{k}$ (initial equilibrium position) to $y=0$ (natural length of spring) is ($A$ represents the amplitude of motion).

• A. $\sqrt{\frac{4m}{3k}}[{\cos }^{-1}\left(\frac{mg}{3kA}\right)-{\cos }^{-1}\left(\frac{4mg}{3kA}\right)]$
• B. $\sqrt{\frac{3k}{4m}}[{\cos }^{-1}\left(\frac{mg}{3kA}\right)-{\cos }^{-1}\left(\frac{4mg}{3kA}\right)]$
• C. $\sqrt{\frac{4m}{6k}}[{\sin }^{-1}\left(\frac{4mg}{3kA}\right)-{\sin }^{-1}\left(\frac{mg}{3kA}\right)]$
• D. $Noneoftheabove$

Answer 1

Answer: (C)

$\sqrt{\frac{4m}{6k}}[{\sin }^{-1}\left(\frac{4mg}{3kA}\right)-{\sin }^{-1}\left(\frac{mg}{3kA}\right)]$

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.

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

$\Rightarrow A=\sqrt{\frac{m{v}_0^2}{12k}+{\left(\frac{mg}{3k}\right)}^2}$
To compute the time taken by the combined mass from $y=mg/k$ to $y=0$, we can either go for equation method or circuit motion projection method.

Required time, $t=\frac{\theta }{\omega }=\frac{\alpha -\beta }{\omega }$

$\cos a=\frac{y-y_0}{A}=\left(\frac{\frac{4mg}{3k}-\frac{4mg}{k}}{A}\right)=\frac{mg}{3kA}$

$\cos \beta =\frac{y}{A}=\frac{4mg}{3kA}$

So, $t=\frac{\cos \left(\frac{mg}{3kA}\right)-{\cos }^{-1}\left(\frac{4mg}{3kA}\right)}{\omega }$

$t=\sqrt{\frac{4m}{3k}}[{\cos }^{-1}\left(\frac{mg}{3kA}\right)-{\cos }^{-1}\left(\frac{4mg}{3kA}\right)]$