Question

Two indentical blocks each of mass $M=9\text{kg}$ are placed on a rough horizontal surface of coefficient of static friction $\mu =0.1$, assume static and kinetic coefficient of friction to be equal. The two blocks are joined by a light spring and block $B$ is in contact with a vertical fixed wall as shown in figure. A bullet of mass $m=1\text{kg}$ and velocity $V_o=10\text{m/s}$ hit the block $A$ and gets embedded in it.
Find the maximum compression of spring (Take spring constant $k=240\text{N/m}$ and $g=10{\text{m/s}}^2$.

• A. $\frac{1}{12}\text{m}$
• B. $\frac{1}{3}\text{m}$
• C. $\frac{1}{6}\text{m}$
• D. $\frac{1}{4}\text{m}$

Answer 1

The correct option is C $\frac{1}{6}\text{m}$

Mass of bullet, $m=1\text{kg}$
Mass of block $A$ and $B$ be $M=9\text{kg}$
Velocity of bullet $V_o=10\text{m/s}$
$\Rightarrow$Let velocity of system block ($A+$ Bullet) after collision is $v\text{m/s}$ towards $+vex-$ direction. Block $B$ will remain at rest since it is supported by wall.
$\Rightarrow$Applying conservation of momentum in $x-$direction on system:
$P_i=P_f$
$m\times V_o+0=(M+m)v$
$1\times 10=(9+1)v$
$\therefore v=1\text{m/s}$
Let $x$ be the maximum compression in the spring after collision of bullet with mass $A$.

Applying Work energy theorem after collision, till the system comes to rest i.e $v_f=0$
$W_{spring}+W_f=\Delta KE...\left(i\right)$
$W_{spring}=-\frac{1}{2}k{x}^2$
$W_f=-fx=-\mu Nx=-\mu (M+m)gx$
since block will slide kinetic friction will act and $f=\mu N=\mu (M+m)g$
$\Delta KE=K{E}_f-K{E}_i=0-\frac{1}{2}(M+m)v^2$
Substituting the works and $\Delta KE$ in Eq $\left(i\right)$ gives:

$-\frac{1}{2}k{x}^2-\mu (M+m)gx=-\frac{1}{2}(M+m)v^2$
$\frac{1}{2}\times 240\times x^2+0.1(1+9)\times 10x=\frac{1}{2}\times 10\times 1^2$
$120x^2+10x-5=0$
$\Rightarrow 120x(x-\frac{1}{6})+30(x-\frac{1}{6})=0$
$\therefore x=\frac{1}{6}\text{m}$
So, maximum compression in spring will be $x=\frac{1}{6}\text{m}$