Question

In the figure shown below, if the block is a cube of side $1\text{m}$, then loss of mechanical energy during impact if the difference in vertical heights between the two horizontal surfaces is very small is:

• A. $5\text{J}$
• B. $6\text{J}$
• C. $6\text{J}$
• D. $8\text{J}$

Answer 1

The correct option is A $5\text{J}$

Given,
mass of the cube, $m=1\text{kg}$
side of the cube, $a=1\text{m}$

After impact, the cube will rotate about B,

From conservation of angular momentum about B,

$\text{Initial angular momentum of the cube about B= Final angular momentum of the cube about B}$

$L_i=L_f$

$\Rightarrow mv{r}_{\perp }=I_B\omega ......\left(1\right)$

Here, $r_{\perp }=OA=\frac{1}{2}\text{m}$

$v=4\text{m/s}$,

$I_B=I_O+m(OB{)}^2$

$=m\frac{a^2}{6}+ma\times (\frac{1}{\sqrt{2}}{)}^2$.

Substituting the values in $\left(1\right)$,

$\Rightarrow 1\times 4\times \frac{1}{2}=[\frac{1\times 1^2}{6}+1\times {\left(\frac{1}{\sqrt{2}}\right)}^2]\omega$

$\Rightarrow 2=\frac{2}{3}\omega$

$\therefore \omega =3\text{rad/s}$

Now, loss of mechanical energy,

$\Delta E=E_i-E_f$

$\Rightarrow \Delta E=\frac{1}{2}m{v}^2-\frac{1}{2}I{w}^2$

$\Rightarrow \Delta E=\frac{1}{2}\times 1\times 4^2-\frac{1}{2}\times [\frac{1\times 1^2}{6}+1\times {\left(\frac{1}{\sqrt{2}}\right)}^2]\times 3^2$

$\Rightarrow \Delta E=8-3=5\text{J}$

Hence, option $\left(a\right)$ is correct option.