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A uniform thin wooden plank $A B$ of length $L$ and mass $M$ is kept on a table with its $B$ end slightly outside the edge of the table. When an impulse $J$ is given to the end $B$, the plank moves up with centre of mass rising a distance $h$ from the surface of the table. Then,
$h > 9 J^{2} / 8 M^{2} g$
$h=J^{2} / 2 M^{2} g$
$J^{2} / 2 M^{2} g < h < 9 J^{2} / 8 M^{2} g$
$h < J^{2} / 2 M^{2} g$
Solution
$(c)$ When an impulse $J$ is given to wooden plank $A B$, then there are two possibilities
$(i)$ Plank started to rotate about $A$ as soon as impulse is imparted.
$(ii)$ Plank moves vertically without rotation.
Now, we consider both cases one by one.
Case $I$
In case rotation occurs about $A$, angular momentum about $A$ is conserved. $\Rightarrow \quad I_{A} \omega=J L$
where, $I_{A}=$ moment of inertia of plank about $A=\frac{M L^{2}}{3}$
and $\omega=$ angular speed of plank about $A$.
$\frac{M L^{2}}{3} \omega=LJ$
$\Rightarrow \quad \omega=\frac{3 J}{M L}$
So, linear velocity of centre of mass,
$v=\left(\frac{L}{2}\right) \times \omega \Rightarrow v=\frac{3 \cdot J}{2 M}$
If $h=$ maximum height attained by centre of mass, then
By energy conservation, we have
$\frac{1}{2} m v^{2}=m g h \Rightarrow h=\frac{v^{2}}{2 g}=\frac{9 J^{2}}{8 M^{2} g}$
Case $II$ If there is no rotation of plank, then by conservation of linear momentum, we have
and from conservation of energy, we have
$\frac{1}{2} m v^{2}=m g h \Rightarrow h=\frac{J^{2}}{2 M^{2} g}$
Clearly, maximum height attained by centre of mass of plank is
$\frac{J^{2}}{2 M^{2} g} < h < \frac{9 J^{2}}{8 M^{2} g}$
