Question

In the figure, pendulum bob on left side is pulled aside to a height h from its initial position. After it is released it collides with the right pendulum bob at rest, which is of same mass. After the collision the two bobs stick together and raise to a height

• A. $\frac{3h}{4}$
• B. $\frac{2h}{3}$
• C. $\frac{h}{2}$
• D. $\frac{h}{4}$

Answer 1

Answer: (D)

When bob $A$ strikes the bob $B$, then
$mu=(m+m)v^{\prime }$
$\Rightarrow v^{\prime }=\frac{u}{2}$
The potential energy of $A$ at height h converts into kinetic energy of this mass, at point $0$, ie
$mgh=\frac{1}{2}m{u}^2$
or $u=\sqrt{2gh}$
$\therefore v^{\prime }=\frac{\sqrt{2gh}}{2}=\sqrt{\frac{gh}{2}}$
Let combined mass moves to a height $h$', then
$2mg{h}^{\prime }=\frac{1}{2}(2m)v^2$
or $g{h}^{\prime }=\frac{gh}{4}$
or $h^{\prime }=\frac{h}{4}$