Question

Two identical blocks P and Q have mass ‘m’ each. They are attached to two identical springs initially unstretched. Now the left spring (along with P) is compressed by $\frac{A}{2}$ and the right spring (along with Q) is compressed by A. Both the blocks are released simultaneously. They collide perfectly inelastically. Initial time period of both the blocks was T. [consider K as spring constant of each spring]

The energy of oscillation of the combined mass is

• A. $\frac{1}{2}K{A}^2$
• B. $\frac{1}{4}K{A}^2$
• C. $\frac{1}{8}K{A}^2$
• D. $\frac{1}{16}K{A}^2$

Answer 1

The correct option is D. $\frac{1}{16}K{A}^2$.

Consider the following diagram,

Applying conservation of linear momentum, the velocity of combined mass just after the collision is, $v=\frac{A\omega }{4}$.

We know that,

$E=\frac{1}{2}M{v}^2$

where M = m + m = 2m

$\Rightarrow E=\frac{1}{2}\left(2m\right)v^2$

$E=\frac{1}{2}\left(2m\right){\left(\frac{A\omega }{4}\right)}^2=\frac{m{\omega }^2{A}^2}{16}$

Since $F=-Kx=m{\omega }^{2}x\Rightarrow K=m{\omega }^2$
$E=\frac{1}{16}K{A}^2$.