Question

The left block in figure collides in-elastically with the right block and sticks to it. Find the amplitude of the resulting simple harmonic motion.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Applying conservation of momentum,

$mv=2m{v}_f$

$v_f=\frac{v}{2}$

$v=\omega \sqrt{A^2-X^2}$

$\omega$ for spring block = $\sqrt{\frac{k}{2m}}$

at $x=0,v=\frac{v}{2}$

$\Rightarrow \frac{v}{2}=\sqrt{\frac{k}{2m}}A$

$A=\left(\sqrt{\frac{m}{2k}}\right)v$

Alternate Solution:-

The common velocity after the collision is $\frac{v}{2}$. The kinetic energy $=\frac{1}{2}\left(2m\right)(\frac{v}{2}{)}^2=\frac{1}{4}m{v}^2$. This is also the total energy of vibration as the spring is un-stretched at this moment. If the amplitude is A, the total energy can also be written as $\frac{1}{2}k{A}^2$ Thus

$\frac{1}{2}k{A}^2=\frac{1}{4}m{v}^2,givingA=\sqrt{\frac{m}{2k}}v$