The figure shows a block of mass m resting on a horizontal smooth floor at a distance $l$ from a rigid wall. Block is pushed toward the right by a distance $\frac{3l}{2}$ and released from rest. When block passes from its mean position, another block of mass $m_1$ is gently placed on it which sticks to it due to friction. The value of $m_1$ so that the combined block just touches the left wall is $\frac{\eta }{4}m$. Find the value of $\mathrm{\eta }$.

When block P is released from rest from a distance $\frac{3l}{2}$ toward right from mean position, this will be the amplitude of oscillation so velocity of block when passing from its mean position is given as
$v=A\omega =\frac{3l}{2}\sqrt{\frac{k}{m}}$
If mass m₁ is added to it and just after if velocity of combined block becomes v₁, from momentum conservation, we have
mv = (m + m₁)v₁
$\therefore v_1=\frac{m}{m+m_1}\cdot \frac{3l}{2}\sqrt{\frac{k}{m}}$
If this is the velocity of combined block at mean position, it is given as
$v_1 =A_1{\mathrm{\omega }}_1$ where ${\omega }_1=\sqrt{\frac{k}{m+m_1}}$
where A₁ and ${\mathrm{\omega }}_1$ are the new amplitude and angular frequency of SHM of the block.
It is given that combined block just reaches the left wall.
Thus, the new amplitude of oscillation must be $l$ . So, we have,
$\frac{m}{m+m_1}\cdot \frac{3l}{2}\sqrt{\frac{k}{m}}=l\sqrt{\frac{k}{m+m_1}}$
On solving, we get
$m_1=\frac{5}{4}m$
$\therefore n=5$