A block A of mass 1 kg is released from rest from a height of 1.8 m on a smooth curved track which ends tangentially on a horizontal surface. At the…

Block $A$ slides down the smooth track, so energy conservation gives its speed at the bottom $v_A=\sqrt{2gh}=\sqrt{2\times10\times1.8}=6\ \mathrm{m\,s^{-1}}$. In a one-dimensional perfectly elastic collision, momentum and kinetic energy are both conserved, and the velocity of the struck body is $v_B=\dfrac{2m_1}{m_1+m_2}v_A=\dfrac23\times6=4\ \mathrm{m\,s^{-1}}$, while $A$ rebounds with $\dfrac{m_1-m_2}{m_1+m_2}v_A=-2\ \mathrm{m\,s^{-1}}$ (it climbs the track, returns at $2\ \mathrm{m\,s^{-1}}$ and stops within $2^2/(2\times2)=1\ \mathrm{m}$ on the rough surface, so it never reaches $B$ again). Block $B$ then decelerates under friction with $a=\mu g=2\ \mathrm{m\,s^{-2}}$, so the stopping distance is $s=\dfrac{v_B^2}{2\mu g}=\dfrac{16}{4}=4\ \mathrm{m}$. The answer is $4$. Note that friction acts only on $B$ after the collision and the mass of $B$ cancels out in the stopping distance, so the result depends only on the speed imparted to it and on $\mu$.