A bob of mass 1.0 kg hangs from a light inextensible string of length 1.2 m. The bob is pulled aside until the string is horizontal and then…

Energy conservation for the bob gives its speed just before impact: $v^2=2gL=2\times9.8\times1.2=23.52\ \mathrm{m^2s^{-2}}$. In a one-dimensional elastic collision with a stationary mass, the second body acquires $v_2=\dfrac{2m_1}{m_1+m_2}v=\dfrac{2\times1}{3}v$, so $v_2^2=\tfrac49\times23.52=10.453\ \mathrm{m^2s^{-2}}$. The block decelerates under kinetic friction with $a=\mu g=0.20\times9.8=1.96\ \mathrm{m\,s^{-2}}$, and the sliding distance follows from $v_2^2=2\mu g\,s$, giving $s=\dfrac{10.453}{3.92}=2.667\ \mathrm{m}\approx2.67\ \mathrm{m}$. The mass ratio is the crucial detail: using equal masses would give $v_2=v$ and a wrong distance of $6.00$ m. The pendulum energy step and the elastic collision step both depend on the mass ratio only through the factor $2m_1/(m_1+m_2)$.