# A man of mass 40 kg is standing on a uniform Planck of mass 60 kg lying on a horizontal friction less ice. The man walks from one end to the other end of the Planck. The distance walked by the man relative to ice is (length of Planck=5 m)

$\begin {array} {1 1} (a)\;2m & \quad (b)\;3m \\ (c)\;5m & \quad (d)\;4m \end {array}$

$d= \large\frac{mL}{m_1+m_2}$
$\quad= \large\frac{40 \times 5 m}{100}$
$\quad=2m$