Kevin S. Huang

2010: Problem 14

Topic: Collisions
Concepts: 1D elastic collision, Energy dissipation

Solution:

By conservation of energy, the velocity of $m_1$ when it reaches $m_2$ is

\frac{1}{2}m_1v_1’^2=\frac{1}{2}m_1v_1^2-\mu m_1gd

v_1’=\sqrt{v_1^2-2\mu gd}

Using the elastic collision equation, the velocity of $m_2$ right after the collision is

v_2=\frac{2m_1}{m_1+m_2}v_1′

We conserve energy again to find how far $m_2$ slides:

\frac{1}{2}m_2v_2^2=\mu m_2gd_2

d_2=\frac{v_2^2}{2\mu g}=\frac{1}{2\mu g}\left(\frac{2m_1}{m_1+m_2}\right)^2(v_1^2-2\mu gd)=\frac{4m_1^2}{(m_1+m_2)^2}\left(\frac{v_1^2}{2\mu g}-d\right)=1.79\,\mathrm{m}

so the answer is B.