Question

A 45.0 kg girl is standing on a 159 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.38 m/s relative to the plank.
(a) What is her velocity relative to the surface of ice?
1 m/s
(b) What is the velocity of the plank relative to the surface of ice?
2 m/s

Answer 1

Answer:

(a) $v_g_i=1.08\frac{m}{s}$

(b) $v_p_i=-0.3\frac{m}{s}$

Explanation:

According to the law of conservation of momentum:

$\Delta p=0\\p_i=p_f$

Initially, the girl and the plank are at rest. So, relative to the ice, we have:

$0=m_gv_g_i+m_pv_p_i\\v_p_i=-\frac{m_gv_g_i}{m_p_i}(1)$

(a) The velocity of the girl relative to the ice is:

$v_g_i=v_g_p+v_p_i(2)$

Here, $v_g_p$ is the velocity of the girl relative to the plank and $v_p_i$ is the velocity of the plank relative to the ice.

Replacing (1) in (2):

$v_g_i=v_g_p-\frac{m_gv_g_i}{m_p_i}\\v_g_i+\frac{m_gv_g_i}{m_p_i}=v_g_p\\v_g_i(1+\frac{m_g}{m_p})=v_g_p\\v_g_i=\frac{v_g_p}{1+\frac{m_g}{m_p}}\\v_g_i=\frac{1.38\frac{m}{s}}{1+\frac{45kg}{159kg}}\\v_g_i=1.08\frac{m}{s}$

(b) According to (2), the velocity of the plank relative to the surface of ice is:

$v_p_i=v_g_i-v_g_p\\v_p_i=1.08\frac{m}{s}-1.38\frac{m}{s}\\v_p_i=-0.3\frac{m}{s}$

The negative sing indicates that the plank is moving to the left.