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

(a)
`v_g_i=1.08(m)/(s)`

(b)
`v_p_i=-0.3(m)/(s)`

Step-by-step 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=-(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-(m_gv_g_i)/(m_p_i)\\v_g_i+(m_gv_g_i)/(m_p_i)=v_g_p\\v_g_i(1+(m_g)/(m_p))=v_g_p\\v_g_i=(v_g_p)/(1+(m_g)/(m_p))\\v_g_i=(1.38(m)/(s))/(1+(45kg)/(159kg))\\v_g_i=1.08(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(m)/(s)-1.38(m)/(s)\\v_p_i=-0.3(m)/(s)`

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