Question

A 46.0-kg girl is standing on a 157-kg plank. Both originally at rest on a frozen lake that constitutes a frictionless, flat surface. The girl begins to walk along the plank at a constant velocity of 1.48î m/s relative to the plank.

(a) What is her velocity relative to the surface of ice?
(b) What is the velocity of the plank relative to the surface of ice?

Answer 1

To solve this problem we will apply the linear motion kinematic equations. Just as we will also find the relative speed of the body through the conservation of momentum. Our data is given as

$M = 157kg$

$m = 46kg$

$v_1 = 1.48m/s$

PART A)

From the conservation of momentum,

$\text{Momentum of Plank+girl}+\text{Mometum of girl} = 0$

$(M+m)v_2+mv_1 = 0$

$(M+m)v_2 = -m_v1$

$v_2 = \frac{-Mv_1}{M+m}$

$v_2 = \frac{-(46)(1.48)}{(157+46)}$

$v_2 = -0.3353m/s$

Since the ice surface is frozen lake and girl is moving on it so the relative velocity will get added. Therefore the velocity of the girl relative to the ice surface is as,

$v_1+v_2 = 1.48+(-0.33353)$

$v_1+v_2 = 1.14647m/s$

The velocity of the girl relative to the ice surface is 1.14647m/s

PART B) The velocity of the plank plus girl is $v_2 = -0.3353m/s$

Since the ice surface is frozen lake and plank is moving with girl on it so the relative velocity will get added. Therefore the velocity of the plank relative to the ice surface is as:

$v_2 = -0.3353m/s$