Question

g A 45.0 - kg girl is standing on a 150. - 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 1.50 m/s to the right relative to the plank. (a) What is her velocity relative to the surface of the ice

Answer 1

Answer:

her velocity relative to the surface of the ice is 1.154 m/s

The positive velocity indicates that her velocity relative to the surface of the ice is in the same direction with her velocity relative to plank.

Explanation:

Given;

mass of the girl, $M_G$ = 45.0 - kg

mass of the plank, $M_P$ = 150 kg

velocity of the girl relative to the plank, $V_G_P$ = 1.5 m/s

Velocity of the plank relative to the ice, $V_P_I$ = ?

Velocity of the girl relative to the ice, $V_G_I$ = ?

Thus, velocity of the girl relative to the ice can be calculated as;

$V_G_I$ = $V_G_P$  + $V_P_I$

$V_G_I$ = 1.5 + $V_P_I$ -----------equation (i)

From the principle of conservation of energy;

$M_GV_{GI} +M_PV_{PI} = 0$

45$V_G_I$ + 150 $V_P_I$ = 0

3$V_G_I$ +  10$V_P_I$ = 0 ------------equation (ii)

From equation (i);  $V_P_I$ = $V_G_I$ - 1.5

Substitute in  $V_P_I$ in equation (ii)

3$V_G_I$ +  10( $V_G_I$ - 1.5) = 0

3$V_G_I$ + 10$V_G_I$ -15 = 0

13$V_G_I$ = 15

$V_G_I$ = 15/13

$V_G_I$ = 1.154 m/s

The positive velocity indicates that her velocity relative to the surface of the ice is in the same direction with her velocity relative to plank.

Answer 2

Answer:

velocity of girl with respect  to surface of ice 1.154 m/s

Explanation:

Plank Mass , M = 150 kg

Girl Mass , m = 45 kg

velocity of the girl with resect to plank, v1 = 1.50 m/s

velocity of the plank + girl = v2

from the conservation of momentum

Momentum (plank+girl) = - momentum of the girl

(M+m)v2 = -mv1

v2 = -(45 kg)/(195 kg) * 1.50

v2 = -0.346 m/s

velocity of girl with respect  to surface of ice =

  v1+v2 = 1.50 + (-0.346) = 1.154 m/s