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

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.

Step-by-step 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

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

Step-by-step 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