Question

A 45.0-kg girl is standing on a 168-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 1.55 m/s to the right relative to the plank.

Required:
What is the velocity of the plank relative to the surface of the ice?

Answer 1

The speed of the plank relative to the ice is:

`v_(p)=-0.33\: m/s`

Step-by-step explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

`m_(g)v_(g)+m_(p)v_(p)=0` (1)

Where:

• m(g) is the mass of the girl
• m(p) is the mass of the plank
• v(g) is the speed of the girl
• v(p) is the speed of the plank

Now, as we have relative velocities, we have:

`v_(g/b)=v_(g)-v_(p)=1.55 \: m/s` (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

`45v_(g)+168v_(p)=0`

`v_(g)-v_(p)=1.55`

`v_(p)=-0.33\: m/s`

I hope it helps you!