# Mechanics Problem

I am stuck on a problem which goes as follows:

A sledge of mass 120kg stands at rest on the horizontal surface of an icy lake. A man of mass 70kg stands at one end of the sledge so that initially the distance from the man to the shore is 20m. The man now walks 3m relative to the sledge, towards the shore. Then he stops. Assuming that the sledge movies without friction on the ice, how far from the shore is the man when he stops?

Any guidance would be helpful. I don't need the answer, just a general way to go about finding distance in terms of mass and conservation of momentum.

## Answers and Replies

Andrew Mason
Science Advisor
Homework Helper
ColdFusion85 said:
I am stuck on a problem which goes as follows:

A sledge of mass 120kg stands at rest on the horizontal surface of an icy lake. A man of mass 70kg stands at one end of the sledge so that initially the distance from the man to the shore is 20m. The man now walks 3m relative to the sledge, towards the shore. Then he stops. Assuming that the sledge movies without friction on the ice, how far from the shore is the man when he stops?

Any guidance would be helpful. I don't need the answer, just a general way to go about finding distance in terms of mass and conservation of momentum.
This is a centre of mass problem. The frictionless surface means that the centre of mass cannot move.

Assume that the sledge has uniform mass/unit length. Where is the centre of mass initially? Where is it in relation to the sledge after he moves? If that centre of mass is in the same position (relative to the ice sheet or the shore), where is the end of the sledge in relation to the shore?

AM

i'm still sorta stuck. the only thing i seem to know is that 70kg*v=120kg*u, and I am guessing that perhaps one has to use the relationship delta x = delta v \ delta t somehow.

Andrew Mason
Science Advisor
Homework Helper
ColdFusion85 said:
i'm still sorta stuck. the only thing i seem to know is that 70kg*v=120kg*u, and I am guessing that perhaps one has to use the relationship delta x = delta v \ delta t somehow.
Speed is irrelevant.

The centre of mass is a point (C) between the man (M) and the centre of the sledge (S) such that:

$70\vec{CM} + 120\vec{CS} = 0$ where CS is the displacement from C to S and CM is the displacement from C to M.

Since C does not change, this is still the relationship between CM and CS when the displacement between M and S decreases by 3. What does that tell you about how CS (and the position of the edge of the sledge) has changed? Since the man is 3 m. from the edge of the sledge, you know how far he is from the shore.

AM