Two people of differing masses on a sheet of ice

1. Jul 7, 2011

aCab

Okay, so I just recently purchased a book from Barnes and Nobles on Physics. I've taken a university level Physics course, so I have a general knowledge of some concepts.

I was reading something that I'm not sure holds water. Here's the problem:

Two men are standing on a sheet of ice. The larger of the two men is a giant weighing 300 kg while the smaller man reaches out and pushes the the larger man with all of his might--a total force of 10 Newtons. What will be the result of this action? Which of the two will slide across the ice and with what velocity?

The book goes on to calculate the accelerations of both parties using F = ma. The acceleration for the bigger man is 0.333 m/s^2 and 0.1 m/s^2 for the smaller man. But they lose me because they use the words acceleration and velocity interchangeably. They say that the "larger man will slide backward with a velocity of 0.333 m/s while the smaller man will slide at 0.1 m/s." If this is the case, wouldn't the larger man slide faster? The conclusion that the book wants to get at is that the smaller man will move much easier on the ice, but how do you justify this using acceleration?

2. Jul 7, 2011

Delphi51

Welcome to PF, aCab.
For starters, the force on the men will be equal (Newton's 3rd law).
The force will act for the same time on both men.
There is no mention of the friction force, so I will use zero friction - not unreasonable for ice. So only the 10 N force on each man.
The impulse formula F*Δt = m*Δv solved for Δv = F*Δt/m lets us compare the velocities attained. F and Δt are the same for both men so only the m varies, and the equation says Δv is larger for the less massive man.

You can see with with acceleration formulas as well. F = m*a,
a = F/m says the guy with the smaller mass has the larger acceleration and so will gain the higher velocity.