Two friends, Al and Jo, have a combined mass of 194 kg. At the ice skating rink, they stand close together on skates, at rest and facing each other. Using their arms, they push on each other for 1 second and move off in opposite directions. Al moves off with a speed of 7.9 m/sec in one direction and Jo moves off with a speed of 6.7 m/sec in the other. You can assume friction is negligible. What is Al's mass? 110.58 What is Jo's mass? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Al on Jo? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Jo on Al?

The mass of Al is 89.027 kilograms.

The mass of Jo is 104.973 kilograms.

The magnitude of the force of Jo on Al is 596.481 newtons.

Explanation:

Given the absence of external forces, this situation can be described will by Principle of Linear Momentum Conservation and Impact Theorem on each skater:

Al:

(1)

Jo:

(2)

Total mass:

Where:

, - Masses of the skaters, in kilograms.

, - Initial and final velocities of Al, in meters per second.

, - Initial and final velocities of Jo, in meters per second.

- Impact force between skaters, in newtons.

- Impact time, in seconds.

If we know that , , , and , then the masses of the skaters are, respectively:

(1b)

(2b)

(2b) in (1b):

And the magnitude of the force is:

The mass of Al is 89.027 kilograms.

The mass of Jo is 104.973 kilograms.

The magnitude of the force of Jo on Al is 596.481 newtons.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.