Question

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?

Answer 1

Answer:

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:

$m_{1}\cdot (v_{1, f}-v_{1, o}) = -F \cdot \Delta t$ (1)

Jo:

$m_{2}\cdot (v_{2,f}-v_{2,o}) = F\cdot \Delta t$ (2)

Total mass:

$m_{1} + m_{2} = 194\,kg$

Where:

$m_{1}$, $m_{2}$ - Masses of the skaters, in kilograms.

$v_{1,o}$, $v_{1,f}$ - Initial and final velocities of Al, in meters per second.

$v_{2,o}$, $v_{2,f}$ - Initial and final velocities of Jo, in meters per second.

$F$ - Impact force between skaters, in newtons.

$\Delta t$ - Impact time, in seconds.

If we know that $v_{1,o} = 0\,\frac{m}{s}$, $v_{1,f} = -7.9\,\frac{m}{s}$, $\Delta t = 1\,s$, $v_{2,o} = 0\,\frac{m}{s}$ and $v_{2,f} = 6.7\,\frac{m}{s}$, then the masses of the skaters are, respectively:

$(194-m_{2})\cdot (-7.9) = -F$ (1b)

$m_{2} \cdot 6.7 = F$ (2b)

(2b) in (1b):

$(194-m_{2})\cdot (-7.9) = -m_{2}\cdot 6.7$

$-1532.6 +7.9\cdot m_{2} = -6.7\cdot m_{2}$

$14.6\cdot m_{2} = 1532.6$

$m_{2} = 104.973\,kg$

$m_{1} = 194\,kg - 104.973\,kg$

$m_{1} = 89.027\,kg$

And the magnitude of the force is:

$F = 6.7\cdot m_{2}$

$F = 596.481\,N$

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.