A TUG-OF-WAR ON THE ICE

James (mass 90.0 kg) and Ramon (mass 60.0 kg) are 20.0 m apart on a frozen pond. Midway between them is a mug of their favorite beverage. They pull on the ends of a light rope stretched between them. When James has moved 6.0 m toward the mug, how far and in what direction has Ramon moved?

Question

A TUG-OF-WAR ON THE ICE

James (mass 90.0 kg) and Ramon (mass 60.0 kg) are 20.0 m apart on a frozen pond. Midway between them is a mug of their favorite beverage. They pull on the ends of a light rope stretched between them. When James has moved 6.0 m toward the mug, how far and in what direction has Ramon moved?

Accepted Answer

IDENTIFY and SET UP:

The surface is horizontal and (we assume) frictionless, so the net external force on the system of James, Ramon, and the rope is zero; their total momentum is conserved. Initially there is no motion, so the total momentum is zero. The velocity of the center of mass is therefore zero, and it remains at rest. Let’s take the origin at the position of the mug and let the +x-axis extend from the mug toward Ramon. Figure 8.31 shows our sketch. We use the first of Eqs. (8.28):

[latex]\begin{aligned}&x_{\mathrm{cm}}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}+\cdots}{m_{1}+m_{2}+m_{3}+\cdots}=\frac{\sum_{i} m_{i} x_{i}}{\sum_{i} m_{i}} \&y_{\mathrm{cm}}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}+\cdots}{m_{1}+m_{2}+m_{3}+\cdots}=\frac{\sum_{i} m_{i} y_{i}}{\sum_{i}m_{i}}\end{aligned}[/latex]

to calculate the position of the center of mass; we ignore the mass of the light rope.

EXECUTE:

The initial x-coordinates of James and Ramon are -10.0 m and +10.0 m, respectively, so the x-coordinate of the center of mass is

[latex]x_{\mathrm{cm}}=\frac{(90.0 \mathrm{~kg})(-10.0 \mathrm{~m})+(60.0 \mathrm{~kg})(10.0 \mathrm{~m})}{90.0 \mathrm{~kg}+60.0 \mathrm{~kg}}=-2.0 \mathrm{~m}[/latex]

When James moves 6.0 m toward the mug, his new x-coordinate is -4.0 m; we’ll call Ramon’s new x-coordinate [latex]x_2[/latex]. The center of mass doesn’t move, so

[latex]\begin{aligned}x_{\mathrm{cm}} &=\frac{(90.0 \mathrm{~kg})(-4.0 \mathrm{~m})+(60.0 \mathrm{~kg})x_{2}}{90.0 \mathrm{~kg}+60.0 \mathrm{~kg}}=-2.0 \mathrm{~m} \x_{2} &=1.0\mathrm{~m}\end{aligned}[/latex]

James has moved 6.0 m and is still 4.0 m from the mug, but Ramon has moved 9.0 m and is only 1.0 m from it.

EVALUATE:

The ratio of the distances moved, [latex](6.0 \mathrm{~m}) /(9.0 \mathrm{~m})=\frac{2}{3}[/latex], is the inverse ratio of the masses. Can you see why? Because the surface is frictionless, the two men will keep moving and collide at the center of mass; Ramon will reach the mug first. This is independent of how hard either person pulls; pulling harder just makes them move faster.

Question 8.14: A TUG-OF-WAR ON THE ICE James (mass 90.0 kg) and Ramon (mass...

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