C Exercise 6.82

Part A = 3.8,5.2 m

Part B = 3.8,5.2 m

Solutions Below:

Part A

Express your answers using two significant figures separated by a comma.

We can solve this using the center mass formula:

cm = (x_{1}m_{1} + x_{2}m_{2}) / (m_{1} + m_{2})

We’ll use the first skater as the frame of reference (but we could use either):

cm = (x_{1}m_{1} + x_{2}m_{2}) / (m_{1} + m_{2})

cm = (0 * 66 + 9 * 48) / (66 + 48)

cm = 432 / 114

cm = 3.79m

so the first skater moves 0 + 3.79m = 3.79m and the second skater moves 9 – 3.79m = 5.21m. The answer needs to be in 2 significant figures, so:

x_{66kg},x_{48kg} = 3.8,5.2m

Note: since each skater moves in proportion to their mass, you could also set up an equation:

66x = 48(9-x)

Solve for x

x = 3.79m (and add/subtract this the same as above to get the same answer)

Part B

The answer will be the same. Since there isn’t any friction, it doesn’t matter who pulls on the rope. Each force produces an equal and opposite force so the 48kg skater pulling the other skater towards him also causes an opposite force pushing him towards the 66kg skater, and they both end up in the same spot as in Part A:

x_{66kg},x_{48kg} = 3.8,5.2m