Mastering Physics Solutions: Exercise 6.82

Mastering Physics Solutions: Exercise 6.82

On January 4, 2012, in Chapter 06: Linear Momentum and Collisions, by Mastering Physics Solutions

C Exercise 6.82

Part A = 3.8,5.2 m Click to use the calculator/solver for this part of the problem
Part B = 3.8,5.2 m

Solutions Below:

Two skaters with masses of 66 kg and 48 kg, respectively, stand 9.0 m apart, each holding one end of a piece of rope.

Part A

If they pull themselves along the rope until they meet, how far does each skater travel? (Neglect friction.)

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

We can solve this using the center mass formula:

cm = (x1m1 + x2m2) / (m1 + m2)

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

cm = (x1m1 + x2m2) / (m1 + m2)
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:

x66kg,x48kg = 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

If only the 48 kg skater pulls along the rope until she meets her friend (who just holds onto the rope), how far does each skater travel?

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:

x66kg,x48kg = 3.8,5.2m

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