Question

Romeo (85.0 kg) entertains Juliet (59.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m away from Juliet, who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo's cheek. How far does the 75.0-kg boat move toward the shore it is facing?

Answer 1

To solve the problem it is necessary to apply the concepts related to the center of mass in a system.

The general formula of the center of mass is given by:

`X_c = \sum\limit^n_(n=1)(X_nM_n)/(M_n)`

Where

`X_n =` Coordenate according the Reference

`M_n =` Mass of objects

For this problem we have three bodies defined: Romeo, Juliet and the Boat. In this way,

`X_c = ((57kg*0)+(75Kg*2.7m)+(77Kg*1.35n))/(57kg+75kg+77kg)`

`X_c = (306.45Kg*m)/(209Kg) = 1.467m`

When they approach to kiss, the movement generates two things, the first is that the center of mass remains at the same distance as it was previously from the shore, however, it generates a substantial change with the back of the boat.

`X_c' = (77kg*1.35m+75kg*2.7m+57kg*2.7m)/(57kg+75kg+77kg)`

`X_c' = (460.35Kg*m)/(209kg)`

`X_c' = 2.2026`

The displacement of the boat would then be given by the change of position which is

`\Delta X = X_c'-X_c`

`\Delta X = 2.202-1.467`

`\Delta X = 0.735m`