The 10-mg barge b supports a 2-mg automobile

a. if someone drives the automobile to the other side of the barge, determine how far the barge moves. neglect the resistance of the water.

The image showing the barge is missing, so i have attached it.

It moves 6.667m in the right direction.

Explanation:

To solve this, we will assume that the barge will move to the left.

If we apply conservation of linear momentum to this question, we'll arrive at ;

MaVa + MbVb = MaVa' + MbVb'

Where;

Ma = mass of automobile

Mb = mass of barge

Va = Initial velocity of automobile

Vb = Initial velocity of barge

Va' = Final velocity of automobile

Vb' = Final velocity of barge

Now;

Mb = 10Mg = 10 x 10³g = 10,000 g

Ma = 2 Mg = 2 x 10³ = 2,000g

Va' = 0 m/s

Vb' = 0 m/s

Thus;

2,000Va + 10,000Vb = 0

2,000Va = - 10,000Vb

Va = - 10,000Vb/2,000

Va = - 5Vb

Now, velocity is distance/time and by experience, with same time, we can also say that;

Sa = -5Sb

From the image i attached, the relative distance between A and B is 40m

Thus, Sa - Sb = 40

From earlierSa = - 5Sb

Thus,

-5Sb - Sb = 40

-6 Sb = 40

Sb = -40/6 = - 6.667m

Since, this is negative it means it is in an opposite direction to the left which we chose.

Thus,

Sb moves 6.667m in the right direction.

are there options? cause you ain’t. real specific

what is the actual question though?