The image showing the barge is missing, so i have attached it.
Answer:
It moves 6.667m in the right direction.
Step-by-step 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.