Two objects are moving in the xy-plane. No external forces are acting on the objects. Object A has a mass of 3.2 kg and has a velocity of Va= (2.3 m/s)i + (4.2m/s)j and object B has a mass of 2.9 kg and has a velocity of Vb= (-1.8m/s)i + (2.7m/s)j. Some time later, object A is seen to have a velocity Va'= (1.7m/s)i + (3.5m/s)j.What is the velocity of object B at that instant?

To find the velocity of object B at the same instant when object A has a velocity of Va'= (1.7m/s)i + (3.5m/s)j, we can use the principle of conservation of momentum.

The principle of conservation of momentum states that the total momentum of a system remains constant if no external forces act on it. In this case, since no external forces act on the objects, the total momentum of the system will remain constant.

The momentum of an object can be calculated by multiplying its mass by its velocity. So, if we denote the velocity of object B at that instant as Vb', the total momentum before and after the velocity change can be equated as:

(mass of object A) * (initial velocity of object A) + (mass of object B) * (initial velocity of object B) = (mass of object A) * (final velocity of object A) + (mass of object B) * (final velocity of object B)

Plugging in the given values:

(3.2 kg) * (2.3 m/s)i + (4.2m/s)j + (2.9 kg) * (-1.8m/s)i + (2.7m/s)j = (3.2 kg) * (1.7m/s)i + (3.5m/s)j + (2.9 kg) * (Vb')i + (Vb')j

Now, we can equate the components of the vector equation separately:

For the i-component:
(3.2 kg) * (2.3 m/s) + (2.9 kg) * (-1.8 m/s) = (3.2 kg) * (1.7 m/s) + (2.9 kg) * (Vb')

Simplifying the equation, we get:
7.36 kg*m/s - 5.22 kg*m/s = 5.44 kg*m/s + 2.9 kg * Vb'

Rearranging and solving for Vb':
(2.9 kg * Vb') = (7.36 kg*m/s - 5.22 kg*m/s) - (5.44 kg*m/s)

Vb' = ((7.36 kg*m/s - 5.22 kg*m/s) - (5.44 kg*m/s)) / (2.9 kg)

Similarly, for the j-component, we get:
(3.2 kg) * (4.2 m/s) + (2.9 kg) * (2.7 m/s) = (3.2 kg) * (3.5 m/s) + (2.9 kg) * (Vb')

Simplifying the equation, we get:
13.44 kg*m/s + 7.83 kg*m/s = 11.2 kg*m/s + 2.9 kg * Vb'

Rearranging and solving for Vb':
(2.9 kg * Vb') = (13.44 kg*m/s + 7.83 kg*m/s) - (11.2 kg*m/s)

Vb' = ((13.44 kg*m/s + 7.83 kg*m/s) - (11.2 kg*m/s)) / (2.9 kg)

Now, we can substitute the values and calculate Vb'.

Vb' = (9.27 kg*m/s) / (2.9 kg)
Vb' ≈ 3.19 m/s

Therefore, the velocity of object B at that instant is approximately Vb' = (3.19 m/s)i + (3.19 m/s)j.