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 given time, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it.
The momentum of an object is defined as the product of its mass and velocity. In this case, the momentum of object A before the time later is given by the equation:
Momentum(A) = mass(A) * velocity(A)
= 3.2 kg * (2.3 m/s)i + (4.2 m/s)j
Similarly, the momentum of object B before the time later is given by:
Momentum(B) = mass(B) * velocity(B)
= 2.9 kg * (-1.8 m/s)i + (2.7 m/s)j
Since there are no external forces acting on the objects, the total momentum before and after the given time later should be the same. Therefore:
Momentum(A) + Momentum(B) = Momentum(A') + Momentum(B')
where,
Momentum(A') = mass(A) * velocity(A')
Momentum(B') = mass(B) * velocity(B')
We can rearrange the equation to solve for velocity(B'):
velocity(B') = (Momentum(A) + Momentum(B) - Momentum(A')) / mass(B)
Now, let's substitute the values given in the question:
mass(A) = 3.2 kg
mass(B) = 2.9 kg
velocity(A) = (2.3 m/s)i + (4.2 m/s)j
velocity(A') = (1.7 m/s)i + (3.5 m/s)j
Plugging these values into the equation, we get:
velocity(B') = [(3.2 kg * (2.3 m/s)i + (4.2 m/s)j) + (2.9 kg * (-1.8 m/s)i + (2.7 m/s)j) - (3.2 kg * (1.7 m/s)i + (3.5 m/s)j)] / 2.9 kg
Simplifying the equation, we find:
velocity(B') ≈ (-0.126 m/s)i + (0.062 m/s)j
Therefore, the velocity of object B at that instant is approximately (-0.126 m/s)i + (0.062 m/s)j.
To find the velocity of object B at the given instant, we can use the principle of conservation of linear momentum. According to this principle, the total linear momentum of a closed system remains constant unless acted upon by external forces.
The linear momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v
Initially, the total momentum of the system, which includes both object A and object B, is given by:
P_initial = m_A * v_A + m_B * v_B
where m_A and m_B are the masses of object A and object B respectively, and v_A and v_B are their initial velocities.
Given:
m_A = 3.2 kg
v_A = (2.3 m/s)i + (4.2 m/s)j
m_B = 2.9 kg
v_B = (-1.8 m/s)i + (2.7 m/s)j
Substituting these values into the equation, we get:
P_initial = (3.2 kg) * ((2.3 m/s)i + (4.2 m/s)j) + (2.9 kg) * ((-1.8 m/s)i + (2.7 m/s)j)
P_initial = (7.36 kg m/s)i + (13.44 kg m/s)j + (-5.22 kg m/s)i + (7.83 kg m/s)j
P_initial = (2.14 kg m/s)i + (21.27 kg m/s)j
Now, let's calculate the final momentum (P_final) using the given velocity of object A at the instant (Va'):
Va' = (1.7 m/s)i + (3.5 m/s)j
P_final = m_A * Va' + m_B * Vb'
where Vb' is the velocity of object B at that instant.
Given:
m_A = 3.2 kg
Va' = (1.7 m/s)i + (3.5 m/s)j
m_B = 2.9 kg
Substituting these values into the equation, we get:
P_final = (3.2 kg) * ((1.7 m/s)i + (3.5 m/s)j) + (2.9 kg) * Vb'
P_final = (5.44 kg m/s)i + (11.2 kg m/s)j + (2.9 kg) * Vb'
Since linear momentum is conserved, P_initial = P_final. Therefore, we can equate the x-components and the y-components separately:
X-component: 2.14 kg m/s = 5.44 kg m/s + (2.9 kg) * Vb'_x
Y-component: 21.27 kg m/s = 11.2 kg m/s + (2.9 kg) * Vb'_y
Solving these equations simultaneously, we can find Vb'_x and Vb'_y, which are the x-component and y-component of the velocity of object B at the given instant.
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