Answer:
Part a)
[tex]10\hat i + 15\hat j = \vec v[/tex]
Part b)
[tex]\Delta K = 500 J[/tex]
Explanation:
As we know that there is no external force on the system of two masses so here total momentum of the system will remains conserved
so we can say
[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]
[tex](2kg)(15\hat i + 30 \hat j) + (2 kg)(-10\hat i + 5\hat j) = 2kg(-5\hat i + 20\hat j) + 2\vec v[/tex]
[tex]5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v[/tex]
[tex]10\hat i + 15\hat j = \vec v[/tex]
Part b)
magnitude of the initial speed of A = [tex]\sqrt{15^2 + 30^2} = 33.54 m/s[/tex]
magnitude of the initial speed of B = [tex]\sqrt{10^2 + 5^2} = 11.18 m/s[/tex]
magnitude of final speed of A = [tex]\sqrt{5^2 + 20^2} = 20.61 m/s[/tex]
magnitude of final speed of B = [tex]\sqrt{10^2 + 15^2} = 18.03 m/s[/tex]
Now change in total kinetic energy is given as
[tex]\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)[/tex]
[tex]\Delta K = (\frac{1}{2}2(33.54)^2 + \frac{1}{2}2(11.18)^2) - (\frac{1}{2}2(20.61)^2 + \frac{1}{2}2(18.03)^2)[/tex]
[tex]\Delta K = 500 J[/tex]
