Answer:
Explanation:
Given
mass of body A [tex]m_a=2 kg[/tex]
mass of body [tex]m_b=2 kg[/tex]
Velocity before Collision is [tex]u_a=15\hat{i}+30\hat{j}[/tex]
[tex]u_b=-10\hat{i}+5\hat{j} m/s[/tex]
after collision [tex]v_a=-5\hat{i}+20\hat{j} m/s[/tex]
let [tex]v_b_x[/tex] and [tex]v_b_y[/tex] velocity of B after collision in x and y direction
conserving momentum in x direction
[tex]m_a\times 15+m_b\times (-10)=m_a\times (-5)+m_b\times (v_b_x)[/tex]
as [tex]m_a=m_b[/tex] thus
[tex]15-10=-5+v_b_x[/tex]
[tex]v_b_x=10 m/s[/tex]
Conserving momentum in Y direction
[tex]m_a\times 30+m_b\times 5=m_a\times 20+m_b\times (v_b_y)[/tex]
[tex]30+5=20+v_b_y[/tex]
[tex]v_b_y=15 m/s[/tex]
thus velocity of B after collision is
[tex]v_b=10\hat{i}+15\hat{j}[/tex]
(b)Change in total Kinetic Energy
Initial Kinetic Energy of A And B
[tex]K.E._a=\frac{1}{2}\times 2(\sqrt{15^2+30^2})^2=1125 J[/tex]
[tex]K.E._b=\frac{1}{2}\times 2(\sqrt{10^2+5^2})^2=125 J[/tex]
Total initial Kinetic Energy =1250 J
Final Kinetic Energy of A And B
[tex]K.E._a=\frac{1}{2}\times 2(\sqrt{5^2+20^2})^2=425 J[/tex]
[tex]K.E._b=\frac{1}{2}\times 2(\sqrt{10^2+15^2})^2=325 J[/tex]
Final Kinetic Energy [tex]=425+325=750 J[/tex]
Change [tex]\Delta K.E.=1250-750=500 J[/tex]
