Answer:
0.22788 J
Explanation:
[tex]m_1[/tex] = Mass of first object= 0.62 kg
[tex]m_2[/tex] = Mass of second object = 0.32 kg
[tex]u_1[/tex] = Initial Velocity of first object = 2.1 m/s
[tex]u_2[/tex] = Initial Velocity of second object = -3.8 m/s
For perfectly elastic collision
[tex]m_{1}u_{1}+m_{2}u_{2}\ =m_{1}v_{1}+m_{2}v_{2}[/tex]
[tex]v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.62-0.32}{0.62+0.32}\times 2.1+\frac{2\times 0.32}{0.62+0.32}\times -3.8\\\Rightarrow v_1=-1.917\ m/s[/tex]
Change in kinetic energy
[tex]\Delta KE=\frac{1}{2}m(u_1^2-v_1^2)\\\Rightarrow \Delta KE=\frac{1}{2}0.62(2.1^2-(-1.917)^2)\\\Rightarrow \Delta KE=0.22788\ J[/tex]
Change in kinetic energy of the 620g object is 0.22788 J
