Answer with explanation:
We are given that
Mass of ball,[tex]m_1=[/tex]75 g=[tex]\frac{75}{1000}=0075kg[/tex]
1 kg=1000 g
Height,[tex]h_1=1.6 m[/tex]
[tex]h_2=0.6 m[/tex]
Horizontal velocity,[tex]v_x=2 m/s[/tex]
Mass of plate[tex]m_2=400 g=\frac{400}{1000}=0.4 kg[/tex]
a.Initial velocity of plate,[tex]u_2=0[/tex]
Velocity before impact=[tex]u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s[/tex]
Where [tex]g=9.8 m/s^2[/tex]
Velocity after impact,[tex]v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s[/tex]
According to law of conservation of momentum
[tex]m_1u_1+m_2u_1=-m_1v_1+m_2v_2[/tex]
Substitute the values
[tex]0.075\times 5.6+0=-0.075\times 3.4+0.4v_2[/tex]
[tex]0.4v_2=0.075\times 5.6+0.075\times 3.4[/tex]
[tex]v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s[/tex]
Velocity of plate=1.69 m/s
b.Initial energy=[tex]\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J[/tex]
Final energy=[tex]\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2[/tex]
Final energy=[tex]\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J[/tex]
Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J
