Answer with Explanation:
We are given that
[tex]m_1=0.3 kg[/tex]
[tex]m_2=0.4 kg[/tex]
Spring constant,[tex]k=200 N/m[/tex]
[tex]h=0.250 m[/tex]
a.Speed of ham before collision
[tex]u=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.25}=2.2m/s[/tex]
Collision is inelastic
According to law of conservation of momentum
[tex]m_1u_1=(m_1+m_2)v[/tex]
[tex]0.3\times 2.2=(0.3+0.4)v[/tex]
[tex]v=\frac{0.3\times 2.2}{0.7}=0.94 m/s[/tex]
Kinetic energy of masses=Spring potential energy at maximum distance
[tex]\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}kA^2[/tex]
[tex]A^2=\frac{(m_1+m_2)v^2}{k}[/tex]
[tex]A=\sqrt{\frac{(m_1+m_2)v^2}{k}}[/tex]
Substitute the values
[tex]A=\sqrt{\frac{(0.3+0.4)\times (0.94)^2}{200}}=0.056 m=0.056\times 100=5.6 cm[/tex]
1 m=100 cm
b.Time period,[tex]T=2\pi\sqrt{\frac{(m_1+m_2)}{k}}[/tex]
[tex]T=2\pi\sqrt{\frac{(0.3+0.4)}{200}}=0.37 s[/tex]
