The **speed **of the** block m1** on the frictionless **table **is** 1.34 m/s.**

The given parameters;

*mass of the first block, m1 = 1.9 kg**mass of the second block, m2 = 2.8 kg**distance of block m1, R = 0.95 m*

The** net torque** on both **blocks **is calculated as;

[tex]\tau _{net} = I \alpha[/tex]

[tex]T_2R- T_1 R_1 = I \alpha \\[/tex]

where;

*T₁ is the tension on first block**I is the moment of inertia of point mass**α is the angular acceleration*

[tex]T_1 = m_1 g + m_1 a\\T_2 = m_2 g - m_2 a[/tex]

The **acceleration **of both **blocks **is calculated as follows;

[tex]R(T_ 2- T_1) = MR^2 \times (\frac{a}{R} )\\R(T_2 -T_1) = MRa\\T_2 - T_1 = Ma\\(m_2g - m_2 a) - (m_1 g + m_1 a) = Ma\\m_2 g - m_1 g - m_2 a - m_1 a = Ma\\g(m_2 - m_1) = Ma + m_2a+ m_1a\\g(m_2 - m_1) = a(M+ m_2 + m_1)\\where;\\M \ is \ mass \ of \ string = 0 \\g(m_2 - m_1) = a (0+ m_2 + m_1)\\g(m_2 - m_1) = a(m_1 + m_2)\\a = \frac{g(m_2 - m_1)}{m_1 + m_2} \\a = 1.88 \ m/s^2[/tex]

The **speed **of the **block m1** is calculated as follows;

[tex]a = \frac{v^2}{r} \\v^2 = ar\\v = \sqrt{a r} \\v = \sqrt{1.88 \times 0.95} \\v = 1.34 \ m/s[/tex]

Thus, the **speed **of the** block m1** on the frictionless **table **is** 1.34 m/s.**