Answer: The velocity of the 2.2 kg block is 5.23 m/s
Explanation:
To calculate the velocity of the 2.2 kg block after the collision, we use the equation of law of conservation of momentum, which is:
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
where,
[tex]m_1[/tex] = mass of block 1 = 5 kg
[tex]u_1[/tex] = Initial velocity of block 1 = 2.3 m/s
[tex]v_1[/tex] = Final velocity of block 1 = 0 m/s
[tex]m_2[/tex] = mass of block 2 = 2.2 kg
[tex]u_2[/tex] = Initial velocity of block 2 = 0 m/s
[tex]v_2[/tex] = Final velocity of block 2 = ?
Putting values in above equation, we get:
[tex](5\times 2.3)+(2.2\times 0)=(5.0\times 0)+(2.2\times v_2)\\\\v_2=\frac{(5\times 2.3)}{2.2}=5.23m/s[/tex]
Hence, the velocity of the 2.2 kg block is 5.23 m/s