Answer: The velocity of concrete block is 0.353 m/s
Explanation:
To calculate the velocity of the concrete block after the collision, we use the equation of law of conservation of momentum, which is:
\(m_1u_1+m_2u_2=m_1v_1+m_2v_2\)
where,
\(m_1\) = mass of rubber bullet = 0.012 kg
\(u_1\) = Initial velocity of rubber bullet = 150 m/s
\(v_1\) = Final velocity of rubber bullet = -100 m/s
\(m_2\) = mass of concrete block = 8.5 kg
\(u_2\) = Initial velocity of concrete block = 0 m/s
\(v_2\) = Final velocity of concrete block = ?
Putting values in above equation, we get:
\((0.012\times 150)+(8.5\times 0)=(0.012\times (-100))+(8.5\times v_2)\\\\v_2=\frac{1.8+1.2}{8.5}=0.353m/s\)
Hence, the velocity of concrete block is 0.353 m/s
