Answer:
[tex]K_2=6.4J[/tex]
Explanation:
According to the principle of conservation of momentum, we have:
[tex]\Delta p=0\\p_i=p_f\\m_1v_1_i +m_2v_2_i=m_1v_1_f+m_2v_2_f[/tex]
Here 1 is for the bullet and 2 is for the block. Since the block is initially at rest [tex]v_i_2=0[/tex]. Solving for [tex]v_2_f[/tex] and replacing the given values:
[tex]v_2_f=\frac{m_1(v_1_i-v_1_f)}{m_2}\\v_2_f=\frac{10*10^{-3}kg(1.8\frac{km}{s}-1\frac{km}{s})}{5kg}\\v_2_f=0.0016\frac{km}{s}*\frac{1000m}{1km}=1.6\frac{m}{s}[/tex]
The kinetic energy of the block is given by:
[tex]K_2=\frac{m_2(v_f_2)^2}{2}\\K_2=\frac{5kg(1.6\frac{m}{s})^2}{2}\\K_2=6.4J[/tex]
