[tex]a)\text{Masa celui de-al doilea fragment este:}\\
m_2=M-m_1 =70-30 =40\ kg\\
\text{Din conservarea impulsului se obtine:}\\
M\cdot v=m_1\cdot v_1+m_2\cdot v_2\\
70\cdot 300=30\cdot 500+40\cdot v_2\\
21000=15000+40\cdot v_2\\
6000=40\cdot v_2\\
\boxed{v_2=150\ \dfrac{m}{s}}[/tex]
[tex]b.\Delta E_c= E_{c1}+E_{c2}-E_c\\
\Delta E_c= \dfrac{m_1\cdot v_1^2}{2}+\dfrac{m_2\cdot v_2^2}{2}-\dfrac{M\cdot v^2}{2}\\
\Delta E_c=\dfrac{30\cdot 250000}{2}+\dfrac{40\cdot 22500}{2}-\dfrac{70\cdot 90000}{2}\\
\Delta E_c=\dfrac{7500000}{2}+\dfrac{900000}{2}-\dfrac{6300000}{2}\\
\Delta E_c= \dfrac{2100000}{2}=1050000\\
\boxed{\Delta E_c= 1,05\ MJ}[/tex]
