A 5.0 kg bomb at rest explodes into three pieces, each of which travels parallel to the ground. The first piece, with a mass of 1.2 kg travels at 5.5 m/s at an angle of 20 degrees south of east. The second piece has a mass of 2.5 kg and travels 4.1 m/s at an angle of 25 degrees north of east. Determine the velocity of the third piece.
\(Cosine\ in \ the\ vector \ triangle\)
\(c^2=a^2+b^2-2ac\cdot cos\gamma\)
\(2P_a=1.2kg\cdot 5.5^2\frac{m^2}{s^2}\)
\(2P_b=2.5kg\cdot 4.1^2\frac{m^2}{s^2}\)
\(\gamma=(180-20-25)°=135°\)
\(4P_c^2=(1.2^2\cdot 5.5^4+2.5^2\cdot 4.1^4-2\cdot 1.2\cdot 5.5^2\cdot 2.5\cdot 4.1^2\cdot cos135°)kg^2\cdot \frac{m^4}{s^4}\)
\(2P_c=\sqrt{5241.184021\ kg^2\cdot \frac{m^4}{s^4}}\)
\(2P_c=72.3960221352 kg\cdot\frac{m^2}{s^2}\)
\(2P_c=m_c\cdot v_c^2\)
\(\large v_c=\sqrt{\frac{2P_c}{m_c}}=\sqrt{\frac{72.3960221352 kg\cdot\frac{m^2}{s^2}}{(5-1.2-2.5)kg}}\)
\(v_c=7.463\ m/s\)
\(The\ third \ piece \ of \ the \ bomb \ is \ removed \ at \ a \ rate \ of \ 7.463 \ m/s.\)
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