Respuesta :
Answer:[tex]v_3=14.61\times 10^6 m/s[/tex]
Explanation:
Given
mass of spaceship [tex]M=2.30\times 10^6 kg[/tex]
speed of spaceship [tex]u=6\times 10^6 m/s[/tex]
After blowing in to three pieces
[tex]m_1=5\times 10^5 kg[/tex]
[tex]v_1=-1.80\times 10^6 m/s[/tex]
[tex]m_2=8.40\times 10^5 kg[/tex]
[tex]v_2=8\times 10^5 m/s[/tex]
[tex]m_3=M-m_1-m_2[/tex]
[tex]m_3=2.30\times 10^6-5\times 10^5-8.40\times 10^5[/tex]
[tex]m_3=9.6\times 10^5 kg [/tex]
consider forward motion to be Positive
Let [tex]v_3[/tex] be the velocity of third particle
conserving momentum as no external Force is applied
[tex]Mu=m_1v_1+m_2v_2+m_3v_3[/tex]
[tex]2.30\times 10^6\times 6\times 10^6=5\times 10^5\cdot (-1.80\times 10^6)+8.40\times 10^5\times 8\times 10^5+9.6\times 10^5\cdot v_3[/tex]
[tex]13.8\times 10^{12}=-0.9\times 10^{12}+0.672\times 10^{12}+9.6\times 10^5\cdot v_3[/tex]
[tex]v_3=\frac{14.028\times 10^{12}}{9.6\times 10^5}[/tex]
[tex]v_3=14.61\times 10^6 m/s[/tex]
