0
$\begingroup$

enter image description here

How he treated the center of mass as if it is still in the same place , is not it supposed to change after explosion ? and is not explosion destroying the closed system? i mean it represents and external force and energy , so how can i use conservation laws ?

$\endgroup$
3
  • $\begingroup$ related (duplicate?): physics.stackexchange.com/q/135208 $\endgroup$ – Kyle Oman Oct 14 '14 at 21:39
  • $\begingroup$ maybe the part of explosion but i can not understand the motion of center of mass after explosion so i posted these Question $\endgroup$ – Mohamed Osama Oct 14 '14 at 21:41
  • $\begingroup$ Unless there is an external force acting on the system, the speed of the center of mass will not change. That's a fundamental property of closed systems that follows directly from Newton's laws. So if it was moving at 150 m/s before the explosion, it is still moving at 150 m/s after the explosion - conservation of momentum demands it. $\endgroup$ – Floris Oct 14 '14 at 22:55
2
$\begingroup$

You can use the conservation of momentum to check their solution. Since no external forces are added, the initial momentum of the spaceship must equal the sum of the momentum of A,B,C in X, Y, and Z directions. The initial x momentum is 30000Ns and y and z momentum are 0.

In the X-Dir: $ p_x = m_Av_{xA} + m_Bv_{xB} + m_Cv_{xC} = 100*555/2.5+60*255/2.5+40*105/2.5 = 30000Ns $

In the Y-Dir: $ p_y = m_Av_{yA} + m_Bv_{yB} + m_Cv_{yC} = 100*-180/2.5+60*0/2.5+40*450/2.5 = 0Ns$

In the Y-Dir: $ p_z = m_Av_{zA} + m_Bv_{zB} + m_Cv_{zC} = 100*240/2.5+60*-120/2.5+40*-420/2.5 = 0Ns$

So their solution checks out. Essentially the above is the same as saying that the center of mass post explosion must move at the same total momentum due to momentum conservation.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.