Question 14.SP.1: A 200-kg space vehicle passes through the origin of a newton......
A 200-kg space vehicle passes through the origin of a newtonian reference frame Oxyz at time t = 0 with velocity v_{0} = (150 m/s)i relative to the frame. Following the detonation of explosive charges, the vehicle separates into three parts A, B, and C, each with a mass of 100 kg, 60 kg, and 40 kg, respectively. At t = 2.5 s, the positions of parts A and B are observed to be A(555, –180, 240) and B(255, 0, –120), where the coordinates are expressed in meters. Determine the position of part C at that time.
STRATEGY: There are no external forces, so the linear momentum of the system is conserved. Use kinematics to relate the motion of the center of mass of the spacecraft and the rectangular coordinates of its position.
MODELING and ANALYSIS: The system is the space vehicle. After the explosion, the system is composed of all three parts: A, B, and C. The mass center G of the system moves with the constant velocity v_{0} = (150 m/s)i. At t = 2.5 s, its position is
{\bar{\mathbf{r}}}=\mathbf{v}_{0}t=(150\ \mathrm{m}/\mathrm{s}){{\mathbf{i}}}(2.5\ \mathrm{s})=(375\ \mathrm{m}){{\mathbf{i}}}
Recalling Eq. (14.12), you have
m \overline{ r }=\sum_{i=1}^n m_i r _i (14.12)
m\mathbf{\bar{r}}=m_{A}\mathbf{{r}}_{A}+m_{B}\mathbf{{r}}_{B}+m_{C}\mathbf{{r}}_{C}
( 200 kg)( 375 m )i = ( 100 kg)[( 555 m )i − ( 180 m )j + ( 240 m )k]+ ( 60 kg)[( 255 m )i − ( 120 m )k] + ( 40 kg)r_{C}
r_{C} = (105 m)i + (450 m)j – (420 m)k ◂
REFLECT and THINK: This kind of calculation can serve as a model for any situation involving fragmentation of a projectile with no external forces present.