Question 14.1: A 200-kg space vehicle is observed at t = 0 to pass through ......
A 200-kg space vehicle is observed at t = 0 to pass through the origin of a newtonian reference frame Oxyz 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, of mass 100 kg, 60 kg, and 40 kg, respectively. Knowing that 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.
Since there is no external force, 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
\overline{ r }=\text v _0 t=(150 ~m / s ) i (2.5 ~s )=(375 ~m ) iRecalling Eq. (14.12),
m \overline{ r }=\sum_{i=1}^n m_i r _i (14.12)
we write
m \overline{ r }=m_A r _A+m_B r _B+m_C 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