A space probe of mass 1000kg, moving at 200m/s, explosively ejects a capsule of mass 300kg. The speed of the probe after the explosion is 250m/s. What is the velocity of the capsule?
To solve this problem, we must apply conservation of momentum. Even though there is kinetic energy being added to the probe-capsule system by the explosion, momentum will always be conserved if there are no external forces. These external forces could be, for example, air resistance or friction, but since we are in space we do not need to consider this.
We first calculate the momentum of the probe-capsule system before the explosion:
pi = 1000kg * 200 m/s = 200,000kgm/s
After ejecting the 300 kg capsule, the probe only weighs 700 kg. The total momentum is therefore:
pf = pprobe + pcapsule = 700kg * 250 m/s + 300 kg * vcapsule
Conservation of momentum requires
pf = pi
and inserting the above results yields:
700kg * 250 m/s + 300kg * vcapsule = 200,000kgm/s
vcapsule = (200,000 – 700*250)/300 m/s = 83.33 m/s
Since we have chosen velocities to be positive along the direction of motion of the probe, this means the probe and capsule must still be moving in the same direction.
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