F mv v_{o}t t_{o}F 10000 lbm110 ft/sec9 secF _{1.2×105}ftlbmsec^{2}m_{G}v_{G}m_{B}v_{B}0orm_{G}v_{G}m_{B}v_{B}Force and MotionMOMENTUM PRINCIPLESRev. 0Page 7CP-03Example:The velocity of a rocket must be increased by 110 ft/sec to achieve proper orbit aroundthe earth. If the rocket has a mass of 5 tons and it takes 9 sec. to reach orbit, calculatethe required thrust (force) to achieve this orbit.Solution:Even though the initial velocity (v ) and final velocity (v) are unknown, we do know theochange in velocity (v-v ), which is 110 ft/sec. Therefore, using Equation 3-4 we can findothe solution.ConservationofMomentumOne of the most useful properties of momentum is that it is conserved. This means that if no netexternal force acts upon an object, the momentum of the object remains constant. UsingEquation 3-6, we can see that if force (F) is equal to zero, then P = 0. It is most important forcollisions, explosions, etc., where the external force is negligible, and states that the momentumbefore the event (collision, explosion) equals the momentum following the event.The conservation of momentum applies when a bullet is fired from a gun. Prior to firing the gun,both the gun and the bullet are at rest (i.e., V and V are zero), and therefore the totalG Bmomentum is zero. This can be written as follows:When the gun is fired, the momentum of the recoiling gun is equal and opposite to themomentum of the bullet. That is, the momentum of the bullet (m v ) is equal to the momentumB Bof the gun (m v ), but of opposite direction.G G