Let's work an example. An arrow of mass 40 g initially moves at a speed of 10 meters per second. It rips through an apple of mass 100 grams and removes a 8-gram piece. The arrow and the apple piece continue together with a speed of 8 meters per second. What is the final speed of the apple? From the law of conservation of momentum we know that the momentum of the system must be the same before and after the collision. Before the collision, the momentum of the arrow is its mass times its velocity; the arrow has a mass of 0.04 kilograms and a velocity of 10 meters per second. The initial momentum p-sub-one-i of the arrow equals 0.4 kilogram meters per second The apple has a mass m-two-i of zero-point-one kilograms and an initial velocity v-sub-two-i of zero. The initial momentum of the apple is therefore zero kilogram-meters per second. Before the collision, the total initial momentum of the system is therefore the initial momentum of the arrow plus the initial momentum of the apple. The total initial momentum of the system, p-sub-i, is equal to 0.4 kilogram-meters per second. After the collision, a 0.008 kilogram piece of apple sticks to the arrow and the arrow and the apple piece continue forward with a velocity of 8 meters per second. After the collision, the momentum of the arrow and apple-piece system is equal to the sum of their masses times the final velocity. Thus their final momentum is 0.384 kilogram-meters per second. The final momentum of the apple is equal to the mass of the apple minus the apple-piece times the apple's final velocity v-sub-two-f. Thus the apple's final momentum is 0.092 kilogram times v-sub-two-f. Solving for v2f yields the final velocity of the apple. In this case, everything moves forward after the collision. The arrow plus the apple chunk moves with a speed of 8 meters per second and the rest of the apple with a speed of about 0.17 meters per second.

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