Conservation of Momentum
Law of Conservation of Momentum
Now that we've talked about momentum in an isolated system, where no external forces act, we can state that momentum is always conserved. Put more simply, in any closed system, the total momentum of the system remains constant.
In the case of a collision or explosion (an event), if you add up the individual momentum vectors of all of the objects before the event, you'll find they're equal to the sum of the momentum vectors of the objects after the event. Written mathematically: . This is a direct outcome of Newton's 3rd Law.
In analyzing collisions and explosions, a momentum table can be a powerful tool for problem solving. To create a momentum table, follow these basic steps:
- Identify all objects in the system. List them vertically down the left-hand column.
- Determine the momenta of the objects before the event. Use variables for any unknowns.
- Determine the momenta of the objects after the event. Use variables for any unknowns.
- Add up all the momenta from before the event, and set them equal to the momenta after the event.
- Solve your resulting equation for any unknowns.
A collision is an event in which two or more objects approach and interact strongly for a brief period of time. Let's look at how our problem-solving strategy can be applied to a simple collision:
Question: A 2000-kg car traveling at 20 m/s collides with a 1000-kg car at rest at a stop sign. If the 2000-kg car has a velocity of 6.67 m/s after the collision, find the velocity of the 1000-kg car after the collision.
Answer: Let's call the 2000-kg car Car A, and the 1000-kg car Car B. We can then create a momentum table as shown below:
Objects Momentum Before (kg*m/s) Momentum After (kg*m/s) Car A 2000*20=40000 2000*6.67=13,340 Car B 1000*0=0 1000*vb Total 40,000 1000vb+13,340
Because momentum is conserved in any closed system, the total momentum before the event must be equal to the total momentum after the event. Therefore, we can write: 40,000=1000vb+13,340. Solving for vb, we find that vb must be equal to 26.67 m/s.
Not all problems are quite so simple, but problem solving steps remain consistent. Let's take a look at another example which emphasizes the vector nature of momentum while examining an explosion. In physics terms, an explosion results when an object is broken up into two or more fragments.
Question: A 4-kilogram rifle fires a 20-gram bullet with a velocity of 300 m/s. Find the recoil velocity of the rifle.
Answer: Once again, we can use a momentum table to organize our problem-solving. To fill out the table, we must realize that the initial momentum of the system is 0, and we can consider the rifle and bullet as a single system with a mass of 4.02 kg:
Objects Momentum Before (kg*m/s) Momentum After (kg*m/s) Rifle 4.02*0=0 4*vr Bullet 300*0.02=6 Total 0 4vr+6
Due to conservation of momentum, we can again state that the total momentum before must equal the total momentum after, or 0=4vr+6. Solving for the velocity of the rifle, we find that vr=-1.5 m/s, where the negative sign indicates the direction of the rifle... if the bullet traveled forward at 300 m/s, the rifle must travel in the opposite direction.
Let's see if you can't run through a problem solo using the momentum table strategy: