# Conservation of Momentum in a explosion

• mvl46566
In summary, the firecracker explodes into two fragments with the first fragment having a mass of 14g and moving in the positive x direction with a velocity of 48m/s. The second fragment has a mass of 41g and moves in the negative x direction with a velocity of 32m/s. The momentum of the system before and after the explosion is conserved, meaning the total momentum is zero, and therefore the second fragment must have a negative x direction to balance out the positive x direction of the first fragment.
mvl46566

## Homework Statement

A firecracker, initially at rest, explodes into two fragments. The first, of mass 14g, moves in the positive x direction at 48m/s. The second moves at 32m/s. Find the mass and direction of its motion.

p=mv

## The Attempt at a Solution

So I know momentum is conserved in this situation. I set m1v1 equal to m2v2 and found the mass of the second fragment to be .041 kg or 41 grams. I think this goes in the negative x direction because the initial momentum is zero and so the sum of all momentums would have to be zero, but I am not sure.

mvl46566 said:

## Homework Statement

A firecracker, initially at rest, explodes into two fragments. The first, of mass 14g, moves in the positive x direction at 48m/s. The second moves at 32m/s. Find the mass and direction of its motion.

p=mv

## The Attempt at a Solution

So I know momentum is conserved in this situation. I set m1v1 equal to m2v2 and found the mass of the second fragment to be .041 kg or 41 grams. I think this goes in the negative x direction because the initial momentum is zero and so the sum of all momentums would have to be zero, but I am not sure.
Correct. In the conservation of momentum, the momentum after the explosion = momentum before the explosion. What is the momentum before the explosion?

Remember that momentum is a vector, which has magnitude and direction. We can have to vectors of the same magnitude, but opposite direction, so the sum is zero.

Your solution is correct. The conservation of momentum principle states that the total momentum before and after a collision or explosion remains constant. In this case, the initial momentum is zero since the firecracker was at rest. Therefore, the total momentum after the explosion must also be zero. Since one fragment is moving in the positive x direction, the other fragment must have an equal but opposite momentum in the negative x direction to balance out the total momentum to zero. This is why the mass of the second fragment is negative and its direction of motion is in the negative x direction. Great job using the conservation of momentum principle to solve this problem!

## 1. What is conservation of momentum in an explosion?

Conservation of momentum in an explosion refers to the principle that the total momentum of a system before and after an explosion remains constant. This means that the combined momentum of all objects involved in the explosion, including the explosive material and any fragments or debris, will remain the same.

## 2. Why is conservation of momentum important in explosions?

Conservation of momentum is important in explosions because it helps us understand and predict the behavior of objects during and after an explosion. It also allows us to analyze the amount of force and energy involved in the explosion, which is crucial for safety and engineering purposes.

## 3. How is conservation of momentum calculated in an explosion?

Conservation of momentum is calculated using the equation: pbefore = pafter, where p represents momentum. This equation takes into account the mass and velocity of each object involved in the explosion and ensures that the total momentum remains constant.

## 4. Does conservation of momentum apply to all types of explosions?

Yes, conservation of momentum applies to all types of explosions, including chemical, nuclear, and mechanical explosions. As long as there is a closed system and no external forces acting on the system, the total momentum will remain constant.

## 5. What factors can affect conservation of momentum in an explosion?

The two main factors that can affect conservation of momentum in an explosion are the mass and velocity of the objects involved. In general, the larger the mass and/or velocity of an object, the greater its momentum and the more it will contribute to the total momentum of the system.

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