# Conservation of Momentum in an Explosion

In summary, during an explosion, a bomb explodes into 3 pieces with masses of 0.8kg, 0.5kg, and an unknown mass. The first two fragments have velocities of 10m/s and 16m/s respectively, while the third fragment has a velocity of 24m/s. Using the conservation of momentum, the direction of the third fragment can be determined to be anti-parallel to the resultant of the momenta of the first and second fragments. This can be visualized in a 2D diagram.

## Homework Statement

During an explosion, a bomb explodes into 3 pieces. Two fragments, whose masses are 0.8kg and 0.5kg fly off with velocities of 10m/s and 16m/s respectively along the paths at right angles to each other. If the third fragment goes off with a velocity of 24m/s, then find its mass and direction w.r.t the first fragment.

## Homework Equations

- Components ( u sin A and u cos A)
- Conservation of momentum. The horizontal components=0 and the vertical components=0

## The Attempt at a Solution

For M1
horizontal component of velocity= 10 cos A
vertical component of velocity= 10 sin A
Horizontal momentum= 0.8 x 10cosA
= 8cosA
Vertical momentum=10sinA x 0.8
= 8sinA

For M2
Horizontal component of velocity=16sinA
Vertical Component of velocity= 16cosA
Horizontal momentum= 8sinA
Vertical momentum= 8cosA

How do I find out the direction in which the third pieces moves in? Without the direction on the diagram I've drawn, I cannot split its velocity in component.

You'll need to either include your diagram, or specify either "clock" or compass directions for us.

How do I find out the direction in which the third pieces moves in?
Momentum is conserved, right? And you know the final momentum of the first two pieces. If you assume that the bomb started at rest, what must be the final momentum of the third piece?

## Homework Statement

During an explosion, a bomb explodes into 3 pieces. Two fragments, whose masses are 0.8kg and 0.5kg fly off with velocities of 10m/s and 16m/s respectively along the paths at right angles to each other. If the third fragment goes off with a velocity of 24m/s, then find its mass and direction w.r.t the first fragment.

## Homework Equations

- Components ( u sin A and u cos A)
- Conservation of momentum. The horizontal components=0 and the vertical components=0

## The Attempt at a Solution

For M1
horizontal component of velocity= 10 cos A
vertical component of velocity= 10 sin A
Horizontal momentum= 0.8 x 10cosA
= 8cosA
Vertical momentum=10sinA x 0.8
= 8sinA

For M2
Horizontal component of velocity=16sinA
Vertical Component of velocity= 16cosA
Horizontal momentum= 8sinA
Vertical momentum= 8cosA

How do I find out the direction in which the third pieces moves in? Without the direction on the diagram I've drawn, I cannot split its velocity in component.
Since momentum is being conserved.The momentum of the third particle will be anti parallel to the resultant of the momenta of the first and second particles.

Ellispson said:
Since momentum is being conserved.The momentum of the third particle will be anti parallel to the resultant of the momenta of the first and second particles.
But why is the momentum anti parallel?

Here's the diagram.

#### Attachments

• 20150802_214931.jpg
26.8 KB · Views: 850
3D or 2D?

But why is the momentum anti parallel?
Because the bomb is initially at rest with a total momentum of zero.The momentum after the explosion should be zero too because there are no external forces acting on it.

Dr. Courtney said:
3D or 2D?
The question doesn't say. But I think it's 2D.

Ellispson said:
Because the bomb is initially at rest with a total momentum of zero.The momentum after the explosion should be zero too because there are no external forces acting on it.
Thank you! I get it now!

## 1. How does conservation of momentum apply in an explosion?

The law of conservation of momentum states that the total momentum of a closed system remains constant. In an explosion, the system is not closed as there is an external force (the explosion) acting on it. However, the law still applies as the total momentum of the system before and after the explosion is equal.

## 2. What factors affect the conservation of momentum in an explosion?

The key factor that affects the conservation of momentum in an explosion is the mass and velocity of the explosive material and the objects it interacts with. Other factors may include the shape and direction of the explosion, as well as any external forces acting on the system.

## 3. How does the direction of an explosion impact the conservation of momentum?

The direction of an explosion plays a crucial role in determining the conservation of momentum. In a symmetrical explosion, where the force is evenly distributed in all directions, the momentum of the system remains constant. However, if the explosion is directional, the momentum of the system will change in the direction of the explosion.

## 4. Can the conservation of momentum be violated in an explosion?

No, the conservation of momentum cannot be violated in an explosion. As long as the system is closed, the total momentum will remain constant. Even in the case of a directional explosion, the change in momentum in one direction will be offset by an equal and opposite change in momentum in the opposite direction.

## 5. How is the conservation of momentum in an explosion used in real-world applications?

The conservation of momentum in an explosion is used in various real-world applications, such as in rocket propulsion and car airbags. By understanding and applying this law, engineers can design systems that efficiently use the momentum generated in an explosion to achieve a desired outcome, such as launching a rocket into space or protecting passengers in a car crash.

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