How Does Conservation of Momentum Determine the Movement of Explosion Fragments?

In summary, Homer the human powder keg, initially at rest, explodes into 3 pieces each with equal mass. One piece moves east at 30 m/s and a second piece moves at 30 m/s south east. To calculate the speed and direction of the third piece, you can use the given directions and find that they are 45º apart. You can also determine the compass bearing of the third piece. The problem can be solved if at least one angle is given.
  • #1
mindhater
6
0
Homer the human powder keg, initially at rest, suddenly explodes into 3 pieces each with = mass. One piece moves east at 30 m/s and a second piece moves at 30 m/s south east

a) calculate the speed of the 3rd piece and direction

Problem I'm having is I can't find the angles because they are not given. With at least one given angle i could solve the problem. Is there a way to find an angle?
 
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  • #2
The angles are given with the directions. East and South East are 45º apart. You can also find a compass bearing of the thrid piece.
 
  • #3



There are a few ways to approach this problem without given angles. One method is to use the conservation of momentum principle, which states that in a closed system, the total momentum remains constant. In this case, the initial momentum of Homer before the explosion is zero since he is at rest, and after the explosion, the total momentum of the 3 pieces must also be zero.

Using this principle, we can set up the following equation:

0 = m1v1 + m2v2 + m3v3

Where m1, m2, and m3 are the masses of the 3 pieces and v1, v2, and v3 are their corresponding velocities. We know that m1 = m2 = m3 = m, and v1 = 30 m/s (east) and v2 = 30 m/s (south east).

Substituting these values into the equation, we get:

0 = m(30 m/s) + m(30 m/s) + m3v3

Simplifying, we get:

0 = 60m + m3v3

Since we don't know the value of m3 or v3, we cannot solve for them directly. However, we can use the Pythagorean theorem to find the magnitude of the third piece's velocity:

v3 = √(v3x² + v3y²)

Where v3x and v3y are the x and y components of the third piece's velocity. We can also use trigonometric ratios to find the relationship between v3x and v3y. For example, if we assume that the angle between the third piece's velocity and the east direction is θ, then we can say:

v3x = v3 cosθ
v3y = v3 sinθ

Substituting these values into the Pythagorean theorem equation, we get:

v3 = √[(v3 cosθ)² + (v3 sinθ)²]
v3 = √(v3²(cos²θ + sin²θ))
v3 = √(v3²)

Simplifying, we get:

v3 = v3

This means that the magnitude of the third piece's velocity is equal to the magnitude of the initial explosion, which is 30 m/s. However, without knowing the angle θ, we cannot determine the direction of the third piece's velocity.
 

FAQ: How Does Conservation of Momentum Determine the Movement of Explosion Fragments?

What is momentum in explosions?

Momentum in explosions refers to the force that is generated by an object as it moves through space. In the context of explosions, momentum is usually measured in terms of the mass and velocity of the exploding object.

How is momentum calculated in explosions?

The momentum of an object in an explosion can be calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity. This formula can be used to calculate the momentum of individual objects or the total momentum of a system of objects involved in an explosion.

What factors affect momentum in explosions?

The main factors that affect momentum in explosions are the mass and velocity of the exploding object(s). Additionally, the direction and angle of the explosion, as well as any external forces acting on the object(s), can also impact the momentum generated.

Why is momentum important in studying explosions?

Momentum is important in studying explosions because it helps us understand the force and impact of the explosion. By analyzing the momentum of an exploding object, we can determine the potential damage and destruction caused by the explosion.

How is momentum conserved in explosions?

According to the law of conservation of momentum, the total momentum of a system before and after an explosion remains constant. This means that the momentum of the exploding objects must be equal to the momentum of the debris and any other objects affected by the explosion.

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