Physics Lesson 6.8.3 - Explosions

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Welcome to our Physics lesson on Explosions, this is the third lesson of our suite of physics lessons covering the topic of Momentum and Impulse in Two Dimensions. Explosions., you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Explosions

Explosions are the reverse phenomenon of inelastic collisions. Thus, in inelastic collisions the kinetic energy converts into other form after the impact, while in explosions, the energy stored in the object converts into kinetic energy, making it move faster than before. Therefore, the equation of explosions is the inverse of that used in inelastic collisions.

where M is the mass of the original object, m₁ and m₂ are the masses of the respective pieces after the explosion, v₀ is the initial velocity of the original object before the explosion, v₁ and v₂ are the velocities of the pieces after the explosion.

The simplest case of explosions is when the original object explodes and splits in two pieces which move in opposite directions as shown in the figure.

In this case, we have only one dimension involved and therefore, a single equation to write (the equation of explosions written above). Let's consider an example to explain this point.

Example 2

A 5 kg ball is moving initially due right at 1 m/s. The ball explodes due to an internal process and as a result, it divides in two pieces at ratio 3:2 between them. After explosion, the smallest piece contine moving in the original direction at 15 m/s. What is the velocity (and the direction) of the largest piece after the explosion?

Solution 2

First let's work out the masses of the two pieces after the explosion. We have the ratio

This means m₁ = 3x and m₂ = 2x (where x is a constant).

Also, we have m₁ + m₂ = 5 kg.

Therfore, we can write

Thus,

and

Now, applying the equation

where M = 5 kg, v⃗₀ = 1 m/s and v⃗₁ = 15 m/s, we find for the velocity v⃗₂ of the second piece after the explosion,

This result means the larger piece will move at 8.33 m/s in the opposite direction as it was before the explosion, i.e. due left.

Explosions in Two Dimensions

In explosions in two dimensions, we use the same approach as for momentum and impulse in two dimensions, i.e. we write the equation of momentum conservation for each direction separately. We have

For illustration, let's consider an example where an object explodes and divides in three pieces.

Example 3

A 12 kg object moving at 3 m/s due right - as shown in the figure below - explodes in three pieces, which have a ratio of 2:1:3 between them.

After the explosion, the first piece moves at 8 m/s up, the second piece (the smallest) moves at 40 m/s due right (in the same direction it was moving before the explosion).

Calculate:

1. The velocity components of the largest piece after the explosion
2. Magnitude of velocity of the largest piece after the explosion
3. The angle formed by the largest piece to the left-right direction after the explosion
4. What happens with the kinetic energy of the object after the explosion? Explain.

Solution 3

First, let's calculate the masses of the three pieces after the explosion. Giving that the total mass of the original object was M = 12 kg, we have

and

From the first equation, we obtain

where x is a constant. Substituting them in the second equation, we obtain

Therefore,

We can take the let-to-right as positive of x-direction and down-to-up as positive of y-direction (as usual). Therefore, the other clues of this problem are:

Applying the law of conservation of momentum for two dimensions,

we obtain after the substitutions

This means the third and largest piece will move at 7.3 m/s left and 5.3 m/s down after the explosion as shown in the figure below.

b. |v⃗₃ | = ?

Magnitude of the velocity of the largest piece after explosion is

c. We have to calculate the size of the wide angle θ formed by the third piece velocity vector and the left-to-right direction. We have:

We consider only the second value as the angle θ is wide as shown in the figure.

d. The initial velocity of the object before explosion was directed only according the x-direction. Therefore, v₀ = v_0x = 3 m/s.

Kinetic energy of the object before the explosion therefore was

On the other hand, giving that v₁ = v_1y = 8 m/s, v₂ = v_2x = 40 m/s and v₃ = 13.6 m/s (as foud earlier), the kinetic energy of each piece after the explosion is

Therefore, the total kinetic energy KE of the system after the explosion is

Obviously, this value is much gretaer than the initial kinetic energy of the object before expoding. This means some of the stored (idle) energy of the object has been activated and it was converted into kinetic energy during the explosion. Therefore, there is no break in the energy conversion law discusses in the previous articles.

You have reached the end of Physics lesson 6.8.3 Explosions. There are 3 lessons in this physics tutorial covering Momentum and Impulse in Two Dimensions. Explosions., you can access all the lessons from this tutorial below.

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