Animated Solution for Physics - System of Particles: A particle of mass $4m$ which is at rest explodes into three fragments. Two of the fragments each of mass $m$ are found to move with a speed $v$ each in mutually perpendicular directions. The total energy released in the process of explosion is ......... .

A Bomb Waiting to Go Off

Imagine a particle of mass $4m$ resting peacefully in space. It has no velocity, and therefore, no kinetic energy and no momentum. Suddenly, an explosion occurs! This explosion isn't caused by anything hitting it from the outside; it's entirely due to internal forces—perhaps a chemical reaction or a sudden release of potential energy.

Because there are no external forces acting on our $4m$ mass, a fundamental law of the universe comes into play: The Conservation of Linear Momentum. This law dictates that whatever the total momentum was before the explosion, it must be exactly the same after the explosion. Since the initial momentum was zero, the vector sum of the momenta of all the resulting fragments must also perfectly cancel out to zero.

Tracking the Pieces

The problem tells us that the particle shatters into three pieces. Two of these fragments, each with a mass of $m$, fly off at exactly $90^{\circ }$ to each other, both with a speed of $v$. Let's map this out on an x-y coordinate system. We can say the first fragment moves along the positive x-axis, giving it a momentum vector of ${\vec{p}}_1=mv\hat{i}$. The second fragment moves along the positive y-axis, giving it a momentum vector of ${\vec{p}}_2=mv\hat{j}$.

To find out what happens to the third piece, we first need to know the combined momentum of the first two. Since they are perpendicular, we use the Pythagorean theorem to find the magnitude of their resultant momentum:

Now, what about the third fragment? First, let's find its mass. The original mass was $4m$, and two pieces of mass $m$ have already been accounted for. By the conservation of mass, the third piece must have a mass of $4m-m-m=2m$.

For the total momentum of the system to remain zero, this $2m$ fragment must act as the perfect counterbalance. Its momentum vector, ${\vec{p}}_3$, must be exactly equal in magnitude and opposite in direction to the resultant of the first two fragments. Therefore, the magnitude of its momentum is also $\sqrt{2}mv$.

Let's call the speed of this third fragment $v^{\prime }$. We can set up a simple equation:

Solving for $v^{\prime }$, we find that the heaviest fragment is moving at a speed of $v^{\prime }=\frac{v}{\sqrt{2}}$.

Tallying the Kinetic Energy

The final question asks for the total energy released in the explosion. Before the explosion, the particle was at rest, so its kinetic energy was zero. After the explosion, all three fragments are moving, meaning they all possess kinetic energy. The "energy released" is simply the total kinetic energy of the system after the explosion, which was converted from the internal potential energy of the original particle.

Kinetic energy is a scalar quantity, so we don't need to worry about directions anymore; we just add them up:

The first two terms easily combine to give $m{v}^2$. For the third term, we substitute the speed $v^{\prime }$ we found earlier:

Adding it all together, we get the grand total of the energy released:

And there we have it! By carefully tracking the momentum vectors and then tallying up the scalar kinetic energies, we've completely decoded the physics of this explosion.