Animated Solution for Physics - System of Particles: A body of mass $1$ kg initially at rest, explodes and breaks into three fragments of masses in the ratio $1:1:3$. The two pieces of equal mass fly-off perpendicular to each other with a speed of $30$ m/s each. What is the velocity of the heavier fragment ?

Imagine a stationary object resting peacefully in space. Suddenly, an internal mechanism triggers, and it violently shatters into multiple pieces. This is the classic "explosion" scenario in physics. At first glance, it seems chaotic and unpredictable. However, beneath the chaos lies one of the most elegant and unbreakable laws of nature: the Conservation of Linear Momentum.

Because the explosion is driven entirely by internal forces—like chemical reactions or spring releases—the net external force acting on the system is exactly zero. According to Newton's Second Law, if no external force acts on a system, its total momentum cannot change.

Before the explosion, our $1$ kg body was at rest. Its initial momentum was zero. Therefore, no matter how many pieces it breaks into, or how fast they fly away, the vector sum of all their momenta must perfectly add up to zero.

The problem tells us the body breaks into three fragments with a mass ratio of $1:1:3$.

Let's assign a variable to represent the base unit of mass. Let the masses of the three fragments be $m$, $m$, and $3m$.

You might be tempted to immediately calculate the exact value of $m$ since the total mass is $1$ kg.

This would give $m=0.2$ kg. However, as we will soon discover, the beauty of this problem is that the actual value of the mass is completely irrelevant! The physics depends only on the ratio of the masses.

We are told that the two lighter pieces, each of mass $m$, fly off perpendicular to each other with an identical speed of $v=30$ m/s.

Let's set up a coordinate system to visualize this. We can align the path of the first piece along the positive x-axis and the second piece along the positive y-axis.

The momentum vector of the first piece is:

The momentum vector of the second piece is:

To find out what the third piece must do, we first need to understand the combined effect of the first two pieces. We calculate their resultant momentum by adding their vectors.

Since these two vectors are perpendicular and equal in magnitude, they form a square. The resultant vector is the diagonal of this square. Using the Pythagorean theorem, the magnitude of this resultant momentum is:

Because the components are equal, this resultant vector points exactly at a $45^{\circ }$ angle between the x and y axes.

Now, we bring in the Conservation of Linear Momentum. The total final momentum must be zero.

We can rewrite this using our resultant vector:

This simple equation tells a profound story. The third fragment, which is the heaviest piece with mass $3m$, must carry a momentum that perfectly cancels out the combined momentum of the first two pieces. It must have the exact same magnitude as the resultant, but point in the exact opposite direction.

Let the velocity of the third fragment be $v^{\prime }$. Its momentum magnitude is its mass times its velocity, which is $(3m)v^{\prime }$.

Equating the magnitudes, we get:

Notice what happens here! The mass variable $m$ appears on both sides of the equation. We can divide both sides by $m$, completely eliminating it from our calculation. This proves that the initial $1$ kg mass was a distractor—a classic trap set by examiners to see if you truly understand the underlying principles.

Solving for $v^{\prime }$, we find:

We are given that the speed of the lighter fragments is $v=30$ m/s. Substituting this value into our equation:

We have found the speed, but velocity is a vector; it needs a direction.

We established earlier that the resultant of the first two pieces points at a $45^{\circ }$ angle in the first quadrant. Because the third piece must move in the exact opposite direction to cancel this momentum, it must fly off into the third quadrant.

Geometrically, this is an angle of $180^{\circ }+45^{\circ }=225^{\circ }$ from the positive x-axis. Alternatively, we can simply state that it moves at a $45^{\circ }$ angle relative to the negative axes, directly opposite to the resultant of the other two fragments.

The heavier fragment sacrifices speed for mass, moving slower than its lighter counterparts, but carrying enough momentum to keep the universe in perfect balance.