Animated Solution for Physics - System of Particles: In a collinear collision, a particle with an initial speed $v_0$ strikes a stationary particle of the same mass. If the final total kinetic energy is $50\%$ greater than the original kinetic energy, the magnitude of the relative velocity between the two particles after collision, is

Imagine a scenario where a moving particle strikes a stationary one, and instead of losing energy to sound or heat, the system actually gains kinetic energy! This might sound like a violation of the laws of physics, but it is entirely possible in what we call a super-elastic or explosive collision. In such events, stored internal potential energy (like a compressed spring or a chemical explosive) is released during the impact, adding to the total kinetic energy.

Let's unravel the mathematics behind this fascinating phenomenon.

No matter what happens to the kinetic energy, as long as there are no external forces acting on the system, the law of conservation of linear momentum holds absolute true.

Let the mass of each particle be $m$. The first particle moves with an initial velocity $v_0$, while the second is at rest. After the collision, let their velocities be $v_1$ and $v_2$.

Equating the initial and final momentum:

Since the masses are identical, we can elegantly divide the entire equation by $m$:

The problem states a crucial condition: the final total kinetic energy is $50\%$ greater than the original kinetic energy. This means the final kinetic energy is $1.5$ times the initial kinetic energy.

Let's translate this into an equation:

Again, we can cancel out the common factor of $\frac{1}{2}m$ from all terms, leaving us with a clean relationship between the squared velocities:

Our ultimate goal is to find the magnitude of the relative velocity between the two particles after the collision, which is $|v_1-v_2|$.

Instead of painfully solving for $v_1$ and $v_2$ individually, we can use a powerful algebraic identity:

We already know $(v_1+v_2{)}^2=v_0^2$ from Equation 1. But we need the value of the cross-term $2v_1{v}_2$. To find it, we simply square Equation 1:

Now, substitute the value of $v_1^2+v_2^2$ from Equation 2:

With all the pieces of the puzzle in hand, let's substitute them back into our algebraic identity:

Taking the square root of both sides gives us the magnitude of the relative velocity:

This elegant result shows that the particles fly apart with a relative speed greater than the initial approach speed, perfectly consistent with the explosive nature of the collision!