Animated Solution for Physics - System of Particles: A body A, of mass $m=0.1\text{ kg}$ has an initial velocity of $3\hat{\mathbf{i}}{\text{ ms}}^{-1}$. It collides elastically with another body, $B$ of the same mass which has an initial velocity of $5\hat{\mathbf{j}}{\text{ ms}}^{-1}$. After collision, A moves with a velocity $\mathbf{v}=4(\hat{\mathbf{i}}+\hat{\mathbf{j}})$. The energy of $B$ after collision is written as $\frac{x}{10}\text{ J}$. The value of $x$ is ……… .

Imagine two bodies, A and B, gliding across a frictionless surface. Both have an identical mass of $0.1\text{ kg}$. Body A is moving purely along the x-axis with a velocity of $3\hat{\mathbf{i}}{\text{ ms}}^{-1}$, while body B is moving purely along the y-axis with a velocity of $5\hat{\mathbf{j}}{\text{ ms}}^{-1}$. They are on a collision course.

When they collide, the forces they exert on each other are internal to the two-body system. Because there are no external forces acting on them, the total linear momentum of the system must remain perfectly conserved. This principle is our master key to unlocking the problem.

Momentum Conservation

Let's calculate the total initial momentum of the system before the collision. Momentum is a vector quantity, so we must preserve its directional components.

Substituting the given values:

After the collision, we are told that body A moves with a new velocity ${\mathbf{v}}_A=4(\hat{\mathbf{i}}+\hat{\mathbf{j}})$. However, the final velocity of body B, let's call it ${\mathbf{v}}_B$, is unknown. We can write the total final momentum as:

By the law of conservation of linear momentum, the initial and final momenta must be exactly equal:

Now, we simply rearrange the terms to isolate the unknown velocity vector ${\mathbf{v}}_B$:

Dividing by $0.1$, we find the final velocity of body B:

The problem asks for the kinetic energy of body B after the collision. The kinetic energy is given by $K_B=\frac{1}{2}{m}_B|{\mathbf{v}}_B{|}^2$. First, we find the square of the magnitude of its velocity:

Plugging this into the kinetic energy formula:

We are given that this energy is equal to $\frac{x}{10}\text{ J}$. Equating the two:

A Fascinating Observation: The problem explicitly stated that the collision was elastic. However, we solved the entire problem without ever using the kinetic energy conservation equation! Why? Because the problem provided the exact final velocity vector of body A. This was enough information to solve for the single unknown vector ${\mathbf{v}}_B$ using only momentum conservation. The word "elastic" was technically redundant information, though it serves as a great consistency check. If you calculate the total initial and final kinetic energies, you will find they both equal $1.7\text{ J}$, proving the collision is indeed perfectly elastic.