Animated Solution for Physics - System of Particles: A mass $m$ moves with a velocity $v$ and collides inelastically with another identical mass. After collision, the 1st mass moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision.

Imagine a mass $m$ moving with a velocity $v$ along the x-axis, heading straight towards another identical mass $m$ that is sitting perfectly at rest. This is our initial setup.

Suddenly, they collide! But this isn't a simple head-on bounce. The problem tells us that after the collision, the first mass shoots off in a direction perpendicular to its original path. If it was originally moving along the x-axis, it is now moving purely along the y-axis with a new speed of $\frac{v}{\sqrt{3}}$.

Because the first mass suddenly gained momentum in the y-direction, the second mass can't just sit there or move straight ahead. It must move diagonally to balance things out. Let's assume the second mass moves with a velocity vector ${\vec{v}}_2=v_x\hat{i}+v_y\hat{j}$. Our goal is to find the magnitude of this final velocity, $v^{\prime }$.

Even though the problem states the collision is inelastic (meaning kinetic energy is lost to heat or deformation), the Law of Conservation of Linear Momentum remains our most powerful tool. Because there are no external forces acting on the two-mass system during the brief moment of impact, the total momentum before the collision must exactly equal the total momentum after the collision.

The beauty of momentum being a vector is that we can break this rule down and apply it independently along the x-axis and the y-axis.

Let's look at the x-direction first. Initially, all the momentum is carried by the first mass moving at speed $v$. After the collision, the first mass is moving purely along the y-axis, meaning its x-velocity is zero. Therefore, all the initial x-momentum must have been transferred to the second mass.

Canceling the mass $m$ from both sides, we get a beautifully simple result for the x-component of the second mass:

Now, let's analyze the y-direction. Initially, neither mass was moving up or down, so the total initial y-momentum is zero. After the collision, the first mass is moving upwards with momentum $m\left(\frac{v}{\sqrt{3}}\right)$. To keep the total y-momentum at zero, the second mass must have an equal and opposite momentum downwards.

Solving for $v_y$, we find:

The negative sign perfectly confirms our intuition: the second mass moves downwards to balance the upward motion of the first mass.

We now have the two perpendicular components of the second mass's velocity: $v_x=v$ and $v_y=-\frac{v}{\sqrt{3}}$. To find the actual speed $v^{\prime }$ (the magnitude of the velocity vector), we simply use the Pythagorean theorem.

Substituting our components into the equation:

Squaring the terms (and remembering that squaring a negative makes it positive), we get:

Finding a common denominator to add the terms inside the square root:

Finally, taking the square root gives us the final speed of the second mass:

This elegant result matches option (c).

Energy Considerations

We solved this entire problem without ever needing to look at energy. However, because it was an inelastic collision, we know that the final kinetic energy of the system is strictly less than the initial kinetic energy. A fantastic exercise to test your deeper understanding would be to calculate exactly what fraction of the initial kinetic energy was lost during this scattering event!