Animated Solution for Physics - Rotational Motion: A smooth sphere $A$ is moving on a frictionless horizontal plane with angular velocity $\omega$ and centre of mass velocity $v$. It collides elastically and head on with an identical sphere $B$ at rest. Neglect friction everywhere. After the collision their angular speeds are ${\omega }_A$ and ${\omega }_B$ respectively. Then,

A Tale of Two Spheres

Imagine a perfectly smooth billiard table. On this frictionless horizontal plane, we have a smooth sphere, let's call it Sphere $A$, gliding forward with a linear center-of-mass velocity $v$. But it's not just sliding; it's also spinning with an angular velocity $\omega$.

Directly in its path lies Sphere $B$, an identical twin to Sphere $A$, but currently resting peacefully with zero linear and zero angular velocity. The problem sets the stage for a classic physics encounter: a perfectly elastic, head-on collision. Our goal is to determine the rotational fate of both spheres after the impact.

A Classic Kinematics Move

Before we dive into the spinning, let's address the linear motion. The problem explicitly states that the collision is elastic and head-on between two identical spheres.

If you recall your kinematics, this specific scenario has a beautifully elegant result: the objects simply exchange their linear velocities. It's like a cosmic baton pass. Sphere $A$, which was moving, transfers all its linear momentum to Sphere $B$ and comes to a complete halt ($v_A=0$). Sphere $B$, previously at rest, inherits Sphere $A$'s initial velocity and darts off ($v_B=v$).

But what happens to the spin? Does Sphere $A$ transfer its rotation to Sphere $B$ as well?

Analyzing the Impact

To solve the rotational mystery, we must zoom in on the exact moment of impact and analyze the forces at play. According to Newton's Second Law for rotation, an object's angular momentum (and thus its angular velocity) will only change if a net external torque acts upon it.

The problem gives us two critical keywords: "smooth sphere" and "frictionless plane". Because the spheres are perfectly smooth, they cannot grip each other. The only force they can exert on one another during the collision is the normal contact force ($N$).

By definition, the normal force between two smooth spheres acts perfectly perpendicular to their surfaces at the point of contact. Geometrically, this means the line of action of this normal force passes directly through the center of mass of both spheres.

Independent Spins

Torque ($\tau$) is calculated as the force multiplied by the perpendicular distance from the axis of rotation ($r_{\perp }$). Since the normal force passes exactly through the center of mass, the perpendicular distance is zero ($r_{\perp }=0$).

Consequently, the torque generated by the collision about the center of mass for both spheres is absolutely zero (${\tau }_{cm}=0$). With no torque to alter their rotational states, the angular momentum of each sphere is strictly conserved.

Sphere $A$ will continue to spin with its initial angular velocity $\omega$, even though it is now linearly stationary. Sphere $B$ will slide forward with velocity $v$, but it will not start spinning.

Therefore, the final angular velocities are ${\omega }_A=\omega$ and ${\omega }_B=0$.