Animated Solution for Physics - System of Particles: A particle of mass $m$ with an initial velocity $u\hat{i}$ collides perfectly elastically with a mass $3m$ at rest. It moves with a velocity $v\hat{j}$ after collision, then $v$ is given by

The Cosmic Billiards

Imagine a game of cosmic billiards. A small particle of mass $m$ is cruising along the x-axis with a velocity $u\hat{i}$. Waiting patiently at the origin is a heavier particle, three times as massive ($3m$), completely at rest.

Suddenly, they collide! But this isn't just any collision; it's a perfectly elastic one. The lighter particle bounces off at a perfect right angle, shooting up the y-axis with a new velocity $v\hat{j}$. To maintain the delicate balance of the universe, the heavier particle must recoil in some other direction. Let's call its recoil velocity ${\mathbf{v}}_2$. Our mission is to find the exact relationship between the initial speed $u$ and the final speed $v$.

Momentum

In the absence of external forces, the universe demands that total momentum remains constant. This is the Principle of Conservation of Linear Momentum.

Before the collision, all the momentum is carried by the small particle: ${\mathbf{p}}_i=mu\hat{i}$. After the collision, the momentum is shared between the two particles: ${\mathbf{p}}_f=mv\hat{j}+3m{\mathbf{v}}_2$. Equating the two, we get:

We can elegantly isolate the velocity of the heavier mass by dividing out the common mass $m$ and rearranging the terms:

To use this in our energy calculations, we need the square of its speed. The magnitude squared of any vector is simply the sum of the squares of its orthogonal components. Thus:

Perfect Elasticity

The phrase "perfectly elastic" is a golden key in physics. It tells us that not a single joule of kinetic energy was lost to heat or sound during the impact. The total kinetic energy before the collision must exactly equal the total kinetic energy after.

Let's write down the energy balance:

We can immediately simplify this by multiplying the entire equation by $\frac{2}{m}$, stripping away the constants to reveal the raw relationship between the speeds:

Bringing It Together

Now, we merge our momentum and energy equations. We substitute the expression for $|{\mathbf{v}}_2{|}^2$ that we found earlier into our simplified energy equation:

The $3$ and the $9$ gracefully cancel, leaving a $3$ in the denominator. To clear the fraction, we multiply everything by $3$:

Grouping the like terms together, we bring the $u$ terms to the left and the $v$ terms to the right:

Dividing by $4$, we find $v^2=\frac{1}{2}{u}^2$. Taking the square root of both sides yields our final, elegant answer:

Through the beautiful interplay of momentum vectors and scalar energy, we've precisely determined the outcome of this 2D collision!