Animated Solution for Physics - System of Particles: A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one-fourth of its original speed. What is the mass of the second body?

Imagine a smooth, frictionless track. A small body of mass $m=2\text{ kg}$ is cruising along with an initial velocity $v$. Up ahead, a second body of unknown mass $M$ is sitting perfectly still, minding its own business.

Suddenly, BAM! A head-on collision occurs.

But this isn't just any collision; the problem explicitly states it's an elastic collision. After the impact, the first body doesn't bounce back. Instead, it continues moving in its original direction, but its speed has been drastically reduced to exactly one-fourth of its initial speed, $v/4$. The second body, having absorbed a massive transfer of momentum, is now moving forward with a new velocity, let's call it $v^{\prime }$. Our mission is to uncover the hidden mass $M$.

In the absence of any external horizontal forces, the universe demands that the total linear momentum of our two-body system remains absolutely conserved.

Let's translate our physical situation into this mathematical law. Before the collision, all the momentum is carried by the first body. After the collision, the momentum is shared between the two bodies.

This equation is a great start, but we have a problem. We have one equation but two unknowns: the mass $M$ and the final velocity $v^{\prime }$. We need another piece of the puzzle.

This is where the word "elastic" saves the day. In a perfectly elastic collision, not only is momentum conserved, but kinetic energy is conserved as well. However, working with kinetic energy involves squaring velocities, which can get algebraically messy.

Instead, we use a brilliant shortcut derived from energy conservation: the Coefficient of Restitution ($e$). For a perfectly elastic collision, $e=1$. This means the relative velocity at which the bodies separate after the collision is exactly equal to the relative velocity at which they approached each other before the collision.

Let's rearrange this to solve for $v^{\prime }$.

We've just discovered that the second body shoots off with a speed that is $1.25$ times the initial speed of the first body!

Now, we take this newfound knowledge of $v^{\prime }$ and plug it back into our momentum conservation equation.

Look closely at this equation. The initial velocity $v$ is present in every single term. This is a profound physical insight: the mass ratio required to achieve this specific outcome is completely independent of how fast the first body was initially moving! We can divide the entire equation by $v$.

Let's group the $m$ terms together. Subtracting $m/4$ from $m$ gives us $3m/4$.

The denominators cancel out beautifully, leaving us with a simple, elegant relationship:

Finally, we solve for our unknown mass $M$.

Since we know the initial mass $m$ is $2\text{ kg}$, we just plug it in.

The mass of the second body is exactly $1.2\text{ kg}$. The physics checks out perfectly!