Animated Solution for Physics - System of Particles: A point mass of $1\text{ kg}$ collides elastically with a stationary point mass of $5\text{ kg}$. After their collision, the $1\text{ kg}$ mass reverses its direction and moves with a speed of $2{\text{ ms}}^{-1}$. Which of the following statement(s) is/are correct for the system of these two masses?

Imagine you are observing a classic physics experiment on a frictionless track. A small, nimble $1\text{ kg}$ mass is hurtling towards a much larger, stationary $5\text{ kg}$ mass. They collide, and because the collision is perfectly elastic, no kinetic energy is lost to heat or sound.

Instead, the lighter mass bounces off the heavier one, reversing its direction completely and retreating with a speed of $2\text{ m/s}$. Meanwhile, the heavier mass, having absorbed a significant impulse, begins to slide forward. Our mission is to dissect this event, uncovering the hidden initial velocity and analyzing the momentum and energy of the system.

The key to unraveling this entire problem lies in finding the initial velocity of the $1\text{ kg}$ mass, let's call it $v_1$. For a perfectly elastic, one-dimensional collision where the target mass is initially at rest, physics provides us with a powerful, ready-to-use formula for the final velocity of the projectile:

Here is where many students fall into a classic trap: sign convention. Because the $1\text{ kg}$ mass reverses its direction, its final velocity $v_1^{\prime }$ is not just $2\text{ m/s}$; it is strictly $-2\text{ m/s}$. Substituting our known masses ($m_1=1\text{ kg}$, $m_2=5\text{ kg}$), we get:

Simplifying the fraction gives us $-2=-\frac{4}{6}{v}_1$, which reduces to $-2=-\frac{2}{3}{v}_1$. The negative signs elegantly cancel out, revealing that the initial velocity $v_1$ was exactly $3\text{ m/s}$.

With the initial velocity in hand, the rest of the problem unfolds beautifully. Let's evaluate the options one by one. Option (a) asks for the total momentum of the system. One of the most sacred laws of physics is the conservation of linear momentum. The total momentum after the collision is exactly the same as the total momentum before the collision.

Substituting our values, we get $P_{\text{system}}=(1)(3)+(5)(0)=3\text{ kg-m/s}$. This confirms that statement (a) is absolutely correct.

What about option (b), the final momentum of the $5\text{ kg}$ mass? To find this, we first need its final velocity, $v_2^{\prime }$. The formula for the target mass is:

Plugging in the numbers, $v_2^{\prime }=\left(\frac{2(1)}{1+5}\right)(3)=\left(\frac{2}{6}\right)(3)=1\text{ m/s}$. Therefore, its final momentum is $P_5^{\prime }=m_2{v}_2^{\prime }=(5)(1)=5\text{ kg-m/s}$. Option (b) claims it is $4\text{ kg-m/s}$, so it is incorrect.

Now, let's delve into the energy of the system. Option (c) asks for the kinetic energy of the center of mass ($K_{CM}$). While you could calculate the velocity of the center of mass and then find its kinetic energy, there is a much more elegant shortcut. The kinetic energy of the center of mass is directly related to the total momentum of the system:

We already know $P_{\text{system}}=3\text{ kg-m/s}$ and the total mass $M_{\text{total}}=1+5=6\text{ kg}$. Substituting these yields:

This perfectly matches option (c), making it a correct statement.

Finally, option (d) asks for the total kinetic energy of the system. Because the collision is elastic, the total kinetic energy is conserved. We can simply calculate the initial kinetic energy:

Option (d) claims the total kinetic energy is $4\text{ J}$, which is incorrect. Thus, our journey concludes with the realization that only statements (a) and (c) hold true.