Animated Solution for Physics - System of Particles: Two balls, having linear momenta ${\mathbf{p}}_1=p\hat{\mathbf{i}}$ and ${\mathbf{p}}_2=-p\hat{\mathbf{i}}$, undergo a collision in free space. There is no external force acting on the balls. Let ${\mathbf{p}}_1^{\prime }$ and ${\mathbf{p}}_2^{\prime }$ be their final momenta. The following option is (are) not allowed for any non-zero value of $p,a_1,a_2,b_1,b_2,c_1$ and $c_2$.

A Collision in Free Space

Imagine two celestial bodies, or perhaps just two simple billiard balls, gliding towards each other in the vast emptiness of free space. There is no gravity pulling them down, no friction slowing them down, and absolutely no external forces interfering with their cosmic dance.

The first ball carries a momentum of ${\mathbf{p}}_1=p\hat{\mathbf{i}}$, moving steadily along the positive x-axis. The second ball mirrors this exactly, carrying a momentum of ${\mathbf{p}}_2=-p\hat{\mathbf{i}}$, moving along the negative x-axis.

What happens when we look at the system as a whole? The total initial momentum is simply the vector sum of the two:

Conservation of Momentum

Because our system exists in free space, isolated from the rest of the universe, it obeys one of the most sacred laws of physics: the Conservation of Linear Momentum. This law dictates that if no net external force acts on a system, its total momentum must remain constant forever.

Therefore, no matter how chaotic or explosive the collision between these two balls might be, the total momentum after the collision must still be exactly zero. If we denote their final momenta as ${\mathbf{p}}_1^{\prime }$ and ${\mathbf{p}}_2^{\prime }$, we arrive at our master equation:

For a three-dimensional vector to be truly zero, it is not enough for just one part of it to vanish. A vector is composed of independent directions: the x-direction ($\hat{\mathbf{i}}$), the y-direction ($\hat{\mathbf{j}}$), and the z-direction ($\hat{\mathbf{k}}$).

If the total final momentum is zero, it means that the sum of the $\hat{\mathbf{i}}$ components must be zero, the sum of the $\hat{\mathbf{j}}$ components must be zero, and the sum of the $\hat{\mathbf{k}}$ components must be zero. They cannot compensate for each other. A movement in the x-direction cannot cancel out a movement in the y-direction.

With this strict mathematical filter, let's interrogate the given options to see which final states are physically allowed, keeping in mind the crucial constraint: all constants ($p,a_1,a_2,b_1,b_2,c_1,c_2$) are strictly non-zero.

Let's add the final momenta given in option (a):

Look closely at the $\hat{\mathbf{k}}$ component. It is simply $c_1$. For this entire vector to be zero, we are forced to conclude that $c_1=0$. However, the problem explicitly forbids any of these constants from being zero! Because this state requires a zero constant to exist, it violates our constraints. Thus, option (a) is not allowed.

Now, let's look at option (b):

For this to be zero, we need $c_1+c_2=0$, which means $c_1=-c_2$. Can two non-zero numbers add up to zero? Absolutely! If $c_1=5$ and $c_2=-5$, the condition is perfectly met without violating the non-zero rule. Therefore, this state is physically allowed.

Adding the vectors in option (c) yields:

The $\hat{\mathbf{k}}$ component beautifully cancels itself out immediately. We are left needing $a_1=-a_2$ and $b_1=-b_2$. Just like in option (b), we can easily find non-zero numbers that satisfy these equations. This state is perfectly allowed.

Finally, let's examine option (d):

Focus on the $\hat{\mathbf{j}}$ component: $2b_1$. For the total momentum to be zero, we must have $2b_1=0$, which inescapably means $b_1=0$. Once again, we hit a brick wall. The problem states $b_1$ must be non-zero. Because this state forces $b_1$ to be zero, it is physically impossible under the given constraints. Thus, option (d) is not allowed.

The question cleverly asks us to identify the options that are not allowed. Through our rigorous application of momentum conservation and vector independence, we discovered that options (a) and (d) force our non-zero constants to become zero. Therefore, they represent impossible physical realities and are the correct answers to this problem.