Consider an elastic collision between two particles $A$ and $B$ of same mass, moving in the same direction. Particle $A$ is moving at speed $v_A$ and particle $B$ is moving at speed $v_B$. In the figures shown, the solid lines represent the motion before the collision and the dotted lines represent the motion after the collision. Which of the following describes the motion of these two particles most accurately?
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For a 1D elastic collision of equal masses, always remember that velocities are swapped.
This means the slopes on an $x-t$ plot and values on a $v-t$ plot are exchanged after collision.
This shortcut helps you identify the correct graph instantly without calculating.
Step 1: Understanding the Question:
This question tests the conceptual understanding of a one-dimensional elastic collision between two identical masses, and how to represent this motion using position-time ($x-t$) and velocity-time ($v-t$) graphs. Step 2: Key Formulas and Approach:
1. When two particles of equal mass ($m_A = m_B$) undergo a perfectly elastic head-on collision, they completely exchange their velocities.
2. Let $v_A$ and $v_B$ be the initial velocities of particles $A$ and $B$, respectively, and $v'_A$ and $v'_B$ be their final velocities post-collision.
3. The exchange of velocities implies:
\[ v'_A = v_B \quad \text{and} \quad v'_B = v_A \]
4. In the $x-t$ graph, the slope of the line represents velocity:
\[ \text{Slope} = \frac{dx}{dt} = v \] Step 3: Detailed Explanation:
• Since particle $A$ collides with particle $B$ while both are moving in the same direction, particle $A$ (which is behind) must have a greater initial velocity than $B$ ($v_A > v_B > 0$).
• Before the collision, the position-time ($x-t$) graph shows a steeper solid line for $A$ (higher slope $v_A$) and a flatter solid line for $B$ (lower slope $v_B$).
• Since $A$ starts from $x=0$ at $t=0$ and $B$ starts from $x > 0$, the two lines intersect at the point of collision.
• After the elastic collision, the two particles exchange their velocities. Therefore, particle $B$ now moves with the higher velocity $v'_B = v_A$, and particle $A$ moves with the lower velocity $v'_A = v_B$.
• This means that after the collision (dotted lines), the slope of $B$'s graph becomes steep, while the slope of $A$'s graph becomes flat.
• Now, looking at the velocity-time ($v-t$) graphs, before the collision, the velocity of $A$ is a constant line at a higher value, and the velocity of $B$ is a constant line at a lower value.
• After the collision, the velocity of $B$ jumps up to the higher value, and the velocity of $A$ drops down to the lower value. This is represented by dotted lines in the graph.
• Analyzing the given options, only option (a) correctly depicts this exchange in both the $x-t$ and $v-t$ graphs. Step 4: Final Answer:
The correct representation of the elastic collision between the two identical masses is given in Option (A).
