Two masses $m_a$ and $m_b$ moving with velocities $v_a$ and $v_b$ in opposite directions collide elastically. After the collision $m_a$ and $m_b$ move with velocities $v_b$ and $v_a$ respectively, then the ratio $m_a : m_b$ is
Show Hint
Velocity exchange is a classic hallmark of perfectly elastic collisions between two objects of equal mass. If you read "velocities are exchanged", you can immediately conclude $m_1 = m_2$.
Step 1: Understand the collision.
Two masses moving in opposite directions collide elastically. Afterward they swap speeds: $m_a$ ends with $v_b$ and $m_b$ ends with $v_a$. We want $m_a:m_b$.
Step 2: Recall a known elastic result.
In a head-on elastic collision, two bodies completely exchange their velocities only when their masses are equal. We can confirm this with momentum.
Step 3: Write conservation of momentum.
Taking one direction as positive: \[ m_a v_a+m_b v_b=m_a v_{a,f}+m_b v_{b,f}. \]
Step 4: Put in the final velocities.
With $v_{a,f}=v_b$ and $v_{b,f}=v_a$: \[ m_a v_a+m_b v_b=m_a v_b+m_b v_a. \]
Step 5: Group the masses.
\[ m_a v_a-m_a v_b=m_b v_a-m_b v_b \] \[ m_a(v_a-v_b)=m_b(v_a-v_b). \]
Step 6: Cancel the common bracket.
Since the two had different speeds, $v_a\neq v_b$, so we can cancel $(v_a-v_b)$: \[ m_a=m_b. \]
Step 7: State the ratio.
\[ \frac{m_a}{m_b}=1, \] which is option (3).
\[ \boxed{1} \]
