Question

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

• A. $\frac{v_a + v_b}{v_a - v_b}$
• B. $\frac{1}{2}$
• C. $1$
• D. $\frac{v_a - v_b}{v_a + v_b}$

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 $m_1 = m_2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a 1D elastic collision between two masses moving in opposite directions. After the collision, they completely exchange their velocities. We need to find the ratio of their masses.

Step 2: Key Formula or Approach:
In a perfectly elastic one-dimensional collision, two bodies will completely exchange their velocities if and only if they possess identical masses. We can also prove this using the conservation of momentum.

Step 3: Detailed Explanation:
Let's use the principle of conservation of linear momentum:
$$m_a v_{a,i} + m_b v_{b,i} = m_a v_{a,f} + m_b v_{b,f}$$
Substitute the final velocities $v_{a,f} = v_b$ and $v_{b,f} = v_a$:
$$m_a v_a + m_b v_b = m_a v_b + m_b v_a$$
Rearrange to group the respective masses together on each side:
$$m_a v_a - m_a v_b = m_b v_a - m_b v_b$$
$$m_a(v_a - v_b) = m_b(v_a - v_b)$$
Assuming they had different initial velocities ($v_a \neq v_b$), we can cancel $(v_a - v_b)$ from both sides:
$$m_a = m_b$$
Therefore, the ratio of their masses is:
$$\frac{m_a}{m_b} = 1$$

Step 4: Final Answer:
The ratio $m_a : m_b$ is $1$, matching option (C).