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Step-by-Step Solution
Step 1: Understand the Problem
An object of mass $m_1$ collides with a stationary object of mass $m_2$. After the collision, they move with the same speed but in opposite directions. We want to find the ratio $m_2 : m_1$.
Step 2: Use Conservation of Momentum
Let the initial speed of $m_1$ before collision be $v_1$. Since $m_2$ is at rest initially, its speed is $0$. After collision, each object moves with speed $v$ in opposite directions. Taking the direction of $m_1$'s initial velocity as positive, we can write momentum conservation as:
$m_1 v_1 = m_2 \, ( + v ) \;-\; m_1 \, ( v )$
The term $m_2 (\!+v)$ is the momentum of the second mass (moving in the original positive direction), and $- m_1 v$ accounts for the first mass moving in the opposite direction.
Step 3: Apply the Coefficient of Restitution (Elastic Collision)
Since the collision is perfectly elastic, the coefficient of restitution $e = 1$. By definition:
$e = \dfrac{\text{relative speed of separation}}{\text{relative speed of approach}} = 1$
The relative speed of separation after the collision is $(v - (-v)) = 2v$, and the relative speed of approach before the collision is $(0 - v_1) = -v_1$. So,
$\dfrac{2v}{-v_1} = 1 \quad\Longrightarrow\quad v_1 = 2v$
Step 4: Substitute Back into Momentum Conservation
Substitute $v_1 = 2v$ into the momentum conservation equation:
$m_1 (2v) = m_2 \, v \;-\; m_1 \, v$
Simplify:
$2m_1 = m_2 - m_1 \quad\Longrightarrow\quad 2m_1 + m_1 = m_2 \quad\Longrightarrow\quad 3m_1 = m_2
Step 5: Conclude the Mass Ratio
From $3m_1 = m_2$, we conclude:
$\dfrac{m_2}{m_1} = 3 : 1$
Hence, the ratio of $m_2 : m_1$ is 3 : 1.
