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Step-by-Step Solution
Step 1: Understand the Problem
A body of mass 2 kg is moving with an initial speed $v_0$. It collides elastically with a second body (initially at rest). After the collision, the first body continues moving in the same direction but with speed $v_0/4$. We need to find the mass of the second body.
Step 2: List the Known Variables
• Mass of the first body, $m_1 = 2\text{ kg}$
• Initial velocity of the first body, $u_1 = v_0$
• Initial velocity of the second body, $u_2 = 0$
• Final velocity of the first body, $v_1 = \frac{v_0}{4}$
• Mass of the second body, $m_2 = m$ (unknown)
• Final velocity of the second body, $v_2 = v$ (unknown)
Step 3: Apply Conservation of Linear Momentum
Momentum before collision = Momentum after collision:
$$
m_1 \, u_1 + m_2 \, u_2 = m_1 \, v_1 + m_2 \, v_2
$$
Since $u_2 = 0$, substitute the values:
$$
2\,v_0 = 2 \Bigl(\frac{v_0}{4}\Bigr) + m\,v
$$
Simplify:
$$
2\,v_0 = \frac{v_0}{2} + m\,v
\quad\Longrightarrow\quad
2\,v_0 - \frac{v_0}{2} = m\,v
\quad\Longrightarrow\quad
\frac{3\,v_0}{2} = m\,v
\quad (1)
$$
Step 4: Use the Condition for a Perfectly Elastic Collision
In an elastic collision, the relative speed of separation equals the relative speed of approach:
$$
(v_2 - v_1) = (u_1 - u_2)
$$
Substitute $v_2 = v$, $v_1 = \tfrac{v_0}{4}$, $u_1 = v_0$, $u_2 = 0$:
$$
v - \frac{v_0}{4} = v_0
\quad\Longrightarrow\quad
v = \frac{5\,v_0}{4}
\quad (2)
$$
Step 5: Solve for the Unknown Mass
Substitute $v = \tfrac{5\,v_0}{4}$ from equation (2) into equation (1):
$$
\frac{3\,v_0}{2} = m \left(\frac{5\,v_0}{4}\right)
$$
Cancel $v_0$ from both sides (assuming $v_0 \neq 0$):
$$
\frac{3}{2} = m \left(\frac{5}{4}\right)
\quad\Longrightarrow\quad
m = \frac{3}{2} \times \frac{4}{5} = \frac{6}{5} = 1.2
$$
Final Answer
The mass of the second body is $1.2\text{ kg}$.
