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Step-by-Step Solution
Step 1: Identify known variables
• Mass of the first body, m_1 = 2\text{ kg}
• Initial speed of the first body, u_1 = 4\text{ m/s}
• Speed of the first body after collision, v_1 = \frac{1}{4} \times 4 = 1\text{ m/s}
• Mass of the second body, m_2 = ? (at rest initially)
• Final speed of the second body after collision, v_2 = ?
Step 2: Apply conservation of momentum
Because momentum is conserved in an elastic collision:
m_1 \, u_1 = m_1 \, v_1 + m_2 \, v_2.
Plugging in known values:
2 \times 4 = 2 \times 1 + m_2 \, v_2.
8 = 2 + m_2 \, v_2.
m_2 \, v_2 = 6. \quad (1)
Step 3: Use the coefficient of restitution (for an elastic collision, e = 1)
The coefficient of restitution e is given by:
e = \frac{v_2 - v_1}{u_1 - 0}.
Since u_1 = 4\text{ m/s} (initial velocity of first body) and v_1=1\text{ m/s} (final velocity of first body) while the second body was initially at rest:
1 = \frac{v_2 - 1}{4}.
Solving for v_2 :
v_2 - 1 = 4 \quad \Rightarrow \quad v_2 = 5 \text{ m/s}.
Step 4: Determine the mass of the second body
Substitute v_2 = 5\text{ m/s} into equation (1):
m_2 \times 5 = 6
m_2 = \frac{6}{5} = 1.2\text{ kg}.
Step 5: Compute the velocity of the center of mass
The velocity of the center of mass, v_{cm} , of the two-body system is:
v_{cm} = \frac{m_1 \, v_1 + m_2 \, v_2}{m_1 + m_2}.
Substitute the known values:
v_{cm} = \frac{(2 \times 1) + (1.2 \times 5)}{2 + 1.2}
= \frac{2 + 6}{3.2}
= \frac{8}{3.2}
= \frac{25}{10} \text{ m/s}.
Hence,
v_{cm} = \frac{x}{10} \text{ m/s} = \frac{25}{10} \text{ m/s}.
Therefore, x = 25.
Final Answer
The value of x is 25.
