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Step-by-Step Solution
Step 1: Apply Conservation of Linear Momentum
Initially, the ball of mass m moves with velocity v , while the pendulum bob of the same mass m is at rest. After a perfectly inelastic collision, the two masses stick together and move with a common velocity v' .
By conservation of linear momentum:
m \cdot v + m \cdot 0 = (m + m) \cdot v' \\
\implies m\,v = 2m\,v' \\
\implies v' = \frac{v}{2}.
Step 2: Convert Kinetic Energy to Potential Energy
Right after the collision, the combined mass 2m moves with velocity v' = \frac{v}{2} . The kinetic energy of this combined mass is:
\text{K.E.} = \frac{1}{2} \times 2m \times \left(\frac{v}{2}\right)^2
= m \frac{v^2}{4}.
As the pendulum rises to its maximum height h , this kinetic energy is entirely converted into the potential energy of the mass 2m at height h :
(2m)g\,h = m \frac{v^2}{4}.
Solving for h :
2m\,g\,h = m \frac{v^2}{4}
\quad \Longrightarrow \quad
2g\,h = \frac{v^2}{4}
\quad \Longrightarrow \quad
h = \frac{v^2}{8g}.
Step 3: Relate to the Given Form h = \frac{v^2}{x\,g}
From the problem statement, h is expressed as \frac{v^2}{x\,g} . We have found h to be \frac{v^2}{8g} . Matching these expressions gives:
\frac{v^2}{x\,g} = \frac{v^2}{8g} \quad \Longrightarrow \quad x = 8.
