Ballistic Pendulum: Inelastic Collision and Energy Conversion Steps:

• Initial momentum of pellet: $m_p v_p = (0.005 \, \text{kg})(200 \, \text{m/s}) = 1 \, \text{kg} \cdot \text{m/s}$ upward (clay at rest).
• Post-collision velocity $v$: Combined mass $0.1 \, \text{kg}$, so $0.1 v = 1$, $v = 10 \, \text{m/s}$ upward.
• Height rise from kinematics: $v_f^2 = v_i^2 - 2 g h$, , . Why B is correct:
• Applies conservation of momentum for inelastic embedding, then mechanical energy conservation (KE to PE) with $g = 10 \, \text{m/s}^2$. Why the others are wrong:
• A: Underestimates post-collision velocity, perhaps halving momentum contribution.
• C: Uses $h = v^2 / g$ (ignores factor of 2 in energy equation).
• D: Applies energy conservation to initial pellet KE without collision energy loss.

Final answer: B