Ballistic Pendulum And Energy Conservation

This problem combines two distinct stages: a perfectly inelastic collision followed by energy conservation during the swing. During the rapid embedding, momentum is conserved, so $m_{b}u=(m_b+M)v$, giving $v=(0.01\times 200)/(2.00)=2/2=1\text{ m/s}$ for the combined mass. Kinetic energy is not conserved in this stage. After impact, mechanical energy is conserved as the system rises, so $\frac{1}{2}(m_b+M)v^2=(m_b+M)gh$, yielding $h=v^2/(2g)=1/(2\times 10)=0.05\text{ m}$. The option 0.5 m wrongly applies energy conservation directly to the bullet's original speed, skipping the collision. The option 0.1 m forgets the factor of two in the height relation. The option 0.025 m halves the correct height through a similar slip. The conceptual trap is to apply a single conservation law across both stages; momentum carries the system through the violent embedding, whereas energy conservation governs only the smooth upward swing that follows. Separating the two stages cleanly is the entire skill the problem is built to test, and it is precisely what historically allowed such pendulums to measure otherwise unmeasurable projectile speeds. This is the NCERT ballistic-pendulum treatment. As a check, the post-collision speed of 1 m/s is far below the bullet's 200 m/s, reflecting the large energy loss during embedding, exactly as expected.