A bullet of mass $10\text{ g}$ moving horizontally with a velocity of $400\text{ m/s}$ strikes a wooden block of mass $3.99\text{ kg}$ suspended by a long string and gets embedded in it. The vertical height to which the block rises is ($g = 10\text{ m/s}^2$):
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For any inelastic ballistic pendulum problem, you can combine momentum and energy conservation into one neat master shortcut formula for height:
$$\mathbf{h = \frac{1}{2g} \left(\frac{m_1 \cdot v_1}{m_1 + m_2}\right)^2}$$
Plugging in the standard variables saves valuable time during exams and prevents multi-step rounding errors!
Step 1: Use one combined shortcut for the height. The bullet sticks in the block, so momentum is shared first, then the joined mass swings up turning its motion into height. Putting both ideas together gives \[ h = \frac{1}{2g}\left(\frac{m_1 v_1}{m_1+m_2}\right)^2 \]
Step 2: Put numbers in standard units. Bullet mass $m_1 = 10\text{ g} = 0.01\text{ kg}$, bullet speed $v_1 = 400\text{ m/s}$, block mass $m_2 = 3.99\text{ kg}$, and $g=10\text{ m/s}^2$. The total mass is $0.01+3.99 = 4\text{ kg}$.
Step 3: Find the shared speed just after the hit. \[ V = \frac{m_1 v_1}{m_1+m_2} = \frac{0.01\times 400}{4} = \frac{4}{4} = 1\text{ m/s} \]
Step 4: Turn that speed into rise height. \[ h = \frac{V^2}{2g} = \frac{1^2}{2\times 10} = \frac{1}{20} = 0.05\text{ m} \] So the block rises 0.05 m, which is option (C). \[ \boxed{0.05\text{ m}} \]
