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IV. Numerical Problems · Q5

Q.A bullet of mass 20 g strikes a pendulum of mass 5 kg. The centre of mass of the pendulum rises through a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed. Take g=10 m s−2g = 10\ \mathrm{m\,s^{-2}}.

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Step 1. Given: m1=20 g=0.02 kgm_1=20\ \mathrm{g}=0.02\ \mathrm{kg} (bullet), m2=5 kgm_2=5\ \mathrm{kg} (pendulum), rise h=10 cm=0.10 mh=10\ \mathrm{cm}=0.10\ \mathrm{m}, g=10 m s−2g=10\ \mathrm{m\,s^{-2}}.

Step 2. After the bullet embeds, the combined mass (m1+m2)(m_1+m_2) moves off with a common velocity vv, then rises height hh under gravity, converting all its kinetic energy to potential energy (energy conservation, since only gravity acts once the bullet is embedded): 12(m1+m2)v2=(m1+m2)gh⇒v=2gh\tfrac12(m_1+m_2)v^2=(m_1+m_2)gh \Rightarrow v=\sqrt{2gh}.

Step 3. v=2×10×0.10=2≈1.4142 m s−1v=\sqrt{2\times10\times0.10}=\sqrt{2}\approx1.4142\ \mathrm{m\,s^{-1}}. …

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