Physics - Center of Mass Question with Solution | TestHub

This is a case of inelastic collision where momentum is conserved. Let $m_b$ be the mass of the bullet and $v_b$ be its initial speed. Let $m_w$ be the mass of the wooden block and $v_w$ be its initial speed. Given:$m_b=20\text{ g}=0.02\text{ kg}$ $m_w=980\text{ g}=0.98\text{ kg}$ $v_w=0\text{ m/s}$ (initially at rest) Final speed of combined system $V=10\text{ m/s}$ By conservation of momentum:$$m_b{v}_b+m_w{v}_w=(m_b+m_w)V$$ $$0.02\times v_b+0.98\times 0=(0.02+0.98)\times 10$$ $$0.02v_b=1.00\times 10$$ $$0.02v_b=10$$ $$v_b=\frac{10}{0.02}=\frac{1000}{2}=500\text{ m/s}$$ The initial speed of the bullet was 500 m/s.