1. **State the problem:** A bullet of mass 10g (0.01 kg) is fired at 100 m/s into a wooden block of mass 190g (0.19 kg) suspended by a string. The bullet embeds in the block, and we want to find how high the block rises after the collision. 2. **Relevant formulas and principles:** - Conservation of momentum for the collision: $$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f$$ where $m_1$ and $v_1$ are the bullet's mass and velocity, $m_2$ and $v_2$ are the block's mass and velocity (initially zero), and $v_f$ is the final velocity of the combined block and bullet immediately after collision. - Conservation of mechanical energy after collision (block + bullet rise): $$\frac{1}{2} (m_1 + m_2) v_f^2 = (m_1 + m_2) g h$$ where $g = 9.8$ m/s$^2$ is acceleration due to gravity and $h$ is the height the block rises. 3. **Calculate final velocity $v_f$ after collision using momentum conservation:** $$0.01 \times 100 + 0.19 \times 0 = (0.01 + 0.19) v_f$$ $$1 = 0.2 v_f$$ $$v_f = \frac{1}{0.2}$$ $$v_f = 5 \text{ m/s}$$ 4. **Calculate height $h$ using energy conservation:** $$\frac{1}{2} \times 0.2 \times 5^2 = 0.2 \times 9.8 \times h$$ $$\frac{1}{2} \times 0.2 \times 25 = 0.2 \times 9.8 \times h$$ $$2.5 = 1.96 h$$ $$h = \frac{2.5}{1.96}$$ $$h \approx 1.28 \text{ meters}$$ **Final answer:** The block will rise approximately 1.28 meters after the bullet embeds in it.