• # question_answer A bullet of mass m moving vertically upwards instantaneously with a velocity 'u' hits the hanging block of mass 'm' and gets embedded in it, as shown in the figure. The height through which the block rises after the collision, (assume sufficient space above block) is:   A) ${{\text{u}}^{2}}/2g$ B) ${{\text{u}}^{2}}/g$ C) ${{\text{u}}^{2}}/8g$ D) ${{\text{u}}^{2}}/4g$

 [c] From momentum conservation $mu=2mv$ $\Rightarrow \,v=\frac{u}{2}$ from energy conservation $\frac{1}{2}\times 2m\times {{\left( \frac{u}{2} \right)}^{2}}=2mgh\,\,\,\,\,\,\,\,\,\,\,\Rightarrow h=\frac{{{u}^{2}}}{8g}$