Import Question JSON

$\text{Hint: Apply conservation of momentum.}$ $\text{Step: Find the value of the } n. \text{ The collision of the particle is shown in the figure below;} \text{By conserving the momentum of the body along } y \text{ direction we get;} m_1u_1 \sin \theta_1 = m_2u_2 \sin \theta_2 \text{ i.e., } mu_1 \sin \theta_1 = 10mu_2 \sin \theta_2 \Rightarrow u_1 \sin \theta_1 = 10u_2 \sin \theta_2 \quad \cdots (1) \text{According to the question;} k_{f_{m_1}} = \frac{1}{2} k_{i_{m_1}} \Rightarrow \frac{1}{2} mu_1^2 = \frac{1}{2} \times \frac{1}{2} mu^2 \Rightarrow u_1 = \frac{u}{\sqrt{2}} \quad \cdots (2) \text{As the collision is elastic in nature i.e., } k_i = k_f \frac{1}{2} mu^2 = \frac{1}{2} mu_1^2 + \frac{1}{2} \cdot 10m \cdot u_2^2 \Rightarrow \frac{1}{2} mu^2 = \frac{1}{2} mu_1^2 + \frac{1}{2} \cdot 10m \cdot u_2^2 \Rightarrow \frac{1}{4} mu^2 = 10 \times mu_2^2 \Rightarrow u_2 = \frac{u}{\sqrt{20}} \quad \cdots (3) \text{By putting values of } u_1 \text{ and } u_2 \text{ in equation } (1) \text{ we get;} \frac{u}{\sqrt{2}} \sin \theta_1 = 10 \cdot \frac{u}{\sqrt{20}} \sin \theta_2 \Rightarrow \sin \theta_1 = \sqrt{10} \sin \theta_2 \Rightarrow n = 10 \text{Hence, option } (2) \text{ is the correct answer.}$