91Ó°ÊÓ

We are given: - Mass of glider 1 and glider 2: \(m_1 = m_2 = 0.10 \ \mathrm{kg}\) - The initial velocity of glider 1: \(v_{1i} = 0.20 \ \mathrm{m/s}\) - The initial velocity of glider 2: \(v_{2i} = 0 \ \mathrm{m/s}\) - The collision is elastic. We need to find the final velocities, \(v_{1f}\) and \(v_{2f}\), of glider 1 and 2 respectively.

In an elastic collision, the total momentum is conserved. So, the momentum before collision equals the momentum after collision. Momentum conservation equation: $$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$ $$0.10(0.20)+0.10(0)=0.10v_{1f}+0.10v_{2f}$$ Simplifying the equation, we get: $$0.02=0.10(v_{1f}+v_{2f}) \Rightarrow 0.2 = v_{1f} + v_{2f} \ \ \ \ (1)$$

Elastic collision implies that the total kinetic energy is conserved. So, the kinetic energy before collision equals the kinetic energy after collision. Kinetic energy conservation equation: $$\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2$$ $$0.5(0.10)(0.20^2)+0.5(0.10)(0)=0.5(0.10)v_{1f}^2+0.5(0.10)v_{2f}^2$$ Simplifying the equation, we get: $$0.1(0.02)=v_{1f}^2+v_{2f}^2 \Rightarrow 0.2(1-v_{1f})=v_{1f}^2+v_{2f}^2 \ \ \ \ (2)$$ Here, we used equation (1) to express \(v_{2f}\) in terms of \(v_{1f}\).

We have the two equations (1) and (2): $$0.2 = v_{1f} + v_{2f}$$ $$0.2(1-v_{1f})=v_{1f}^2+v_{2f}^2$$ Now, square equation (1) and equate to equation (2): $$(0.2)^2 = (v_{1f} + v_{2f})^2$$ $$0.2(1-v_{1f})= v_{1f}^2+v_{2f}^2$$ Substituting \(v_{2f}=0.2-v_{1f}\) in the equation, we have: $$0.2(1-v_{1f})=v_{1f}^2+(0.2-v_{1f})^2$$ Solve for \(v_{1f}\): $$0.2-0.2v_{1f}=v_{1f}^2+0.04-0.4v_{1f}+v_{1f}^2$$ $$0.4v_{1f}-0.2=v_{1f}^2$$ $$0.2 = 0.4v_{1f}-v_{1f}^2$$ $$1 = 2v_{1f}-v_{1f}^2$$ Solving for \(v_{1f}\), we get: \(v_{1f} = 2(1-v_{1f})\) \(v_{1f} = 1\) Now, use equation (1) to find \(v_{2f}\): $$0.2 = 1 + v_{2f}$$ $$v_{2f} = -0.8$$ So, after the collision, glider 1 has a final velocity of \(1 \ \mathrm{m/s}\) and glider 2 has a final velocity of \(-0.8 \ \mathrm{m/s}\).