Question

Two gliders are moving on a horizontal frictionless air track. Glider 1 has mass \(m_{1}=160.1 \mathrm{~g}\) and is moving to the right (positive \(x\) -direction) with a speed of \(2.723 \mathrm{~m} / \mathrm{s}\). Glider 2 has mass \(m_{2}=354.1 \mathrm{~g}\) and is moving to the left (negative \(x\) -direction) with a speed of \(3.515 \mathrm{~m} / \mathrm{s}\). The gliders undergo a totally elastic collision. What is the velocity of glider 1 after the collision?

Short answer

Based on the given information and solution steps: Glider 1 mass: \(m_1=0.1601\,\text{kg}\) Glider 1 initial velocity: \(v_{1i}=2.723\,\text{m/s}\) Glider 2 mass: \(m_2=0.3541\,\text{kg}\) Glider 2 initial velocity: \(v_{2i}=-3.515\,\text{m/s}\) The calculated final velocity of glider 1 after an elastic collision is approximately \(v_{1f} \approx -0.656\,\text{m/s}\), indicating that it will be moving to the left (negative \(x\)-direction) after the collision.

Step-by-step solution

Understand the Problem and Given Values

We are given that glider 1 has a mass \(m_1 = 160.1\,\text{g}\) and an initial velocity \(v_{1i} = 2.723\,\text{m/s}\), and glider 2 has a mass \(m_2 = 354.1\,\text{g}\) and an initial velocity \(v_{2i} = -3.515\,\text{m/s}\). The negative sign indicates that glider 2 is moving to the left. The collision is elastic, and we need to find the final velocity of glider 1, \(v_{1f}\).

Convert Grams to Kilograms

First, we need to convert the given masses from grams to kilograms (the SI unit for mass) by dividing by 1000: $$m_1 = \frac{160.1}{1000}\,\text{kg} = 0.1601\,\text{kg}$$ $$m_2 = \frac{354.1}{1000}\,\text{kg} = 0.3541\,\text{kg}$$

Apply Conservation of Linear Momentum

Before the collision, the total momentum is given by the sum of the initial momenta of glider 1 and glider 2: $$p_{totali} = m_1v_{1i} + m_2v_{2i}$$ After the collision, the total momentum will be the sum of the final momenta of gliders 1 and 2: $$p_{totalf} = m_1v_{1f} + m_2v_{2f}$$ Since the momentum is conserved, the initial and final total momenta will be equal: $$p_{totali} = p_{totalf}$$ $$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$

Apply Conservation of Kinetic Energy

In an elastic collision, total kinetic energy is conserved. Before the collision, the total kinetic energy is given by the sum of the initial kinetic energies of gliders 1 and 2: $$KE_{totali} = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2$$ After the collision, the total kinetic energy will be the sum of the final kinetic energies of gliders 1 and 2: $$KE_{totalf} = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$$ Since kinetic energy is conserved, the initial and final total kinetic energies will be equal: $$KE_{totali} = KE_{totalf}$$ $$\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$$

Solve for the Final Velocity of Glider 1

We can now solve for the final velocity of glider 1, \(v_{1f}\), by solving the conservation of momentum equation for \(v_{1f}\): $$v_{1f} = \frac{m_1v_{1i} + m_2v_{2i} - m_2v_{2f}}{m_1}$$ Using the expression for \(v_{2f}\) from the conservation of kinetic energy equation and substituting into the conservation of momentum equation: $$v_{1f} = \frac{m_1v_{1i} + m_2v_{2i} - m_2\sqrt{v_{1i}^2 + v_{2i}^2 - v_{1f}^2}}{m_1}$$ We can solve this equation for \(v_{1f}\) using a numerical calculation: $$v_{1f} \approx -0.656\,\text{m/s}$$

Interpret the Result

The final velocity of glider 1 after the elastic collision is approximately \(-0.656\,\text{m/s}\), which means it is moving to the left (negative \(x\)-direction) after the collision.