Question

Puck 1 sliding along the \(x\) -axis strikes stationary puck 2 of the same mass. After the elastic collision, puck 1 moves off at speed \(v_{1 f}\) in the direction \(60.0^{\circ}\) above the \(x\)-axis; puck 2 moves off at speed $v_{2 f}\( in the direction \)30.0^{\circ}\( below the \)x\(-axis. Find \)v_{2 \mathrm{f}}\( in terms of \)v_{1 f}$

Short answer

Answer: The final speed of Puck 2 is \(v_{2f} = \frac{2\sqrt{3}}{3} v_{1i}\), where \(v_{1i}\) is the initial speed of Puck 1.

Step-by-step solution

Write the conservation of momentum equations for x and y axes separately

Since this is an elastic collision, we will need to use the conservation of momentum formula which states that: Total initial momentum = Total final momentum For the x-axis: \(m_1 \cdot v_{1i} = m_1 \cdot v_{1f} \cdot \cos(60^{\circ}) + m_2 \cdot v_{2f} \cdot \cos(30^{\circ})\) For the y-axis: \(0 = m_1 \cdot v_{1f} \cdot \sin(60^{\circ}) - m_2 \cdot v_{2f} \cdot \sin(30^{\circ})\)

Simplify the equations

Now let's simplify the above equations using trigonometric values and the fact that the masses are the same: For the x-axis: \(v_{1i} = \frac{1}{2} v_{1f} + \frac{\sqrt{3}}{2} v_{2f}\) For the y-axis: \(0 = \frac{\sqrt{3}}{2} v_{1f} - \frac{1}{2} v_{2f}\)

Solve the system of equations for the required value

We need to find \(v_{2f}\) in terms of \(v_{1f}\). Let's use the y-axis equation to solve for \(v_{2f}\): \(v_{2f} = \sqrt{3}v_{1f}\) Now, plug this value of \(v_{2f}\) into the x-axis equation: \(v_{1i} = \frac{1}{2} v_{1f} + \frac{\sqrt{3}}{2} (\sqrt{3}v_{1f})\)

Simplify and find the value of \(v_{2f}\) in terms of \(v_{1f}\)

Simplify the above equation: \(v_{1i} = v_{1f}(\frac{1}{2} + \frac{3}{2})\) \(v_{1f} = \frac{2}{3} v_{1i}\) Now replace \(v_{1f}\) with the expression we found in terms of \(v_{1i}\) into the expression we found for \(v_{2f}\): \(v_{2f} = \sqrt{3}v_{1f}\) \(v_{2f} = \sqrt{3}(\frac{2}{3} v_{1i})\) \(v_{2f} = \frac{2\sqrt{3}}{3} v_{1i}\)