Question

Two pendulum bobs have equal masses and lengths \((5.1 \mathrm{m}) .\) Bob \(\mathrm{A}\) is initially held horizontally while bob \(\mathrm{B}\) hangs vertically at rest. Bob A is released and collides elastically with bob B. How fast is bob B moving immediately after the collision?

Short answer

Answer: The speed of Bob B immediately after the elastic collision is given by the formula: \(v_{B2} = \sqrt{2(PE/m)}\) where \(PE\) is the potential energy of Bob A initially and \(m\) is the mass of the bobs.

Step-by-step solution

Determine the initial potential energy of Bob A

Before the collision, Bob A has an initial potential energy at its horizontal position. We will use the gravitational potential energy formula to find this value: \(PE = mgh\) where \(m\) is the mass of Bob A (which is equal to the mass of Bob B), \(g\) is the acceleration due to gravity (approximately \(9.81 \, m/s^2\)), and \(h\) is the difference in height between the horizontal position of Bob A and its resting position. We can find the value of \(h\) using the Pythagorean theorem: \(h = \sqrt{L^2 - (L/2)^2} = \sqrt{5.1^2 - (5.1/2)^2}\) Note that since the two bobs have equal masses and lengths, they will have the same potential energy when they reach the same height.

Calculate the initial kinetic energy of Bob A

As Bob A is initially held horizontally, it has zero initial kinetic energy. As it descends, its potential energy converts into kinetic energy. We will use the conservation of energy principle to find the kinetic energy of Bob A just before the collision. Since the collision is elastic, the mechanical energy of the system will be conserved: \(KE_A = PE - KE_B\) Since Bob B is initially at rest, its kinetic energy before the collision is \(0\). So, \(KE_A = PE\)

Calculate the initial velocities of the bobs before the collision

Now that we have the initial potential energy and kinetic energy for Bob A just before the collision, we can find its initial velocity. We will use the following formula to find the velocity of Bob A: \(KE_A = \frac{1}{2}mv_A^2\) Solving for \(v_A\), we get: \(v_A = \sqrt{2(PE/m)}\) Bob B is initially at rest, so its initial velocity is \(0\).

Apply the conservation of linear momentum to find the final velocity of Bob B

Now we will apply the conservation of linear momentum principle to find the final velocity of Bob B: \(m_Av_{A1} + m_Bv_{B1} = m_Av_{A2} + m_Bv_{B2}\) Since Bob B is initially at rest, \(v_{B1} = 0\). We also know that the masses are equal, so we can simplify the equation: \(m(v_{A1} - v_{A2}) = mv_{B2}\) Since \(v_{A2} = v_{A1}\) (due to the collision being elastic), we have: \(v_{B2} = v_{A1}\) Now, we have all we need to find the final velocity of Bob B immediately after the collision: \(v_{B2} = \sqrt{2(PE/m)}\)