Problem 8-34 Elastic collision - Part 4 - B
A superball of mass \(22\; g\) rolls with a speed of \(3.5\; m/s\) toward another (stationary) superball of mass \(27\; g\). If the balls have a head-on elastic collision, what are the velocities magnitudes and directions) of the balls after the collision?
[Ans. \(3.1\; m/s\) forward and \(0.4\; m/s\) backward]
Accumulated Solution
\(m_1\; v_1 = m_1\; v{_1}'+ m_2\; v{_2}' \\ 3.5 = v{_1}' + 1.227\; v{_2}' \; \text{Eqn. #1}\\ m_1\; v{_1}{^2} = m_1\; v{_1}'{^2} + m_2\; v{_2}'{^2}\)
Correct!
In fact both expressions are correct. In (A) the superfluous factor \(½\) has been canceled.
Let's put in the numbers we know:
\((22)(3.5)2 = 22\; v{_1}'{^2} + 27\; v{_2}'{^2} \\ 3.52 = v{_1}'v^2 + 1.227\; v{_2}'{^2} \; \text{ Eqn. #2}\)
\(\text{Eqns. #1}\) and 2 are 2 equations in 2 unknowns. There are several ways to go about the solution.
Let's eliminate \(v{_1}'{^2}\) by solving \(\text{ Eqn.#1}\) for \(v{_1}'\) and then squaring it and substituting it into \(\text{Eqn.#2}\)
Pause here and do this before continuing.