multiple choice question\ntwo identical steel balls move straight towards each other at the same speed. what…

Explanation:

Step1: Define variables for collision

Let mass of each ball $m_1=m_2=m$, initial velocity of ball 1 $v_{1i}=v$, initial velocity of ball 2 $v_{2i}=-v$ (opposite direction).

Step2: Apply conservation of momentum

Conservation of momentum: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$ Substitute values: $m(v) + m(-v) = mv_{1f} + mv_{2f}$ Simplify: $0 = m(v_{1f}+v_{2f}) \implies v_{1f} = -v_{2f}$

Step3: Apply elastic collision KE conservation

Kinetic energy conservation: $\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$ Substitute values: $\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = \frac{1}{2}mv_{1f}^2 + \frac{1}{2}mv_{2f}^2$ Simplify: $mv^2 = \frac{1}{2}m(v_{1f}^2 + v_{2f}^2)$ Substitute $v_{1f}=-v_{2f}$: $mv^2 = \frac{1}{2}m(v_{2f}^2 + v_{2f}^2) = mv_{2f}^2$ Solve: $v_{2f}=v$, $v_{1f}=-v$ (magnitude equal to initial speed, opposite direction)

Answer:

They both bounce straight back with the same initial speed.