Answer to Question #192506 in Physics for Xairis Vitug

Question

A 0.3-kg block is moving to the right on a horizontal, frictionless surface with a speed of 0.6 m/s. It makes a head-on collision with a 0.2-kg block that is moving to the left with a speed of 1.2 m/s. Find the final velocity (magnitude and direction) of each block if the collision is elastic. (Since the collision is head-on, all motion is along a line.)

Answer 1

For an elastic collision

$v_1=\frac{m_1-m_2}{m_1+m_2}\cdot u_1+\frac{2m_2}{m_1+m_2}\cdot u_2=$

$=\frac{0.3-0.2}{0.3+0.2}\cdot 0.6+\frac{2\cdot0.2}{0.3+0.2}\cdot (-1.2)=-0.84\ (m/s)$ to the left

$v_2=\frac{m_2-m_1}{m_1+m_2}\cdot u_2+\frac{2m_1}{m_1+m_2}\cdot u_1=$

$=\frac{0.2-0.3}{0.3+0.2}\cdot (-1.2)+\frac{2\cdot0.3}{0.3+0.2}\cdot 0.6=0.96\ (m/s)$ to the right