Answer to Question #231039 in Physics for Ian

Question

Two masses collide on a frictionless horizontal floor and in perfectly inelastic collision. Mass 1 is 4 times that of mass 2. Velocity of mass 1 = 10 m/s to the right while the velocity of mass 2 = 20 m/s to the left. What is the velocity and direction of the resulting combined mass?

Answer 1

Let's choose the rightwards as the positive direction. Then, we can find the velocity and direction of the resulting combined mass from the law of conservation of momentum:

$m_1v_1-m_2v_2=(m_1+m_2)v_c,$ $v_c=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.$

Since $m_1=4m_2$ we get:

$v_c=\dfrac{4m_2v_1-m_2v_2}{4m_2+m_2}=\dfrac{m_2(4v_1-v_2)}{5m_2}=\dfrac{4v_1-v_2}{5},$ $v_c=\dfrac{4\cdot10\ \dfrac{m}{s}-20\ \dfrac{m}{s}}{5}=4\ \dfrac{m}{s}.$

The sign plus means that the velocity of the resulting combined mass directed to the right.