Answer to Question #231056 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

According to the momentum conservation law:

$$m_1v_1 - m_2v_2 = (m_1 + m_2)V$$

where $m_1 = 4m_2$ are the masses, $v_1 = 10m/s, \space v_2 = 20m/s$ are the velocities before the collision, and $V$ is the velocity after the collision. The direction to the right is taken as positive one. Sign '-' in the left hand side is due to the fact that $v_1$ and $v_2$ are opposite in direction.

Solving for $V$, obtain:

$$V = \dfrac{m_1v_1 - m_2v_2}{m_1 + m_2} = \dfrac{m_1(v_1 - 4v_2)}{5m_1} = \dfrac{v_1 - 4v_2}{5}\\ V = \dfrac{10 - 4\cdot 20}{5} = -14m/s$$

Sign '-' denotes the fact, that $V$ is directed to the left (opposite to the positive direction).

Answer. 14 m/s, to the left.