Introduction
In physics, the Law of Conservation of Linear Momentum states that for an isolated system, the total momentum before a collision is equal to the total momentum after the collision. This principle is a fundamental tool for solving collision mechanics.
The Problem
A ball A (mass $m_A = 0.1 \text{ kg}$) moving at $u_A = 6 \text{ ms}^{-1}$ hits a stationary ball B (mass $m_B = 0.2 \text{ kg}$).
- Find the common velocity ($v$) if they move together after the collision.
- Find the new velocity of B ($v_B$) if ball A rebounds at $v_A' = -2 \text{ ms}^{-1}$.
Part 1: Perfectly Inelastic Collision (Moving Together)
When two objects move together, they share a common velocity, $v$.
Conservation of Momentum: $m_A u_A + m_B u_B = (m_A + m_B)v$
- $m_A = 0.1 \text{ kg}$, $u_A = 6 \text{ ms}^{-1}$
- $m_B = 0.2 \text{ kg}$, $u_B = 0 \text{ ms}^{-1}$ (at rest)
$(0.1 \times 6) + (0.2 \times 0) = (0.1 + 0.2)v$ $0.6 = 0.3v$ $v = \frac{0.6}{0.3} = 2 \text{ ms}^{-1}$
Result: The common velocity is $2 \text{ ms}^{-1}$.
Part 2: Elastic-style Collision (Rebound)
Here, ball A bounces back. We must treat the velocity as a vector, so the rebound velocity is negative.
Conservation of Momentum: $m_A u_A + m_B u_B = m_A v_A' + m_B v_B'$
- $v_A' = -2 \text{ ms}^{-1}$ (due to opposite direction)
Substitute the known values: $(0.1 \times 6) + 0 = (0.1 \times -2) + (0.2 \times v_B')$ $0.6 = -0.2 + 0.2v_B'$ $0.8 = 0.2v_B'$ $v_B' = \frac{0.8}{0.2} = 4 \text{ ms}^{-1}$
Result: The new velocity of ball B is $4 \text{ ms}^{-1}$ in the original direction of ball A.
Intuition
- Vector Nature: Always remember that velocity has a direction. When an object reverses its motion, you must flip the sign of its velocity.
- Total Momentum: In the first scenario, kinetic energy is lost as the balls stick together. In the second, the energy distribution changes as A transfers momentum to B while reversing its own path.