The momentum, $p$, of a body of mass $m$ which is moving with a velocity $v$ is $p = m \times v = mv$. The units are $\mathrm{Ns}$.

The impulse of a force is $I = Ft$ - when a constant force $F$ acts for a time $t$. The units are $\mathrm{Ns}$.

The Impulse-Momentum Principle says $I = mv - mu$ which is final momentum - initial momentum so Impulse is the change in momentum.

The principle of states that total momentum before impact is equal to total momentum after impact, $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.$

\begin{align*} p &= mv,\\ I & = Ft, \\ I & = mv - mu, \\ m_1 u_1 + m_2 u_2 & = m_1 v_1 + m_2 v_2. \end{align*}

Worked Example: Conservation of momentum

Example

An $110 \mathrm{kg}$ rugby player travelling with a velocity of $7 \mathrm{ms^{-1} }$ collides head on with a $90 \mathrm{kg}$ player travelling at $8 \mathrm{ms^{-1} }$. Suppose that the two players entangle and continue travelling together as a unit after the collision, what is their combined velocity?

Solution

We can use the law of conservation of momentum to solve this problem. We have that $m_1 = 110 \mathrm{kg}, m_2 = 90 \mathrm{kg}, u_1 = 7 \mathrm{ms^{-1} }$ and $u_2 = - 8 \mathrm{ms^{-1} }$ as the second player is travelling in the opposite direction. We want to find their combined velocity after the collision, $v \ \mathrm{ms^{-1} }$ as shown in the diagram below.

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We have \begin{align} m_1 u_1 + m_2 u_2 & = (m_1 + m_2) v, \\ v & = \frac{m_1 u_1 + m_2 u_2}{m_1+m_2}, \\ & = \frac{ 110(7) + 90(-8)}{110 + 90}, \\ & = \frac{50}{200}, \\ & = 0.25 \mathrm{ms^{-1} }. \end{align} Here, our value for $v$ is positive meaning that the direction of travel is in the direction we originally took as the positive direction. This is the $110 \mathrm{kg}$ player's original direction of travel. So after the collision the pair travel as a unit at a speed of $0.25 \mathrm{ms^{-1} }$ in the $110 \mathrm{kg}$ player's original direction of travel.