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*}

A ball of mass $400\mathrm{g}$ hits a fixed vertical wall at a right angle with a speed of $5 \mathrm{ms^{-1} }$, the ball rebounds with a speed of $3 \mathrm{ms^{-1} }$. What is the impulse exerted on the wall by the ball?

We can find the impulse exerted on the *ball* by the *wall* and then use 's 3rd Law (every action has an equal and opposite reaction) to deduce the impulse exerted on the *wall* by the *ball* - as these will be equal.

If we take the rebound direction of the ball as the positive direction, and convert the mass of the ball to $\mathrm{kg}$, we have that \begin{align} I & = mv - mu, \\ & = \left(0.4 \times 3 \right) - \left(0.4 \times (-5) \right), \\ & = 1.2 - (-2), \\ & = 3.2 \mathrm{N s}. \end{align} Note that the initial velocity is in the negative direction. So therefore by 's 3rd Law, the magnitude of the impulse exerted on the wall by the ball is $3.2 \mathrm{Ns}$.

Two particles $P$ and $Q$ of mass $5 \mathrm{kg}$ and $8\mathrm{kg}$ respectively are moving towards each other along the same straight line. The surface is smooth and horizontal and the particles move at speeds $4 \mathrm{ms^{-1} }$ and $2 \mathrm{ms^{-1} }$ respectively. $P$ and $Q$ collide. The magnitude of the impulse due to the collision is $8 \mathrm{Ns}$. What is the speed and direction of each particle after the collision?

First draw a diagram which shows the velocities and impulses with arrows.

Here we have guessed that the particles will change direction on collision - they may not! To find the velocities after collision we consider each particle one at a time, choose a positive direction and apply the Impulse-Momentum Principle.

For particle $P$ we will take the direction $\left( \leftarrow \right)$ as the positive. \begin{align} I & = mv - mu,\\ 8 & = 5(v_1 - (-4)), \\ 8 & = 5v_1 + 20, \\ - 12 & = 5 v_1, \\ -2.4 \mathrm{ms^{-1} } & = v_1, \end{align} We have that $v_1 < 0$ this means our guess for the direction of $P$ was incorrect - this is okay, it just means that after the collision $P$ was travelling in the direction $(\rightarrow)$ at a speed of $2.4 \mathrm{ms^{-1} }$ so it doesn't change direction.

For particle $Q$ we will take the direction $\left(\rightarrow \right)$ as the positive. \begin{align} I & = mv - mu,\\ 8 & = 8(v_2 - (-2)), \\ 8 & = 8v_2 +16, \\ -8 & = 8 v_2, \\ -1 \mathrm{ms^{-1} } & = v_2. \end{align} We have that $v_2 <0$ this means our guess for the direction of $Q$ was also incorrect - this is okay, it means that after the collision $Q$ didn't change direction either, and was travelling in the direction $\left( \leftarrow \right)$ at a speed of $1 \mathrm{ms^{-1} }$.

Try our Numbas test on impulse and momentum: Impulse and Momentum