2.4 Impulse

We have seen that an object has a linear momentum given by $\boldsymbol{p}=m\boldsymbol{v}$ . How does the momentum change under the action of a force? Notice that

\frac{\operatorname{\mathrm{d}}\!\boldsymbol{p}}{\operatorname{\mathrm{d}}\!t}% =\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!t}m\boldsymbol{% v}=m\boldsymbol{a}=\boldsymbol{F},

(2.30)

so we have a new way to write Newton’s second law. Let’s now integrate this equation over an arbitrary time interval:

	$\displaystyle\int_{t_{1}}^{t_{2}}\frac{\operatorname{\mathrm{d}}\!\boldsymbol{% p}}{\operatorname{\mathrm{d}}\!t}\operatorname{\mathrm{d}}\!t$	$\displaystyle=\boldsymbol{p}(t_{2})-\boldsymbol{p}(t_{1})$		(2.31)
		$\displaystyle=\Delta\boldsymbol{p}=\int_{t_{1}}^{t_{2}}\boldsymbol{F}(t)% \operatorname{\mathrm{d}}\!t.$		(2.32)

We call the integral of a force over a time interval the impulse.

Definition 2.2.

The impulse of a force $\boldsymbol{F}$ over a time interval $t_{1}\leq t_{2}$ is defined as

\boldsymbol{J}=\int_{t_{1}}^{t_{2}}\boldsymbol{F}(t)\operatorname{\mathrm{d}}% \!t.

(2.33)

It has units of Newton second (N $\cdot$ s). As we have seen, the impulse is equal to the change in momentum.

\Delta\boldsymbol{p}=\boldsymbol{J}.

(2.34)

This result is sometimes called the impulse-momentum theorem.

If we had a constant force, then the impulse would be $\boldsymbol{J}=\boldsymbol{F}\Delta t$ ( $\Delta t=t_{2}-t_{1}$ ). In most problems we want to solve this will not be the case. However, we can define the average force such that

\boldsymbol{J}=\int_{t_{1}}^{t_{2}}\boldsymbol{F}(t)\operatorname{\mathrm{d}}% \!t=\boldsymbol{F}_{\text{avg}}\Delta t.

(2.35)

This is quite useful because if we have a short interaction, we can simply consider the average force over the interval which is a good approximation.

For a general system of $N$ objects, the total impulse on the system over a time interval is

\boldsymbol{J}=\int_{t_{1}}^{t_{2}}\sum_{i=1}^{N}\boldsymbol{F}_{i}(t)% \operatorname{\mathrm{d}}\!t=\int_{t_{1}}^{t_{2}}\boldsymbol{F}_{\text{ext}}% \operatorname{\mathrm{d}}\!t.

(2.36)

By equation 2.17 in the previous section,

$\displaystyle\boldsymbol{J}=\int_{t_{1}}^{t_{2}}\boldsymbol{F}_{\text{ext}}% \operatorname{\mathrm{d}}\!t$	$\displaystyle=\int_{t_{1}}^{t_{2}}M\boldsymbol{a}_{\text{COM}}\operatorname{% \mathrm{d}}\!t$	(2.37)
	$\displaystyle=M\boldsymbol{v}_{\text{COM}}(t_{2})-M\boldsymbol{v}_{\text{COM}}% (t_{1})$	(2.38)
	$\displaystyle=\sum_{i=1}^{N}\left(m_{i}\boldsymbol{v}_{i}(t_{2})-m_{i}% \boldsymbol{v}_{i}(t_{1})\right)$	(2.39)
	$\displaystyle=\sum_{i=1}^{N}\left(\boldsymbol{p}_{i}(t_{2})-\boldsymbol{p}_{i}% (t_{1})\right)$	(2.40)
	$\displaystyle=\boldsymbol{P}(t_{2})-\boldsymbol{P}(t_{1})=\Delta\boldsymbol{P},$	(2.41)

where $\boldsymbol{P}$ denotes the total momentum of the system. So the impulse-momentum theorem still holds for composite systems.

Example 2.5.

Consider a baseball of mass $m=$ 0.3kg being thrown at a speed of 15ms^-1. If the batter bats the ball at a speed of 25ms^-1 and the bat is in contact with the ball for 0.005s, what is the impulse imparted to the ball? What is the average force exerted on the ball? What is the average acceleration of the ball?