Impulse

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What is impulse?

Impulse is the product of average force and the time for which the force acts during an impact. When a cricket ball is hit by a bat or a nail is struck into wood with a hammer, the force of impact

acts for a short time and
is usually not constant throughout the duration of the impact i.e. it varies with time.

The forces which act on bodies for a short time are called impulsive forces.

An impulsive force does not remain constant but changes first from zero to maximum and then from maximum to zero. Thus it is not possible to measure easily the value of impulsive force because it changes with time. In such cases, we measure the total effect of the force, called impulse.

Hence, the impulse of a force is a measure of the total effect of the force. It is given by the product of force and the time for which the force acts on the body; i.e.,

$\text{Impulse} = \text{force} \times \text{time}$

According to Newton’s Second law of motion,

$\vec{F} = \dfrac{d\vec{p}}{dt}$

$\therefore \quad \vec{F} \, dt = d\vec{p}$

Integrating both sides within the limits indicated, we get

$ \displaystyle \int_{0}^{t} \vec{F} \, dt = \int_{\vec{p_1}}^{\vec{p_2}} d\vec{p}$

Where $\vec{p_1}$is the initial linear momentum at $t = 0$ and $\vec{p_2}$ is the final linear momentum at time $t$.

If $\vec{F}_{\text{av}}$is the (constant) force during this time, we may rewrite (11) as

$ \displaystyle \vec{F}_{\text{av}} \times t = \vec{p_2} – \vec{p_1}$

i.e. $\text{Impulse}, \quad \vec{I} = \vec{F}_{\text{av}} \times t = \vec{p_2} – \vec{p_1}$

Hence, impulse of a force, which is the product of average force during impact and the time for which the impact lasts, is measured by the total change in linear momentum produced during the impact.

Impulse of a force is positive, negative, or zero according as the momentum of the body increases, decreases, or remains unchanged under the effect of force.

Impulse is a vector quantity. Its direction is the same as that of the force.

The dimensional formula of impulse is the same as that of linear momentum i.e. $[M^1 L^1 T^{-1}]$.

The SI units of impulse are $(\text{N} \cdot \text{s})$ and $kg \cdot m \cdot s^{-1}$ and the c.g.s. units of impulse are $ \displaystyle \text{dyne-sec and }g\text{ }cm\text{ }{{s}^{{-1}}}$.

If we plot a graph between average force and time, the area under the curve and time axis gives the value of impulse. In Fig. 3(a)4, if the areas under the curves (a) and (b) are equal, they represent equal impulse.

From the graph we note that the force ($ \displaystyle {{F}_{1}}$) for the curve (a) is greater than the force ($ \displaystyle {{F}_{2}}$) for the curve (b). The time $ \displaystyle {{t}_{2}}$ for which the force ($ \displaystyle {{F}_{2}}$) acts is greater in case of curve (b) than the time $ \displaystyle {{t}_{1}}$ in case of curve (a).

Hence, a given change in linear momentum can be produced by applying a larger force for a smaller time or by applying a smaller force for larger time. This is sometimes called impulse-momentum theorem.

Sample Problems

A hammer of mass $1 \, \text{kg}$moving with a speed of $6 \, \text{m/s}$strikes a wall and is brought to rest in $1/10 \, \text{s}$. Calculate (i) the impulse required to stop the hammer, (ii) the retarding force of the hammer, and (iii) the retarding force on the hand of the person.

Solution:

$
\text{Here, } m = 1 \, \text{kg}, \quad u = 6 \, \text{m/s}
$

$
\text{Impulse} = F \times t = m (v – u)
$

$
= 1 \times (0 – 6) = -6 \, \text{Ns}
$

Retarding force that stops the hammer:

$
F = \dfrac{\text{Impulse}}{t} = \dfrac{-6}{\dfrac{1}{10}} = -60 \, \text{N}
$

Retardation of hammer:

$
a = \dfrac{F}{m} = \dfrac{-60}{1} = -60 \, \text{m/s}^2
$

A cricket ball of mass $150 \, \text{g}$is moving with a velocity of $12 \, \text{m/s}$and is hit by a bat so that the ball is turned back with a velocity of $20 \, \text{m/s}$. If the duration of contact between the ball and the bat is $0.01 \, \text{s}$, find the impulse and the average force exerted by the bat.

Solution:

$
\text{Impulse, } I = \text{change in momentum}
$

$$
\begin{array}{l}=m(v-u)=\dfrac{{150}}{{1000}}\times (20-(-12))\\=\dfrac{{150}}{{1000}}\times 32=4.8\text{Ns}\end{array}
$$

Average force:

$
F = \dfrac{\text{Impulse}}{t} = \dfrac{4.8}{0.01} = 480 \, \text{N}
$

Applications of The Concept of Impulse

A cricket player lowers his hands while catching a ball to avoid injury. In this way, he increases the time of catch to reduce the rate of change in momentum to zero. As a result, less force acts on his hands, and injury is minimized.
When a person falls from a certain height on a cemented floor, the floor does not yield. The change in momentum occurs in a very short interval of time. Therefore, as explained above, the floor exerts a much larger force. Due to it, a person receives more injury.
On the other hand, when a person falls on a heap of sand, the sand yields. The same change in linear momentum is produced in a much longer time. The average force exerted on the person by the heap of sand is, therefore, much smaller and the person is not hurt.
The vehicles like scooters, cars, buses, trucks, etc., are provided with shockers. When they move over an uneven road, impulsive forces are exerted by the road. The function of shockers is to increase the time of impact. This would reduce the force/jerk experienced by the rider of the vehicle.
Bogies of a train are provided with buffers. They avoid severe jerks during shunting of the train.
It is difficult to catch a cricket ball than to catch a tennis ball. The cricket ball being heavier has much larger momentum and therefore exerts a much larger force on the hands during catch in comparison to the force exerted by the tennis ball.
In a head-on collision between two vehicles, the change in linear momentum is equal to the sum of the linear momenta of the two vehicles. As the time of impact is very small, this large force develops, which causes damage to the vehicles.
While running, an athlete is advised to come to a stop slowly, after finishing a fast race. This way, the time of stop increases, and hence the force experienced by him decreases.