Impulse-Momentum Theorem

In Physics, momentum is associated with dynamic objects. The momentum objects the product of its mass with velocity. The impulse of an object is the product of force with time. Both the physical quantities play an important role in kinematics and are related to each other by Impulse-Momentum Theorem. According to the impulse-momentum theorem, an object will experience an impulse that is equal to the change in its momentum.

In this article, we can now see how the impulse-momentum theorem demonstrates how a smaller net force applied over a prolonged period of time can result in the same velocity change as a big net force delivered over a brief period of time.

History of Jean Buridan

Jean Buridan

Name: Jean Buridan

Year of Birth: 1301

Year of Death: 1358

Field: Physics

Discovery: He was the very first person to discover momentum which proved to be a very important concept in Physics.

Impulse-Momentum Formula

The formula used for the impulse-momentum theorem is

$F.\Delta t=m({{v}_{f}}-{{v}_{i}})$

\[F\] = Force applied to the object

$\Delta t$ = Time duration

$m$ = Mass of the object

${{v}_{f}}$ = Final velocity of the object

${{v}_{i}}$ = Initial velocity of the object

Statement of the Impulse-Momentum Theorem

According to the impulse-momentum theorem, the change in momentum of an object will be equal to the impulse that was applied to it.

Proof of the Impulse-Momentum Theorem

According to Newton’s Second Law,

$F=\dfrac{dP}{dt}$

F is the force applied and $\dfrac{dP}{dt}$ is the change in momentum.

By cross-multiplying, we get

$F.dt=dP$

Integrating both sides, we get

$\int\limits_{0}^{t}{F.dt}=\int\limits_{{{p}_{i}}}^{{{p}_{f}}}{dP}$

$Ft={{p}_{f}}-{{p}_{i}}$

$Ft=\Delta P$

From Newton’s Second Law of Motion, we know that $P=mv$.

As impulse is represented as $F\times t$, so we can replace it by ‘J’.

J =$Ft=m{{v}_{i}}-m{{v}_{f}}$

J = $Ft=m({{v}_{i}}-{{v}_{f}})$

Hence proved.

Diagram of Impulse Momentum Theorem

Impulse Momentum Theorem

This graph is a depiction of the Impulse Momentum Theorem and we can see that impulse is the area under the curve of the Force-Time graph from the equation. Also, we can see that $J=\int{P.dt}$.

Application of the Impulse-Momentum Theorem

• Car airbags lessen the effect of a crash by extending the time that the force has to act. Whether an airbag is present or not, a car's change in momentum after an abrupt stop is the same. Without an airbag, the force that stops the person moving in a very brief period of time could cause fatal injuries like head injuries, broken necks, and spinal cord injuries.

• Longer time periods are needed in sports like pole vaulting, gymnastics, and boxing to reduce the force's impact. For instance, pole vaulters frequently come to rest on their backs. Only the force and time interval may alter because there is no change in momentum as they land.

• Bungee jumping spreads the force of the impulse over a longer period of time using a long, elastic line, which decreases the force's impact on the jumper. If a regular rope is utilised, the user will abruptly stop, which could result in catastrophic injuries.

Solved Examples

1. A marble of mass 0.1 kg falls off a bed with a height of 1.2m. What is the impulse on the marble as it hits the ground?

Ans: Impulse is written as $J=F\Delta t=m({{v}_{f}}-{{v}_{i}})$

There is no loss of energy, hence we can apply the Law of Conservation of Energy:

$E={{U}_{i}}+{{K}_{i}}={{U}_{f}}+{{K}_{f}}$

We can assume the final height to be zero. Hence, ${{K}_{i}}=0$.and ${{U}_{f}}=0$

$mg{{h}_{i}}=(\dfrac{1}{2}m{{v}_{f}}^{2})$

Cancelling m from both sides and rearranging the equations, we get

${{v}_{f}}=\sqrt{2gh}$

${{v}_{f}}=\sqrt{2(9.8)(1.2)}$

${{v}_{f}}=4.84m/s$

Calculating Impulse, we get

$J=m({{v}_{f}}-{{v}_{i}}){{m}^{2}}/s$

$J=0.1(4.84-0){{m}^{2}}/s$

$J=0.484{{m}^{2}}/s$

Hence, the impulse on the ball when it hits the ground is $0.484{{m}^{2}}/s$.

2. A 150g baseball is thrown with a speed of 40m/s. If it takes 0.7s for the baseball to come to rest in the catcher's glove, what is the average force the catcher experiences due to the ball?

Ans: Impulse is written as

$J=F\Delta t=m({{v}_{f}}-{{v}_{i}})$

Rearranging the equation, we get

$F=\dfrac{m({{v}_{f}}-{{v}_{i}})}{t}$

$F=\dfrac{(0.15kg)(40m/s)}{0.7s}=8.57N$

Hence, the average force the catcher will experience is $8.57N$.

3. A soccer ball of 0.45kg is heading toward a wall at a speed of 20 metres per second. After hitting the wall, the ball bounces back at a speed of 25 metres per second. The ball was in contact with the wall for 0.003 seconds. What is the average force the wall exerted on the ball?

Ans: Initial velocity of the ball = $20m/s$

Final velocity of the ball = $-25m/s$

Sign convention: +ve X-axis is taken as the +ve direction and vice-versa.

$J=F\Delta t=m({{v}_{f}}-{{v}_{i}})$

Rearranging the equation, we get

$F=\dfrac{m({{v}_{f}}-{{v}_{i}})}{t}$

$F=\dfrac{(0.45kg)(-25m/s-20m/s)}{0.003}$

$F=\dfrac{(0.45kg)(-45m/s)}{0.003}$

$F=-6750N$

Important Points to Remember

• Impulse is denoted as ‘\[J\]’.

• Unit of impulse and unit of acceleration are the same, i.e., \[\dfrac{{{m}^{2}}}{s}\].

• Dimensional formula of impulse is $\left[ {{M}^{1}}{{L}^{1}}{{T}^{-1}} \right]$.

Conclusion

The article explains crucial Physics concepts like momentum and impulse. The impulse-momentum theorem explains how impulse and momentum are related. It asserts that the impulse is equal to the body's change in momentum. Impulsive force is a force that affects a body for a short period of time. The force exerted on a body during a collision that causes a change in momentum is the most typical illustration of an impulsive force.