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The impulse on a particle due to a force acting on it during a given time interval is equal to the change in its

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NDA I 2016 GAT Previous Year Paper (17-Apr-2016)
The correct answer is

Momentum

Understanding Impulse and Momentum Change

The question asks about the relationship between the impulse on a particle and its state of motion during a given time interval. Specifically, it asks what quantity the impulse is equal to the change in.

What is Impulse?

Impulse is a term used in physics that describes the change in momentum of an object when a force is applied to it over a period of time. Mathematically, impulse (\(\vec{J}\)) is defined as the integral of a force (\(\vec{F}\)) over the time interval (\(\Delta t\)) it acts:

\(\vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt\)

If the force is constant over the time interval, the formula simplifies to:

\(\vec{J} = \vec{F} \Delta t\)

Impulse is a vector quantity, meaning it has both magnitude and direction. Its direction is the same as the direction of the force.

The Impulse-Momentum Theorem

One of the fundamental principles in mechanics is the impulse-momentum theorem. This theorem states that the impulse applied to an object is equal to the change in its momentum.

Momentum (\(\vec{p}\)) is defined as the product of an object's mass (\(m\)) and its velocity (\(\vec{v}\)):

\(\vec{p} = m\vec{v}\)

The change in momentum (\(\Delta \vec{p}\)) is the final momentum minus the initial momentum:

\(\Delta \vec{p} = \vec{p}_{\text{final}} - \vec{p}_{\text{initial}} = m\vec{v}_{\text{final}} - m\vec{v}_{\text{initial}}\)

The impulse-momentum theorem can be derived from Newton's second law of motion (\(\vec{F} = m\vec{a} = m \frac{d\vec{v}}{dt} = \frac{d(m\vec{v})}{dt} = \frac{d\vec{p}}{dt}\)), by integrating force with respect to time:

\(\int_{t_1}^{t_2} \vec{F} \, dt = \int_{t_1}^{t_2} \frac{d\vec{p}}{dt} \, dt = [\vec{p}]_{t_1}^{t_2} = \vec{p}(t_2) - \vec{p}(t_1)\)

So, we have:

\(\vec{J} = \Delta \vec{p}\)

This equation directly tells us that the impulse on a particle is equal to the change in its momentum during the time the force acts.

Analyzing the Options

Let's examine the given options in light of the impulse-momentum theorem:

  1. Force: Impulse involves force acting over time. It is not equal to the force itself, but rather the effect of the force integrated over time.
  2. Momentum: The impulse-momentum theorem explicitly states that impulse is equal to the change in momentum, not the momentum itself. However, among the choices, "Momentum" is the concept directly related to the outcome of impulse as per the theorem regarding "change in".
  3. Work done: Work done is related to force and displacement, and corresponds to the change in kinetic energy (Work-Energy Theorem). While force is involved in both work and impulse, they represent different physical quantities and relationships.
  4. Energy: Similar to work done, energy is a different physical concept related to the capacity to do work. Impulse is related to changes in momentum, which is distinct from energy changes.

Based on the impulse-momentum theorem, the impulse acting on a particle is equal to the change in its momentum.

Conclusion

The impulse on a particle due to a force acting on it during a given time interval is equal to the change in its momentum. This is a direct consequence of Newton's second law of motion and is formally stated as the impulse-momentum theorem.

Concept Definition/Formula Relationship with Impulse
Impulse (\(\vec{J}\)) \(\int \vec{F} \, dt\) or \(\vec{F}\Delta t\) (constant F) The impulse itself
Momentum (\(\vec{p}\)) \(m\vec{v}\) Impulse equals the change in momentum (\(\Delta \vec{p}\))
Force (\(\vec{F}\)) \(m\vec{a}\) or \(d\vec{p}/dt\) Impulse is the integral of force over time
Work Done (\(W\)) \(\int \vec{F} \cdot d\vec{r}\) or \(\vec{F} \cdot \Delta \vec{r}\) (constant F) Related to change in kinetic energy, not impulse
Energy (\(E\)) Various forms (Kinetic, Potential, etc.) Total mechanical energy change is related to work done by non-conservative forces; not directly equal to impulse.

Revision Table: Impulse and Momentum Key Concepts

Term Symbol Definition Units Relationship
Force \(\vec{F}\) Push or pull; causes acceleration Newtons (N) Impulse = \(\int \vec{F} \, dt\)
Time Interval \(\Delta t\) Duration over which force acts Seconds (s) Impulse = \(\vec{F}\Delta t\) (constant F)
Impulse \(\vec{J}\) Effect of force over time Newton-seconds (N·s) or kg·m/s \(\vec{J} = \Delta \vec{p}\)
Momentum \(\vec{p}\) Mass in motion (\(m\vec{v}\)) kg·m/s or N·s Change in momentum (\(\Delta \vec{p}\)) = \(\vec{J}\)
Change in Momentum \(\Delta \vec{p}\) \(\vec{p}_{\text{final}} - \vec{p}_{\text{initial}}\) kg·m/s or N·s \(\Delta \vec{p} = \vec{J}\)

Additional Information: Conservation of Momentum

The impulse-momentum theorem is closely related to the principle of conservation of momentum. If the net external force acting on a system is zero (i.e., the net impulse is zero), then the total momentum of the system remains constant.

Consider a system of particles. The total momentum of the system is the vector sum of the momenta of individual particles:

\(\vec{P}_{\text{total}} = \sum \vec{p}_i\)

If \(\vec{F}_{\text{net, external}} = 0\), then the total impulse on the system is zero:

\(\vec{J}_{\text{total}} = \int \vec{F}_{\text{net, external}} \, dt = 0\)

According to the impulse-momentum theorem applied to the system:

\(\vec{J}_{\text{total}} = \Delta \vec{P}_{\text{total}}\)

So, if \(\vec{J}_{\text{total}} = 0\), then \(\Delta \vec{P}_{\text{total}} = 0\), which means the total momentum does not change. This is the principle of conservation of total momentum.

This principle is particularly useful in analyzing collisions and explosions where external forces are negligible compared to internal forces between the particles.

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Important Questions from Conservation of Linear Momentum

  1. A particle of mass $6m$ at rest suddenly breaks on its own into three fragments.
    Two fragments of mass $m$ and $2m$ move along mutually perpendicular directions with speeds $2v$ and $v$ respectively.
    The energy released during the process is,
  2. Body A of mass $m$ moving with speed $u$ collides with another body B of mass $3m$, at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is:
  3. What are the dimensions of angular momentum?

  4. The propulsion of a rocket is based on which of Newton's laws of motion?

  5. For a system of interacting particles, which condition is fundamental for the conservation of its total linear momentum $\vec{P}$?

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