Question

The unit of impulse per unit area is the same as that of:
A. Moment force
B. Linear momentum
C. Rate of change of linear momentum
D. Force

Answer 1

The application of Newton's second law to changing mass allows for the use of impulse and momentum as analytical tools for jet or rocket-powered vehicles. In the case of rockets, the impulse imparted may be normalised by the unit of propellant consumed, resulting in a performance metric known as specific impulse. The Tsiolkovsky rocket equation, which links the vehicle's propulsive change in velocity to the engine's specific impulse (or nozzle exhaust velocity) and the propellant-mass ratio, is derived using this knowledge.

Complete step by step solution:
In classical mechanics, the integral of a force, F, over the time interval, t, over which it acts is referred to as impulse. In the same way that force is a vector quantity, so is impulse. When an item receives an impulse, it experiences a vector change in its linear momentum in the opposite direction. The Newton second (Ns) is the SI unit of impulse, while the kilogramme metre per second (kgm/s) is the dimensionally equivalent unit of momentum.
For as long as it acts, a resultant force generates acceleration and a change in the velocity of the body. As a result, a resulting force delivered over a longer time causes a larger change in linear momentum than a force applied quickly: the change in momentum equals the sum of the average force and duration. A little force applied for a long time, on the other hand, generates the same change in momentum—the same impulse—as a greater force applied for a short time.
$J={{F}_{\text{average}}}({{t}_{2}}-{{t}_{1}})$
The integral of the resulting force (F) with respect to time is the impulse:
$J=\int{F}\text{d}t$
From Newton's second law, force is related to momentum p by
$\mathbf{F}=\frac{\text{d}\mathbf{p}}{\text{d}t}$
Hence
$\mathbf{J}\text{ }=\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{\frac{\text{d}\mathbf{p}}{\text{d}t}}\text{d}t=\int\limits_{{{\mathbf{p}}_{1}}}^{{{\mathbf{p}}_{2}}}{\text{d}}\mathbf{p}={{\mathbf{p}}_{2}}-{{\mathbf{p}}_{1}}=\Delta \mathbf{p}$
where $\Delta p$denotes the linear momentum shift from time ${{t}_{1}}$ to time ${{t}_{2}}$. The impulse-momentum theorem is a popular name for this. As a result, an impulse may also be thought of as a change in the momentum of an object to which a force is applied as a result. When the mass is constant, the impulse may be stated more simply.
Momentum and impulse have the same units and dimensions ($ML{{T}^{-1}}$). These are kg m/s = Ns in the International System of Units.
Hence option B is correct.

Note:
A fast-acting force or impact is sometimes referred to as "impulse." This sort of impulse is frequently idealised such that the force's change in momentum occurs with no change in time. This is a significant alteration that is not physically conceivable. This model, on the other hand, is useful for calculating the consequences of perfect collisions (such as in game physics engines). In addition, the term "total impulse" is widely used in rocketry and is regarded identical with the term "impulse."