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Question

Question: The rate of change of momentum represents A. Pressure B. Force C. Work D. Kinetic energy...

The rate of change of momentum represents
A. Pressure
B. Force
C. Work
D. Kinetic energy

Explanation

Solution

Hint: Go with the basic definition of momentum and take the rate of change of momentum with respect to time in mathematical terms to derive the result.
Formula used: The formula from the basic definition of momentum is used.
momentum=mass×velocity\text{momentum}=\text{mass}\times \text{velocity}

Complete step-by-step answer:
Momentum for an object is defined as the quantity of its motion. The momentum of an object is the virtue of its mass and the velocity of its motion.
For ease of solving the question, let us denote the momentum of the object as pp, let the mass of the object be mm, and the velocity at any instant be vv.
The question requires a change in momentum with respect to time tt. Let us consider a small change in the system and thus take a differential change.
Therefore, the expression takes the form:
dpdt\Rightarrow \dfrac{dp}{dt}
But the formula for momentum is known to us.
momentum=mass×velocity p=mv \begin{aligned} & \text{momentum}=\text{mass}\times \text{velocity} \\\ & p=mv \\\ \end{aligned}
Thus, the expression becomes:
dpdt=d(mv)dt=mdvdt\Rightarrow \dfrac{dp}{dt}=\dfrac{d(mv)}{dt}=m\dfrac{dv}{dt}
Look at the expression, the term dvdt\dfrac{dv}{dt} is nothing but the expression of instantaneous acceleration.
And we know that mass times acceleration is force. Where, force is a push or pull on an object resulting from the object's interaction with another object.
Therefore,
mdvdt=ma=Force\Rightarrow m\dfrac{dv}{dt}=ma=\text{Force}
Thus, the answer to this question is option B. force.

Note: Firstly, while taking the derivative of momentum, it is an assumption that the mass of the object remains constant. Therefore, differentiation does not apply on mm. Secondly, one another way to solve this problem can be by performing dimensional analysis on the quantity asked in the question and matching it with the options.