Problem 62 - Single Variable Calculus

Question

Consider a particle moving along a straight line. Newton's Second Law of Motion states that the external force \(F\) acting on the particle is equal to the rate of change of its momentum. Thus, $$F=\frac{d}{d t}(m v)$$ where \(m\), the mass of the particle, and \(v\), the velocity of the particle, are both functions of time. a. Use the Product Rule to show that $$F=m \frac{d v}{d t}+v \frac{d m}{d t}$$ b. Use the results of part (a) to show that if the mass of a particle is constant, then \(F=m a\), where \(a\) is the acceleration of the particle.

Step-by-Step Solution

Answer

In this problem, we first proved the formula for the external force acting on a particle using the product rule, which yielded \(F = m\frac{dv}{dt} + v\frac{dm}{dt}\). Then, we showed that when the mass of a particle is constant (\(\frac{dm}{dt} = 0\)), the formula for force simplifies to \(F = ma\), where \(a\) is the acceleration of the particle.

Step 1: (a) Prove the force formula using the product rule

To prove the given formula for force, we must use the product rule for differentiation, which states that the derivative of the product of two functions is equal to the derivative of the first function multiplied by the second function plus the first function multiplied by the derivative of the second function: \[\frac{d(uv)}{dt} = u\frac{dv}{dt} + v\frac{du}{dt}\] Applying the product rule to \(F = \frac{d(mv)}{dt}\): \[\frac{d(mv)}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}\] Given that \(F = \frac{d(mv)}{dt}\), we can thus conclude that: \[F = m\frac{dv}{dt} + v\frac{dm}{dt}\]

Step 2: (b) Show that when mass is constant, \(F = ma\)

Given the result from part (a): \[F = m\frac{dv}{dt} + v\frac{dm}{dt}\] When the mass of a particle is constant, we can say that its derivative with respect to time is zero, that is: \[\frac{dm}{dt} = 0\] Now substitute this value into the force formula: \[F = m\frac{dv}{dt} + v(0)\] Simplifying the expression, we get: \[F = m\frac{dv}{dt}\] Recall that the acceleration of a particle, \(a\), is the time derivative of its velocity: \[a = \frac{dv}{dt}\] Substitute this definition of acceleration into the force formula: \[F = m \cdot a\] This confirms that when the mass of a particle is constant, the formula for force simplifies to \(F = ma\).

Key Concepts

Product Rule

The Product Rule is a fundamental concept in calculus, especially when dealing with the differentiation of products of two functions. Imagine two functions, say \( u(t) \) and \( v(t) \), where both depend on time. When you're asked to find the derivative of their product \( uv \), the Product Rule tells us how to do it efficiently. It states that the differentiation of the product of two functions is given by:

• The derivative of the first function times the second function
• Plus the first function times the derivative of the second function

In mathematical terms, it is expressed as:\[\frac{d(uv)}{dt} = u \frac{dv}{dt} + v \frac{du}{dt}\]This concept is crucial because it allows us to break down complex expressions into more manageable parts, ultimately making differentiation much simpler. For example, in our original exercise, we applied the Product Rule to find the rate of change of momentum.

Differentiation

Differentiation is one of the cornerstones of calculus and helps us understand how a function changes. In simple terms, if you want to know how something like velocity changes over time, you'd use differentiation. And from our exercise, we learn that when a particle's momentum \((mv)\) is being examined, differentiation helps us find how quickly this momentum is changing.
Differentiation is not just about finding rates of change. It's a tool that uncovers information about the movement, speed, and direction of objects. For Newton's Second Law, this becomes particularly important as it deals with forces and motion.

• It reveals the instantaneous rate at which something changes
• Helps us transition from understanding position to understanding velocity and acceleration

So, using differentiation, we can determine the force acting on a particle based on its changing momentum.

Momentum

Momentum is a key concept in physics that describes the motion of an object and is defined as the product of an object's mass and its velocity. Consider momentum as a measure of how much "push" an object has in its motion. The greater the mass or velocity of an object, the greater its momentum.
In mathematical terms, momentum \( p \) is given by:\[ p = mv \]The exercise highlighted that the force \( F \) acting on a particle was related to the rate of change of its momentum. Understanding momentum leads to insights on how external forces cause changes in an object's motion.

• Momentum connects speed and mass in a single framework
• It's crucial for understanding collisions and motion

By differentiating momentum with respect to time, you can derive how forces influence a particle's velocity and mass over time.

Acceleration

Acceleration describes how quickly an object's velocity changes over time. It's another vital concept from the exercise because it's directly related to force when mass is held constant.
In mathematical terms, acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \):\[ a = \frac{dv}{dt} \]This definition helps us relate the changing motion of an object back to the forces applied to it. When we say "force equals mass times acceleration" (\( F = ma \)), we're linking the concepts beautifully.

• Acceleration deals with changes in speed
• Allows us to track how quickly a body responds to force

This exercise teaches us that understanding acceleration is crucial for deriving force, particularly when considering constant mass scenarios in motion analysis.