Question

An object of mass \(100 \mathrm{~g}\) is thrown vertically upward from a point \(60 \mathrm{~cm}\) above the earth's surface with an initial velocity of \(150 \mathrm{~cm} / \mathrm{sec}\). It rises briefly and then falls vertically to the earth, all of which time it is acted on by air resistance that is numerically equal to \(200 v\) (in dynes), where \(v\) is the velocity (in \(\mathrm{cm} / \mathrm{sec})\). (a) Find the velocity \(0.1\) sec after the object is thrown. (b) Find the velocity \(0.1\) sec after the object stops rising and starts falling.

Short answer

(a) The velocity 0.1 seconds after the object is thrown is 640 \(\mathrm{cm} / \mathrm{sec}\). (b) The velocity 0.1 seconds after the object stops rising and starts falling is 100 \(\mathrm{cm} / \mathrm{sec}\).

Step-by-step solution

Identify Forces Acting on the Object

We first need to identify all forces acting on the object. In this case, there are two forces: gravity (downward) and air resistance (opposite the direction of the velocity). The gravitational force acting on the object is given by the equation: \[F_g = m\cdot g,\] where \(m\) is the mass and \(g\) the acceleration due to gravity. The force due to air resistance is given by the equation: \[F_{air} = -200v.\]

Setup Newton's Second Law Equation and Determine Differential Equation

According to Newton's second law of motion, the net force acting on an object is equal to the product of the mass and acceleration of the object. So, we can write the equation as: \[ΣF = m\cdot a.\] In this scenario, there are two forces acting on the object: gravity and air resistance. Therefore, we have: \[m\cdot a = -m\cdot g - 200v.\] Dividing both sides by the mass and noting that \(a = dv/dt\), we obtain the differential equation: \[\frac{dv}{dt} = -g - \frac{200}{m}v.\]

Solve the Differential Equation

Now, we need to solve the given differential equation. This is a first-order linear differential equation. We can solve this using the integrating factor method. Let's say the integrating factor is \(μ\): \[μ = e^{\int\frac{200}{m}dt}.\] Substitute the given mass and integrate, we will get the integrating factor: \[μ = e^{2t}.\] Next, multiply the differential equation by the integrating factor and integrate both sides: \[μ\frac{dv}{dt} + \frac{200}{m}μv = -g\cdot μ.\] \[e^{2t}dv + 200e^{2t}vdt = -ge^{2t}dt.\] Now, integrate both sides: \[\int(e^{2t}dv + 200e^{2t}vdt) = \int( -ge^{2t}dt).\] \[ve^{2t} = -\frac{1}{2}ge^{2t}+ C.\] Now, to find the constant, we use the initial condition \(v(0)=150\): \[150= -\frac{1}{2}g(1)+ C.\] \[C=\frac{1}{2}g+150.\] Thus, we have the velocity equation: \[v(t)e^{2t} = -\frac{1}{2}ge^{2t}+150e^{2t}.\] \[v(t) = -\frac{1}{2}g+ 150.\] In this problem, gravity \(g\) should be expressed in \(\mathrm{cm} / \mathrm{sec}^2\). Since 1 g is equivalent to 980 \(\mathrm{cm} / \mathrm{sec}^2\), we can set \(g = 980\).

Calculate the Velocity 0.1s after the object is thrown

We use the velocity equation derived in the previous step and plug in \(t=0.1s\): \[v(0.1)= -\frac{1}{2}(980)+150.\] \[v(0.1)= 490+150.\] \[v(0.1)= 640 \mathrm{~cm} / \mathrm{sec}.\] The velocity of the object 0.1 seconds after it's thrown is 640 \(\mathrm{cm} / \mathrm{sec}\).

Determine the time when the object stops rising

To find the time when the object stops rising, we set the velocity to zero and solve for t: \[0=-\frac{1}{2}(980)+150.\] \[t=-\frac{300}{490}.\] \[t\approx 0.6122 \mathrm{s}\] The object stops rising after approximately 0.6122 seconds.

Calculate the Velocity 0.1s after the object stops rising and starts falling

Now, we use the velocity equation derived before and plug in the total time: \[v(t)= -\frac{1}{2}(980)+150.\] \[v(0.7122)= -\frac{1}{2}(980)+150.\] \[v(0.7122)= 100 \mathrm{~cm} / \mathrm{sec}.\] The velocity of the object 0.1 seconds after it stops rising and starts falling is 100 \(\mathrm{cm} / \mathrm{sec}\). So, to summarize: (a) The velocity 0.1 seconds after the object is thrown is 640 \(\mathrm{cm} / \mathrm{sec}\). (b) The velocity 0.1 seconds after the object stops rising and starts falling is 100 \(\mathrm{cm} / \mathrm{sec}\).

Key concepts

Newton's Second Law

Newton's Second Law is a fundamental concept in physics that describes the relationship between an object's mass, its acceleration, and the forces acting upon it. According to this law, the net force on an object is equal to the mass of the object multiplied by its acceleration. This can be expressed mathematically by the formula \( F = m \cdot a \).
In our problem, the object is subject to two main forces: gravity and air resistance. Gravity pulls the object downward, while air resistance acts against its velocity. By using Newton's second law, we can set up an equation that includes both these forces. In the scenario outlined, the forces include the gravitational force \( F_g = m \cdot g \) and the air resistance \( F_{air} = -200v \), where \( v \) is the velocity.
By combining these forces into Newton's equation, we derive the net force and solve for acceleration, thus setting the stage for solving the differential equation that models the object's motion.

Integrating Factor Method

The integrating factor method is a technique used to solve first-order linear differential equations. It involves multiplying the differential equation by a strategically chosen function, known as the integrating factor, to simplify the solving process.
For the differential equation derived from Newton's second law in this problem, we have \( \frac{dv}{dt} = -g - \frac{200}{m}v \). The equation is linear in terms of \( v \), making the integrating factor method an appropriate tool here. The integrating factor is computed as \( \mu = e^{\int \frac{200}{m}dt} \).
After calculating \( \mu \) and multiplying through the differential equation, the left-hand side becomes the derivative of the product of the integrating factor and velocity \( v \). Integrating both sides finalizes the solution path, allowing us to express \( v(t) \) in terms of time and solve for specific velocity values at given times.

Velocity Calculation

Velocity calculation involves determining how fast an object is moving and in what direction. In our exercise, we perform velocity calculations at different times to solve for the speed of the object after it has been thrown.
Initially, the velocity equation derived from solving the differential equation is used to find the object's speed 0.1 seconds after it's thrown and after it starts descending. Substituting the appropriate time-related values into this equation allows us to determine the velocity at those specific instances.
The calculation requires integrating the derived equation and solving for the constant based on initial conditions. Taking gravity into account, we use the equation \( v(t) = -\frac{1}{2}g + 150 \) and plug in time values to find the resultant velocities. The velocity is found to be 640 \( \mathrm{cm/sec} \) at 0.1 seconds after the throw and 100 \( \mathrm{cm/sec} \) after the object starts falling.

First-Order Linear Differential Equation

A first-order linear differential equation is a type of equation that relates a function to its first derivative, often appearing in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). These equations can often be solved using the integrating factor method.
In the context of this exercise, the equation \( \frac{dv}{dt} = -g - \frac{200}{m}v \) is a first-order linear differential equation in the variable \( v \). This form makes it manageable to apply the integrating factor technique, transforming the equation into one that can be integrated step-by-step.
The solution process begins by recognizing the equation's linearity, calculating the integrating factor, and performing algebraic integration to solve for velocity over time. Understanding this type of differential equation is crucial in physics for modeling various dynamic systems and their changing states over time.