Calculating Impulse with Calculus: Videos & Practice Problems

Calculating Impulse with Calculus focuses on finding impulse when the applied force changes with time. For a force-time function, impulse is found with the definite integral \(J=\int_{t_1}^{t_2} F(t)\\,dt\) . This integral represents the area under the force-versus-time curve, and it is the correct method when the graph is not made of simple straight-line regions.

A central idea is that impulse equals change in momentum: \(J=\Delta p=p_f-p_i\) . If mass is constant, this can also be written as \(J=m\Delta v\) . This lets you connect an integrated force-time function to momentum or velocity changes, and in some cases determine an unknown time by setting the integral equal to the required momentum change.

Calculating Impulse by Integrating Force-Time Functions

Impulse is a key concept in physics that represents the change in momentum of an object when a force is applied over a period of time. When given a force versus time graph with simple straight lines, impulse can be calculated by finding the area under the curve. However, when the force varies continuously with time and is described by a function, calculus is required to determine the impulse accurately.

In such cases, impulse J is calculated as the definite integral of the force function F(t) over the time interval from t₁ to t₂:

\[J = \int_{t_1}^{t_2} F(t) \, dt\]

This integral represents the exact area under the force-time curve between the specified times, capturing the total effect of the varying force.

For example, consider a rocket engine exerting a force described by the function F(t) = 6000 t^2 Newtons, where t is time in seconds. To find the impulse delivered between 1.5 seconds and 3.5 seconds, the integral is evaluated as follows:

\[J = \int_{1.5}^{3.5} 6000 t^2 \, dt = 6000 \int_{1.5}^{3.5} t^2 \, dt = 6000 \left[ \frac{t^3}{3} \right]_{1.5}^{3.5} = 2000 \left(3.5^3 - 1.5^3\right)\]

Calculating the values yields an impulse of approximately 79,000 kilogram meters per second (kg·m/s).

Impulse is directly related to the change in momentum of the object, expressed by the equation:

\[J = \Delta p = m \Delta v\]

where m is the mass of the object, and Δv is the change in velocity. Using the impulse calculated and the rocket’s mass of 3000 kg, the change in velocity over the 2-second interval can be found by rearranging the formula:

\[\Delta v = \frac{J}{m} = \frac{79,000}{3000} \approx 26.3 \, \text{m/s}\]

This approach highlights how integrating a time-dependent force function allows for precise calculation of impulse and subsequent changes in velocity, which is essential in understanding motion dynamics in real-world applications such as rocket propulsion.

A puck is initially at rest on an icy surface. At $t=0$, a horizontal force given by $F\left(t\right)=12-3t^2$ begins to move it. The force acts until its magnitude is zero. a) What is the magnitude of the impulse on the puck between $t=0.5s$ and $t=1.25s$? b) What is the change in momentum of the puck between $t=0$ and the instant at which $F=0$?

• A. a) $7.17N\cdot s$; b) $16N\cdot s$
• B. a) $8.25N\cdot s$; b) $12N\cdot s$
• C. a) $7.17N\(\cdot\) s$; b) $24N\cdot s$
• D. a) $5.88N\cdot s$; b) $16N\(\cdot\) s$

Answer: (A)

Calculating Impulse by Integrating Force-Time Functions Example 1

To determine the time at which an object changes its momentum from eastward to westward, we analyze the impulse-momentum relationship. Initially, the object has a momentum of +90 kg·m/s (eastward), and later it has a momentum of -60 kg·m/s (westward). The change in momentum, Δp, is calculated as the final momentum minus the initial momentum, resulting in Δp = -60 - 90 = -150 kg·m/s.

Impulse (J) is defined as the integral of the force over time and is equal to the change in momentum:

\[J = \int_{t_1}^{t_2} F(t) \, dt = \Delta p\]

Given the force function \( F(t) = -8t \) and the initial time \( t_1 = 0 \), the goal is to find the time \( t_2 = T \) when the impulse equals -150 kg·m/s. Setting up the integral:

\[\int_0^T -8t \, dt = -150\]

Evaluating the integral:

\[\left[ -4t^2 \right]_0^T = -4T^2 = -150\]

Solving for \( T \):

\[-4T^2 = -150 \implies 4T^2 = 150 \implies T^2 = \frac{150}{4} = 37.5\]\[T = \sqrt{37.5} \approx 6.12 \text{ seconds}\]

Since time cannot be negative, the positive value \( T = 6.12 \) seconds is the moment when the object’s momentum becomes westward at -60 kg·m/s. This calculation demonstrates how impulse, as the integral of force over time, directly relates to changes in momentum, allowing determination of specific time intervals for momentum changes under variable forces.

Here's what students ask on this topic:

How do you calculate impulse when the force varies with time using calculus?

When force varies with time, impulse is calculated by integrating the force function over the given time interval. Mathematically, impulse $J={\int }_t^{t}F\left(t\right)dt$, where $t=t_1$ and $t_2$ are the start and end times. This integral represents the area under the force-versus-time curve. Unlike simple cases with constant or piecewise constant forces, calculus allows you to handle continuously changing forces by summing infinitesimal contributions over time. This method is essential when the force function is given explicitly, such as $F\left(t\right)=6000t^2$. You compute the definite integral between the specified times to find the total impulse delivered.

What is the relationship between impulse and change in velocity for an object with constant mass?

Impulse is directly related to the change in momentum of an object. For an object with constant mass $m$, the impulse $J$ equals the mass times the change in velocity $\mathrm{\Delta v}$. This is expressed as $J=m\mathrm{\Delta v}$. Since impulse is the integral of force over time, calculating $J$ allows you to find how much the velocity changes during the force application. This is useful in problems where you know the force-time function and want to determine the velocity change without separately calculating initial and final velocities.

How do you evaluate the definite integral of a force-time function to find impulse?

To evaluate the definite integral of a force-time function $F\left(t\right)$ over the interval from $t_1$ to $t_2$, you first find the antiderivative of $F\left(t\right)$. For example, if $F\left(t\right)=6000t^2$, the antiderivative is $2000t^3$. Then, evaluate this antiderivative at the upper and lower limits and subtract: $J=\; igl[2000\; t^\{3\}igr]\_\{t\_1\}^\{t\_2\}\; =\; 2000\; t\_2^\{3\}\; -\; 2000\; t\_1^\{3\}$. This result gives the total impulse delivered during the time interval.

Why is calculus necessary for calculating impulse when the force-time graph is not composed of straight lines?

When the force-time graph consists of straight lines or constant segments, impulse can be found by calculating the area of simple geometric shapes like rectangles or triangles. However, if the force varies continuously and the graph is curved, these simple geometric methods are inaccurate. Calculus is necessary because it allows you to sum infinitesimally small contributions of force over time by integrating the force function. This integral accurately captures the total area under the curve, representing the impulse. Without calculus, you cannot precisely calculate impulse for complex force-time functions that change non-linearly.

How can you use impulse to find the change in velocity of a rocket firing its engines?

For a rocket with mass $m$, the impulse delivered by its engines over a time interval is the integral of the force function $F\left(t\right)$ during that interval. Once you calculate the impulse $J$, you can find the change in velocity $\mathrm{\Delta v}$ using $\mathrm{\Delta v\; =\; rac\{J\}\{m\}}$. For example, if the impulse is 79,000 kg·m/s and the rocket mass is 3000 kg, then $\mathrm{\Delta v\; =\; rac\{79000\}\{3000\}\; =\; 26.3}$ m/s. This method directly links the integrated force over time to the rocket's velocity change.