Impulse Integrals & Variable Mass

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Visualization of a rocket launching upward, leaving a trail of glowing exhaust and mathematical differential equations.

Newton’s Second Law ( $\Sigma F = ma$ ) assumes mass is constant. When a rocket burns fuel, we must use the true form: $\Sigma F = dp/dt$ .

In AP Physics 1, we defined Impulse ( $\vec{J}$ ) as the average force multiplied by the time interval ( $\vec{J} = \vec{F}_{avg} \Delta t$ ). However, during real-world collisions—like a baseball bat hitting a ball—the force is not constant. It starts at zero, spikes to a massive peak, and returns to zero within milliseconds.

1. Impulse as an Integral

To find the true Impulse (and therefore the change in momentum) of a varying force, we must integrate the Force function over the time interval of the collision.

$\vec{J} = \int_{t_1}^{t_2} \vec{F}(t) dt = \Delta \vec{p}$

The Impulse is exactly equal to the area under the curve of a Force vs. Time graph.

A Force vs Time graph showing a bell-shaped collision curve. The area under the curve is shaded and labeled 'Impulse = Integral of F dt'.

Whether you are given an equation to integrate or a graph to calculate the geometric area, the result gives you the exact change in momentum ( $\Delta p$ ).

2. Variable Mass Systems (The Rocket Equation)

What happens if the mass of the system changes while it moves? A rocket is the classic example: it accelerates by violently throwing mass (exhaust gas) backward. As it burns fuel, the rocket’s total mass decreases, making it easier to accelerate!

We cannot use $F = ma$ . We must start from the conservation of momentum for the system (Rocket + Exhaust). The resulting differential equation gives us the thrust force:

$F_{thrust} = v_e \frac{dm}{dt}$

Where $v_e$ is the exhaust velocity (relative to the rocket) and $dm/dt$ is the fuel burn rate (kg/s).

By integrating this relationship, we get the ideal Rocket Equation, which determines the final velocity of the rocket after burning its fuel:

$v_f = v_0 + v_e \ln\left(\frac{m_0}{m_f}\right)$

Concept First: Notice the curve in the simulator. As the rocket burns fuel, it becomes lighter. A lighter rocket being pushed by a constant thrust force results in an increasing acceleration. The velocity-time graph curves upward until burnout!

3. Quick AP Practice

📚 Unit C4 Mastery Challenge

1. A 2 kg object is at rest. A force given by $F(t) = 6t^2$ is applied for 2 seconds. What is the final velocity of the object?

Check Answer

First, find the Impulse by integrating the force:
$J = \int_{0}^{2} 6t^2 dt = \left[ 2t^3 \right]_0^2 = 2(8) = 16 \text{ N}\cdot\text{s}$ .

Impulse equals the change in momentum ( $J = \Delta p = m v_f - m v_0$ ). Since $v_0 = 0$ :
$16 = 2 \cdot v_f \Rightarrow \mathbf{v_f = 8 \text{ m/s}}$ .

2. A rocket has a total initial mass of $M_0$ . It burns fuel at a constant rate $R$ (so $dm/dt = R$ ). If the exhaust velocity is $v_e$ , what is the instantaneous acceleration of the rocket in deep space (ignoring gravity) as a function of time $t$ ?

Check Answer

Start with Newton’s Second Law: $F_{net} = m(t) \cdot a(t)$ .
The net force is the thrust: $F_{thrust} = v_e \cdot R$ .
The mass is decreasing over time: $m(t) = M_0 - Rt$ .

Substitute these in: $v_e \cdot R = (M_0 - Rt) \cdot a(t)$ .
Solve for acceleration: $\mathbf{a(t) = \frac{v_e R}{M_0 - Rt}}$ .

1. Impulse as an Integral

2. Variable Mass Systems (The Rocket Equation)

3. Quick AP Practice

📚 Unit C4 Mastery Challenge

🔗 Next Steps