🎯⭐ INTERACTIVE LESSON

Momentum and Collisions

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Momentum and Collisions - Complete Interactive Lesson

Part 1: Linear Momentum

🎯 Linear Momentum

Part 1 of 7 — Momentum and Its Conservation

What Is Momentum?

$\vec{p} = m\vec{v}$

Quantity	Symbol	Units
Momentum	$\vec{p}$	$\text{kg}\cdot\text{m/s}$

🔑 Momentum is a vector — it has both magnitude and direction.

Newton's Second Law in Terms of Momentum

$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$

For constant mass: $\vec{F} = m\vec{a} = m\frac{d\vec{v}}{dt} = \frac{d(m\vec{v})}{dt}$

This more general form handles cases where mass changes (like rockets).

Conservation of Momentum

When no external forces act on a system:

$\vec{p}_{\text{initial}} = \vec{p}_{\text{final}}$

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

This is valid for any collision or interaction within an isolated system.

📝 Worked Example — Momentum from a Time-Dependent Velocity

A particle of mass $m = 3\,\text{kg}$ moves along the $x$ -axis with velocity $v(t) = (4t^2 - 2t)\,\text{m/s}$ . Find the on the particle at using the momentum form of Newton's second law.

Concept Check 🎯

Part 2: Impulse

💥 Impulse

Part 2 of 7 — Impulse-Momentum Theorem

Impulse Defined

$\vec{J} = \int_{t_1}^{t_2} \vec{F}\,dt = \Delta \vec{p}$

Part 3: Collisions in 1D

💫 Collisions in One Dimension

Part 3 of 7 — Elastic and Inelastic Collisions

Types of Collisions

Type	Momentum Conserved?	KE Conserved?
Elastic	✅ Yes	✅ Yes
Inelastic	✅ Yes	❌ No
Perfectly Inelastic	✅ Yes	❌ No (maximum KE loss)

Perfectly Inelastic Collision

Objects stick together after collision:

$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f$

Part 4: Collisions in 2D

🎱 Collisions in Two Dimensions

Part 4 of 7 — Vector Conservation of Momentum

2D Momentum Conservation

Momentum is conserved independently in each direction:

$\text{x-direction:}\quad m_1 v_{1xi} + m_2 v_{2xi} = m_1 v_{1xf} + m_2 v_{2xf}$

Part 5: Center of Mass

⚖️ Center of Mass

Part 5 of 7 — Center of Mass Motion

Center of Mass Position

For discrete masses:

$x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i} = \frac{m_1 x_1 + m_2 x_2 + \cdots}{m_1 + m_2 + \cdots}$

Part 6: Problem-Solving Workshop

🛠️ Momentum Workshop

Part 6 of 7 — AP Physics C Problem Strategies

Types of Momentum Problems on AP Physics C

Problem Type	Key Approach
Impulse calculation	$J = \int F\,dt$ or $J = \Delta p$

Part 7: Review & Applications

📋 Momentum Review

Part 7 of 7 — Comprehensive Review

Key Formulas

Formula	Name
$\vec{p} = m\vec{v}$

Step 1 — Write momentum as a function of time.

$p(t) = m\,v(t) = 3(4t^2 - 2t) = (12t^2 - 6t)\,\text{kg}\cdot\text{m/s}$

Step 2 — Differentiate to get the net force.

Because $F_{\text{net}} = \dfrac{dp}{dt}$ , we differentiate term by term:

$F_{\text{net}}(t) = \frac{d}{dt}\left(12t^2 - 6t\right) = 24t - 6$

Step 3 — Evaluate at $t = 2\,\text{s}$ .

$F_{\text{net}}(2) = 24(2) - 6 = 42\,\text{N}$

🔑 Even with constant mass, $F = \dfrac{dp}{dt}$ and $F = ma$ agree: here $a(t) = \dfrac{dv}{dt} = 8t - 2$ , so $ma = 3(8t-2) = 24t - 6$ . The momentum form is just the more fundamental statement.

Variable	Meaning
$\vec{J}$	Impulse ( $\text{N}\cdot\text{s} = \text{kg}\cdot\text{m/s}$ )
$\vec{F}$	Force (may vary with time)
$\Delta\vec{p}$	Change in momentum

For constant force: $\vec{J} = \vec{F}\,\Delta t$

Impulse-Momentum Theorem

$\vec{J} = \vec{p}_f - \vec{p}_i = m\vec{v}_f - m\vec{v}_i$

Example: A $0.15\,\text{kg}$ baseball at $40\,\text{m/s}$ is hit and leaves at $50\,\text{m/s}$ in the opposite direction.

$J = m(v_f - v_i) = 0.15\big(50 - (-40)\big) = 0.15(90) = 13.5\,\text{N}\cdot\text{s}$

🔑 The area under the $F$ vs. $t$ curve equals the impulse.

📝 Worked Example — Impulse from a Time-Varying Force

During a $0.20\,\text{s}$ collision a force $F(t) = (600t - 3000t^2)\,\text{N}$ acts on a $0.50\,\text{kg}$ ball (with $t$ in seconds). The ball is initially at rest. Find the impulse delivered and the final speed.

Step 1 — Impulse is the time integral of force.

$J = \int_{0}^{0.20} F(t)\,dt = \int_{0}^{0.20} (600t - 3000t^2)\,dt$

Step 2 — Integrate term by term.

$J = \left[\,300t^2 - 1000t^3\,\right]_{0}^{0.20}$

Step 3 — Evaluate the bounds.

$J = 300(0.20)^2 - 1000(0.20)^3 = 300(0.04) - 1000(0.008) = 12 - 8 = 4\,\text{N}\cdot\text{s}$

Step 4 — Apply the impulse-momentum theorem.

$J = \Delta p = m\,v_f$ (since $v_i = 0$ ), so .

🔑 When force varies in time, you integrate — the area under the $F$ – $t$ curve. A constant "average force" of $J/\Delta t = 4/0.20 = 20\,\text{N}$ would give the same impulse.

Concept Check 🎯

$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$

Both momentum AND kinetic energy are conserved. For 1D elastic collisions:

$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}\,v_{1i} + \frac{2m_2}{m_1 + m_2}\,v_{2i}$

$v_{2f} = \frac{2m_1}{m_1 + m_2}\,v_{1i} + \frac{m_2 - m_1}{m_1 + m_2}\,v_{2i}$

🔑 In an elastic collision between equal masses with one at rest, the first stops and the second moves off with the original velocity (the velocities swap).

📝 Worked Example — Elastic Collision and KE Loss

A $2\,\text{kg}$ cart moving at $3\,\text{m/s}$ strikes a stationary $4\,\text{kg}$ cart head-on. (a) Find the final velocities for an elastic collision. (b) Compare with the kinetic energy lost in a perfectly inelastic collision.

Part (a) — Elastic. With $m_1 = 2$ , $m_2 = 4$ , $v_{1i} = 3$ , $v_{2i} = 0$ :

$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}\,v_{1i} = \frac{2 - 4}{6}(3) = -1\,\text{m/s}$

$v_{2f} = \frac{2m_1}{m_1 + m_2}\,v_{1i} = \frac{2(2)}{6}(3) = 2\,\text{m/s}$

Check momentum: $p_i = 2(3) = 6$ ; $p_f = 2(-1) + 4(2) = 6\,\text{kg}\cdot\text{m/s}$ . ✅ The lighter cart rebounds.

Part (b) — Perfectly inelastic. The carts stick:

$v_f = \frac{m_1 v_{1i}}{m_1 + m_2} = \frac{2(3)}{6} = 1\,\text{m/s}$

Initial KE: $K_i = \tfrac{1}{2}(2)(3)^2 = 9\,\text{J}$ . Final KE: .

KE lost $= 9 - 3 = 6\,\text{J}$ — about $67\%$ of the original kinetic energy converts to heat and deformation.

🔑 Momentum is conserved in both collisions, but only the elastic case conserves kinetic energy.

Concept Check 🎯

$\text{y-direction:}\quad m_1 v_{1yi} + m_2 v_{2yi} = m_1 v_{1yf} + m_2 v_{2yf}$

Strategy for 2D Collision Problems

Choose a coordinate system.
Resolve all velocities into $x$ and $y$ components.
Apply conservation of momentum in each direction independently.
If the collision is elastic, also apply conservation of kinetic energy.

🔑 Treat each dimension separately — just like projectile motion.

📝 Worked Example — A Glancing (2D) Collision

A $1\,\text{kg}$ disk moves east at $4\,\text{m/s}$ and strikes a stationary $1\,\text{kg}$ disk. After the collision, the first disk moves at $30^\circ$ north of east with speed $v_1$ , and the second moves at $60^\circ$ south of east with speed $v_2$ . Find $v_1$ and $v_2$ .

Step 1 — Conserve $x$ -momentum. Initial: $p_x = 1(4) = 4\,\text{kg}\cdot\text{m/s}$ .

$4 = v_1\cos 30^\circ + v_2\cos 60^\circ = 0.866\,v_1 + 0.5\,v_2$

Step 2 — Conserve $y$ -momentum. Initial $p_y = 0$ (the second disk goes south, so its $y$ -component is negative):

$0 = v_1\sin 30^\circ - v_2\sin 60^\circ = 0.5\,v_1 - 0.866\,v_2$

Step 3 — Solve the system. From the $y$ -equation, $v_1 = \dfrac{0.866}{0.5}\,v_2 = 1.732\,v_2$ . Substitute into the -equation:

$4 = 0.866(1.732\,v_2) + 0.5\,v_2 = 1.5\,v_2 + 0.5\,v_2 = 2\,v_2$

So $v_2 = 2\,\text{m/s}$ and $v_1 = 1.732(2) \approx 3.46\,\text{m/s}$ .

🔑 The two final paths are $30^\circ + 60^\circ = 90^\circ$ apart. For an elastic collision of equal masses with one initially at rest, the outgoing velocities are always perpendicular.

Concept Check 🎯

m1+m2+⋯m1x1+m2x2+⋯

For continuous mass distributions:

$x_{\text{cm}} = \frac{1}{M} \int x\,dm$

Center of Mass Velocity

$v_{\text{cm}} = \frac{\sum m_i v_i}{M} = \frac{p_{\text{total}}}{M}$

🔑 The center of mass of an isolated system moves at constant velocity (even during collisions), because $\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}$ .

📝 Worked Example — Center of Mass of a Non-Uniform Rod

A thin rod of length $L$ lies along the $x$ -axis from $x = 0$ to $x = L$ . Its linear mass density increases as $\lambda(x) = \lambda_0\dfrac{x}{L}$ . Find the center of mass.

Step 1 — Set up the mass element. A slice of width $dx$ has mass $dm = \lambda(x)\,dx = \lambda_0\dfrac{x}{L}\,dx$ .

Step 2 — Total mass.

$M = \int_0^L \lambda_0\frac{x}{L}\,dx = \frac{\lambda_0}{L}\left[\frac{x^2}{2}\right]_0^L = \frac{\lambda_0 L}{2}$

Step 3 — Apply the center-of-mass integral.

$x_{\text{cm}} = \frac{1}{M}\int_0^L x\,dm = \frac{1}{M}\int_0^L x\left(\lambda_0\frac{x}{L}\right)dx = \frac{\lambda_0}{ML}\int_0^L x^2\,dx$

Step 4 — Evaluate.

$x_{\text{cm}} = \frac{\lambda_0}{ML}\cdot\frac{L^3}{3} = \frac{\lambda_0 L^2}{3M} = \frac{\lambda_0 L^2}{3}\cdot\frac{2}{\lambda_0 L} = \frac{2L}{3}$

🔑 The center of mass sits at $\dfrac{2L}{3}$ , shifted toward the dense end — exactly what intuition predicts when more mass is concentrated near $x = L$ .

Concept Check 🎯

Worked Example: Ballistic Pendulum

A bullet (mass $m = 0.01\,\text{kg}$ , speed $v_0 = 400\,\text{m/s}$ ) embeds in a block (mass $M = 2\,\text{kg}$ ) hanging from strings. How high does the block + bullet swing?

Step 1 (conservation of momentum during the collision):

$V = \frac{0.01 \times 400}{2.01} \approx 1.99\,\text{m/s}$

Step 2 (conservation of energy during the swing):

$\tfrac{1}{2}(m+M)V^2 = (m+M)gh$

$h = \frac{V^2}{2g} = \frac{(1.99)^2}{2(9.8)} \approx 0.20\,\text{m}$

📝 Worked Example — Variable Mass (the Rocket Equation)

A rocket ejects fuel at constant exhaust speed $u$ relative to itself. Starting from $\vec{F} = \dfrac{d\vec{p}}{dt}$ applied to the rocket-plus-fuel system in free space, we can derive how the rocket's speed grows.

Step 1 — Set up momentum conservation over a small time $dt$ . In $dt$ the rocket of mass $m$ expels $-dm$ of fuel (mass decreases, so $dm < 0$ ) at speed backward relative to the rocket. With no external force, total momentum is unchanged, which leads to:

$m\,dv = -u\,dm$

Step 2 — Separate variables and integrate.

$\int_{v_i}^{v_f} dv = -u\int_{m_i}^{m_f}\frac{dm}{m}$

Step 3 — Evaluate the integral.

$v_f - v_i = -u\big[\ln m\big]_{m_i}^{m_f} = u\,\ln\!\left(\frac{m_i}{m_f}\right)$

This is the Tsiolkovsky rocket equation. The thrust is $F_{\text{thrust}} = u\left|\dfrac{dm}{dt}\right|$ .

Numeric check: If $u = 2500\,\text{m/s}$ and the rocket burns from $m_i = 3000\,\text{kg}$ to $m_f = 1000\,\text{kg}$ , then .

🔑 Variable-mass problems require the general law $F = dp/dt$ — you cannot just use $F = ma$ with constant $m$ .

Concept Check 🎯

📝 Worked Example — Impulse–Momentum with Calculus

A $0.40\,\text{kg}$ ball traveling in $+x$ at $5\,\text{m/s}$ is struck so that a force $F_x(t) = (200 - 4000t)\,\text{N}$ acts on it while $t$ runs from $0$ to $0.05\,\text{s}$ . Find the ball's final velocity.

Step 1 — Compute the impulse by integration.

$J_x = \int_0^{0.05}(200 - 4000t)\,dt = \left[\,200t - 2000t^2\,\right]_0^{0.05}$

Step 2 — Evaluate.

$J_x = 200(0.05) - 2000(0.05)^2 = 10 - 5 = 5\,\text{N}\cdot\text{s}$

Step 3 — Apply the impulse-momentum theorem.

$J_x = m\,v_{xf} - m\,v_{xi}\;\Rightarrow\; v_{xf} = v_{xi} + \frac{J_x}{m} = 5 + \frac{5}{0.40} = 5 + 12.5 = 17.5\,\text{m/s}$

🔑 This single problem ties together the integral definition of impulse and the impulse-momentum theorem — a very common AP Physics C free-response combination.

Concept Check 🎯

v_f = \dfrac{J}{m} = \dfrac{4}{0.50} = 8\,\text{m/s}

K_f = \tfrac{1}{2}(6)(1)^2 = 3\,\text{J}

\Delta v = 2500\,\ln(3) \approx 2747\,\text{m/s}

[200t−2000t2]00.05