Momentum & Collisions

Momentum & Collisions

Part 1 of 4 — Momentum & Impulse

Momentum: Inertia in Motion

$p = mv$

Momentum is a vector (direction matters — assign signs on a line) with units $\text{kg}\cdot\text{m/s}$. A slow truck and a fast baseball can carry the same momentum; what differs is how hard each is to stop quickly.

Impulse: How Momentum Changes

$J = F\Delta t = \Delta p = m\Delta v$

Impulse (units $\text{N}\cdot\text{s}$, identical to $\text{kg}\cdot\text{m/s}$) is force integrated over time. On a force-time graph, impulse is the area under the curve — the exact analog of displacement under a $v$-$t$ curve.

The rearrangement the MCAT tests relentlessly:

$F = \frac{\Delta p}{\Delta t}$

For a FIXED momentum change, force and collision time are inversely related. Stretch out $\Delta t$ and the force shrinks:

• Airbags and crumple zones extend the stopping time of your body, cutting peak force
• Bending your knees on landing extends $\Delta t$
• Following through on a swing lengthens contact time, boosting $\Delta p$ for a given force

None of these change the momentum you must lose — only how violently you lose it.

Sign Care: Reversals Count Double

An object that bounces changes momentum MORE than one that stops. Coming in at $+mv$ and leaving at $-mv$ gives $|\Delta p| = 2mv$, versus $mv$ for sticking. Walls hit by bouncy balls feel larger impulses than walls hit by clay.

Worked Example — Bat Meets Ball

A $0.15\;\text{kg}$ baseball arrives at $40\;\text{m/s}$ and leaves the bat at $30\;\text{m/s}$ in the opposite direction. Bat-ball contact lasts $5\;\text{ms}$.

Step 1 — Set a sign convention. Let the ball's outgoing direction be positive. Then $v_0 = -40\;\text{m/s}$ and $v_f = +30\;\text{m/s}$.

Step 2 — Momentum change.

$\Delta p = m(v_f - v_0) = 0.15(30 - (-40)) = 0.15(70) = 10.5\;\text{kg}\cdot\text{m/s}$

The reversal makes the speeds ADD — treating it as $0.15(40 - 30) = 1.5$ is the classic sign error, off by a factor of seven.

Step 3 — Average force.

$F = \frac{\Delta p}{\Delta t} = \frac{10.5}{0.005} = 2100\;\text{N}$

Roughly $1400\times$ the ball's own weight ($1.5\;\text{N}$) — collision forces dwarf gravity, which is why we can usually ignore gravity during the brief contact.

Step 4 — Airbag contrast. If the same $10.5\;\text{kg}\cdot\text{m/s}$ were removed over $0.5\;\text{s}$ instead (a hundred times longer), the force would drop a hundredfold to $21\;\text{N}$. Same impulse, same shaded area on the $F$-$t$ graph — just low and wide instead of tall and narrow.

Momentum & Impulse 🎯

Key Takeaways — Part 1

• $p = mv$ is a vector — set a sign convention before touching numbers
• Impulse $J = F\Delta t = \Delta p$; on an $F$-$t$ graph it is the area under the curve
• Fixed $\Delta p$: force and stopping time trade off inversely (airbags, crumple zones, bent knees)
• Bouncing doubles the momentum change relative to sticking ($2mv$ vs. $mv$)
• Collision forces dwarf gravity, so gravity is negligible during brief impacts
• Reversal problems: speeds ADD in $\Delta v$ when direction flips

Momentum & Collisions

Part 2 of 4 — Conservation of Momentum & Collision Types

Why Momentum Is Conserved

During a collision, the objects exert third-law force pairs on each other — equal, opposite, and acting for the same contact time. The impulses they trade are equal and opposite, so the momentum one gains the other loses. If no net external force acts on the system:

$p_{total,\;before} = p_{total,\;after}$

$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$

This holds for EVERY collision — sticky, bouncy, or explosive. Friction and gravity are usually negligible over the brief contact, which is why "conservation of momentum" is the default tool the instant two things hit.

The Collision Taxonomy (High Yield)

The discrimination the MCAT tests: momentum is conserved in all three; kinetic energy only in elastic collisions. "Perfectly inelastic" does not mean all KE vanishes — the stuck pair usually still moves (momentum demands it). It means the most KE is lost that momentum conservation allows.

Truly elastic collisions are rare macroscopically (billiard balls approximately; gas molecules genuinely). Car crashes, tackles, and bullets embedding in blocks are perfectly inelastic.

One Special Case Worth Memorizing

Equal masses, head-on elastic collision, one initially at rest: they exchange velocities. The mover stops dead; the target departs at the incoming speed — the Newton's-cradle result.

Worked Example — A Perfectly Inelastic Collision, Fully Audited

A $2\;\text{kg}$ cart moving at $6\;\text{m/s}$ hits a stationary $4\;\text{kg}$ cart and couples to it.

Step 1 — Momentum before.

$p = (2)(6) + (4)(0) = 12\;\text{kg}\cdot\text{m/s}$

Step 2 — Final velocity of the pair.

$v' = \frac{p}{m_{total}} = \frac{12}{6} = 2\;\text{m/s}$

Step 3 — Kinetic energy audit.

$KE_{before} = \tfrac{1}{2}(2)(6)^2 = 36\;\text{J}, \qquad KE_{after} = \tfrac{1}{2}(6)(2)^2 = 12\;\text{J}$

Two-thirds of the kinetic energy ($24\;\text{J}$) became heat, sound, and deformation. Momentum survived intact; kinetic energy did not — the defining fingerprint of an inelastic collision.

The classic trap: setting $\tfrac{1}{2}m_1v_1^2 = \tfrac{1}{2}(m_1+m_2)v'^2$ for a sticking collision. That equation assumes elastic behavior and gives $v' = \sqrt{12} \approx 3.5\;\text{m/s}$ — wrong. When objects stick, momentum is the ONLY conserved quantity you may use across the impact.

Sanity habit: the stuck pair must move slower than the incoming object (mass grew, momentum did not), and its velocity must lie between the two initial velocities.

Conservation & Collision Types 🎯

Key Takeaways — Part 2

• With no external force, total momentum is conserved in EVERY collision — the third law guarantees it
• Kinetic energy is conserved only in elastic collisions; perfectly inelastic (sticking) loses the maximum KE allowed
• Perfectly inelastic: one final velocity, $v' = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$
• Never write a KE-conservation equation across a sticking collision — momentum only
• Equal masses, elastic, target at rest: velocities exchange (Newton's cradle)
• Push-offs and recoils are collisions run backward: total momentum stays zero

Momentum & Collisions

Part 3 of 4 — Recoil, Explosions & the Ballistic Pendulum

Explosions: Collisions Run Backward

A system at rest has zero total momentum — and keeps it, even when it blows apart. Rifles recoiling, astronauts throwing wrenches, radioactive nuclei emitting particles: in every case the fragments' momenta must sum to zero:

$m_1 v_1 = m_2 v_2 \quad \text{(opposite directions)}$

The lighter fragment always moves faster, in inverse proportion to its mass. Kinetic energy is NOT conserved in an explosion — it is created from chemical or nuclear potential energy. Momentum bookkeeping still works perfectly.

The Ballistic Pendulum: Two Regimes, Two Laws

A bullet embeds in a hanging block, which then swings upward. This is the MCAT's favorite momentum problem because it forces you to switch tools mid-problem:

1. The collision (bullet embeds): perfectly inelastic. Use momentum conservation only — kinetic energy is destroyed wholesale here.
2. The swing (block rises): no collision anymore, just gravity acting on a moving pendulum. Use energy conservation: $\tfrac{1}{2}(M+m)v'^2 = (M+m)gh$.

The trap — equating the bullet's initial KE to the final potential energy $(M+m)gh$ — skips the collision's enormous energy loss and wildly overestimates the height (or underestimates the bullet speed). Draw a mental fence at the moment of impact: momentum crosses the fence, kinetic energy does not.

Two Dimensions: Conserve Each Component

Momentum is a vector, so in 2-D collisions, $p_x$ and $p_y$ are conserved independently. A car heading east and a car heading north that lock together leave the crash moving northeast — each component of the wreck's momentum equals the corresponding incoming component.

Worked Example — Ballistic Pendulum, Start to Finish

A $0.02\;\text{kg}$ bullet strikes a $1.98\;\text{kg}$ block hanging from strings and embeds in it. The block-plus-bullet swings up $0.2\;\text{m}$. Find the bullet's speed. ($g = 10\;\text{m/s}^2$)

Step 1 — Work the swing BACKWARD with energy conservation.

$\tfrac{1}{2}v'^2 = gh \Rightarrow v' = \sqrt{2gh} = \sqrt{2(10)(0.2)} = \sqrt{4} = 2\;\text{m/s}$

This is the speed of the $2\;\text{kg}$ combination just after impact.

Step 2 — Work the collision with momentum conservation.

$mv = (M+m)v' \Rightarrow (0.02)v = (2)(2) \Rightarrow v = \frac{4}{0.02} = 200\;\text{m/s}$

Step 3 — Audit the energy (the point of the problem).

$KE_{bullet} = \tfrac{1}{2}(0.02)(200)^2 = 400\;\text{J}, \qquad KE_{after} = \tfrac{1}{2}(2)(2)^2 = 4\;\text{J}$

99% of the bullet's kinetic energy became heat and wood-splitting. Had you equated $400\;\text{J}$ to $(M+m)gh$, you would predict a rise of $20\;\text{m}$ — a hundred times the real $0.2\;\text{m}$.

Worked Example — Rifle Recoil

A $4\;\text{kg}$ rifle fires a $0.01\;\text{kg}$ bullet at $400\;\text{m/s}$:

$v_{rifle} = \frac{(0.01)(400)}{4} = 1\;\text{m/s}$

Same magnitude of momentum, but the bullet carries $\frac{1}{2}(0.01)(400)^2 = 800\;\text{J}$ against the rifle's $2\;\text{J}$ — the lighter partner always hoards the kinetic energy ($KE = p^2/2m$ at fixed $p$).

Recoil & the Ballistic Pendulum 🎯

Key Takeaways — Part 3

• Explosions and push-offs conserve momentum around zero: $m_1v_1 = m_2v_2$, opposite directions
• The lighter fragment moves faster AND carries more kinetic energy ($KE = p^2/2m$)
• Ballistic pendulum = two fenced-off regimes: momentum across the collision, energy across the swing
• Never carry kinetic energy across an embedding impact — most of it dies there
• Head-on collisions: assign signs first; momenta subtract when directions oppose
• 2-D collisions: conserve $p_x$ and $p_y$ separately, as independent equations

Momentum & Collisions

Part 4 of 4 — MCAT Integration: Center of Mass, 2-D Collisions & Passage Strategy

Center of Mass: The System's Representative Point

$x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$

The center of mass sits closer to the heavier object, and it moves as if the total mass were concentrated there with only EXTERNAL forces acting. Internal forces — collisions, explosions, people pulling ropes — cannot budge it:

• A shell explodes mid-flight: the fragments scatter, but their center of mass continues along the original parabola (until pieces land)
• Two skaters pull on a rope between them: they meet AT the center of mass, which never moves
• A person walks toward the bow of a free-floating canoe: the canoe slides backward so the system's center of mass stays put

This is momentum conservation restated: constant total momentum means constant center-of-mass velocity.

Choosing the Tool (the Real MCAT Skill)

Impulse in Physiology and Safety

Force plates in gait labs record the $F$-$t$ curve of each footfall; the area is the impulse, fixed by the runner's momentum change. Cushioned shoes, bent knees, and gymnastic mats cannot alter that area — they stretch it in time, lowering the peak force that bone and cartilage must bear. Passage answers about padding, airbags, or landing technique are all the same sentence: same $\Delta p$, longer $\Delta t$, smaller peak $F$.

Worked Example — A 2-D Perfectly Inelastic Crash

A $1000\;\text{kg}$ car heading east at $20\;\text{m/s}$ collides at an intersection with a $1000\;\text{kg}$ car heading north at $20\;\text{m/s}$. They lock together.

Step 1 — Conserve each component separately.

$p_x = (1000)(20) = 2 \times 10^4\;\text{kg}\cdot\text{m/s (east)}$

$p_y = (1000)(20) = 2 \times 10^4\;\text{kg}\cdot\text{m/s (north)}$

Step 2 — Final velocity components. Total mass $2000\;\text{kg}$:

$v_x = 10\;\text{m/s}, \qquad v_y = 10\;\text{m/s}$

Step 3 — Direction and speed. Equal components mean $45^\circ$ north of east, with speed

$v = \sqrt{10^2 + 10^2} = 10\sqrt{2} \approx 14\;\text{m/s}$

Note the wreck does NOT leave at $20\;\text{m/s}$: each car's momentum had to be shared with the other car's mass in a direction it was not moving.

Worked Example — Center of Mass Bookkeeping

A $60\;\text{kg}$ skater and a $40\;\text{kg}$ skater, $10\;\text{m}$ apart on frictionless ice, pull on a rope. Where do they meet?

$x_{cm} = \frac{(60)(0) + (40)(10)}{100} = 4\;\text{m from the heavier skater}$

No matter who pulls harder or when, the meeting point is fixed at the center of mass — internal forces cannot move it. The heavier skater travels $4\;\text{m}$; the lighter one travels $6\;\text{m}$: displacements in inverse ratio to mass, exactly like recoil speeds.

MCAT Integration: Momentum 🎯

Key Takeaways — Part 4

• The center of mass moves under external forces only; collisions, explosions, and rope-pulling cannot shift it
• Exploding projectile: the fragments' center of mass finishes the original parabola
• 2-D collisions: conserve $p_x$ and $p_y$ as separate equations; sticking crashes leave at the momentum-vector angle
• Tool choice is the tested skill: "stick" means momentum only; "elastic" adds KE; "how high after" chains momentum into energy; "average force" means impulse
• Padding and cushioning never reduce $\Delta p$ — they stretch $\Delta t$ to shrink peak force
• Displacements toward a shared center of mass scale inversely with mass, just like recoil speeds