PHY131H1/Unit 7

Unit 7 · Introduction to Physics I

Momentum & Collisions

The second conserved quantity, and the one that handles collisions. Momentum survives every collision without exception; kinetic energy usually does not. Knowing which is conserved when is the entire unit.

By the end you can

• Relate impulse to change in momentum.
• Say precisely when momentum is conserved.
• Classify collisions as elastic, inelastic or perfectly inelastic.
• Solve two-dimensional collisions by components.

Why it matters & where it comes from

Why it exists

During a collision the forces are enormous, brief and unknown. Momentum conservation sidesteps them entirely: you need only the states before and after, never what happened in the microseconds between.

In the real world

Crumple zones extend the collision time, which reduces the force for the same momentum change. Rockets work by momentum conservation and need nothing to push against, which is why they function in vacuum.

Story & open questions

Descartes proposed a conserved "quantity of motion" but used speed rather than velocity, so his version failed for head-on collisions. Huygens fixed it in the 1660s by making it a vector, and the sign is what makes the whole thing work.

Momentum and impulse

Definition

Momentum is $\vec{p} = m\vec{v}$ — a vector, in the direction of motion, measured in kg·m/s.

Impulse is $\vec{J} = \vec{F}\Delta t$ for a constant force, or the area under a force-time graph in general.

Theorem Impulse-momentum theorem

$\vec{J} = \Delta\vec{p}$: impulse equals change in momentum.

This is Newton's second law rewritten. From $F = ma = m\Delta v/\Delta t$, multiplying through by $\Delta t$ gives $F\Delta t = m\Delta v = \Delta p$.

Example Why crumple zones work

A $1200$ kg car at $20$ m/s stops in a crash. The momentum change is fixed at $24{,}000$ kg·m/s whatever happens.

Rigid car, stopping in $0.1$ s: $F = 24{,}000/0.1 = 240{,}000$ N.

Crumple zone, stopping in $0.5$ s: $F = 48{,}000$ N — a fifth as much.

The impulse is the same; spreading it over more time reduces the peak force. Airbags, helmets, landing mats and bending your knees all exploit exactly this.

Conservation of momentum

Rule When it holds

The total momentum of a system is conserved when the net external force is zero:

$\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}$

Internal forces cannot change total momentum, because by the third law they come in equal and opposite pairs that cancel in the sum.

During a collision, external forces are negligible

Gravity and friction act during a car crash, but the collision forces are thousands of times larger and last only milliseconds. So momentum is treated as conserved through the collision even though it is not conserved over the following seconds.

Example Recoil

A $60$ kg skater throws a $2$ kg ball at $10$ m/s. How fast does she recoil?

Before, total momentum is zero. After, it must still be zero.

$0 = (2)(10) + (60)v$, so $v = -0.33$ m/s — opposite to the throw.

This is how rockets work. Nothing is pushed against; the exhaust carries momentum one way and the rocket the other.

Types of collision

Definition

Elastic. Both momentum and kinetic energy are conserved. Rare in practice — billiard balls and atoms come close.

Inelastic. Momentum conserved, kinetic energy not. Most real collisions.

Perfectly inelastic. The bodies stick together and move as one. This loses the most kinetic energy possible while still conserving momentum.

Rule Momentum always, energy sometimes

Momentum is conserved in every collision. Kinetic energy is conserved only in elastic ones. Lost kinetic energy becomes heat, sound and permanent deformation.

Example Perfectly inelastic

A $2$ kg cart at $6$ m/s hits a stationary $4$ kg cart and they stick.

Momentum: $(2)(6) = (6)v$, so $v = 2$ m/s.

Kinetic energy before: $\tfrac12(2)(36) = 36$ J. After: $\tfrac12(6)(4) = 12$ J.

Two thirds lost — and yet momentum is exactly conserved. The two quantities behave completely differently.

Example Elastic, equal masses

A moving ball strikes an identical stationary one elastically, head-on.

Momentum: $mv = mv_1 + mv_2$. Energy: $v^2 = v_1^2 + v_2^2$.

The only solution with the balls not passing through each other is $v_1 = 0$, $v_2 = v$: the first stops dead and the second leaves at the original speed.

This is the Newton's cradle result, and it needs both conservation laws. Momentum alone would allow them to move off together.

Example Two dimensions

Momentum conserves separately along each axis.

A $3$ kg puck at $4$ m/s east strikes a $2$ kg puck at rest; the first deflects $30$ degrees north of east at $2$ m/s. Find the second puck's velocity.

East: $(3)(4) = (3)(2\cos 30) + 2v_{2x}$, so $v_{2x} = (12 - 5.2)/2 = 3.4$ m/s.

North: $0 = (3)(2\sin 30) + 2v_{2y}$, so $v_{2y} = -1.5$ m/s — south, as it must be to cancel the first puck's northward momentum.

Common pitfalls

• Assuming kinetic energy is conserved. Only in elastic collisions.
• Dropping the sign on momentum. It is a vector; opposite directions must carry opposite signs.
• Adding momenta in two dimensions as magnitudes. Resolve into components first.
• Forgetting that "sticking together" is a physical statement giving both bodies the same final velocity — that is the extra equation.
• Applying conservation when a large external force acts, such as a ball bouncing off a wall bolted to the Earth.

Practice

Work each fully before opening the answer. Original exercises written for this guide in the style of the course's problem sets.

Practice 1 · Impulse

A $0.15$ kg ball arrives at $30$ m/s and leaves at $20$ m/s the other way. Find the impulse.

Answer

$\Delta p = 0.15(-20 - 30) = -7.5$ kg·m/s. The sign change is the whole point — this is larger than for merely stopping the ball.

Practice 2 · Sticking carts

A $1$ kg cart at $4$ m/s hits a stationary $3$ kg cart and sticks. Find the final speed.

Answer

$v = 4/4 = 1$ m/s.

Practice 3 · Energy lost

How much kinetic energy did that collision lose?

Answer

Before $\tfrac12(1)(16) = 8$ J, after $\tfrac12(4)(1) = 2$ J. Six joules lost.

Practice 4 · Recoil

A $5$ kg rifle fires a $0.01$ kg bullet at $600$ m/s. Find the recoil speed.

Answer

$v = -(0.01)(600)/5 = -1.2$ m/s.

Practice 5 · Which is conserved

Two lumps of clay collide and stick. Which quantities are conserved?

Answer

Momentum, yes. Kinetic energy, no — this is perfectly inelastic and loses the maximum possible. Total energy is conserved, with the difference appearing as heat.

Practice 6 · Ballistic pendulum

A $0.02$ kg bullet embeds in a $2$ kg block hanging on a string, which then swings up $0.1$ m. Find the bullet's speed.

Solution

Two stages, and they need different laws. The swing is energy: $v_{\text{after}} = \sqrt{2(9.8)(0.1)} = 1.4$ m/s. The embedding is momentum: $0.02v = (2.02)(1.4)$, so $v = 141$ m/s. Using energy for the collision would be wrong, since energy is lost there.

Practice 7 · Two dimensions

Why must you resolve into components?

Answer

Because momentum is a vector and is conserved independently along each axis. Adding magnitudes would lose the directional information the problem depends on.

Practice 8 · Explosion

A stationary $6$ kg object explodes into a $2$ kg piece at $9$ m/s and a $4$ kg piece. Find the second piece's velocity.

Solution

Total momentum stays zero: $0 = (2)(9) + 4v$, so $v = -4.5$ m/s, opposite. Kinetic energy increased from zero — supplied by the explosive — which shows energy need not be conserved even when momentum is.

Expanded topic checklist

Course coverage. This unit is part of the complete PHY131H1 topic map. It should cover the following items before you move on:

• Momentum and impulse. Definitions and the impulse-momentum theorem.
• Conservation. The condition for it, and why internal forces cannot break it.
• Collision types. Elastic, inelastic, perfectly inelastic, and what each conserves.
• Two dimensions. Independent conservation along each axis.

Textbook/source alignment

The calendar lists "momentum" among the core topics. Check the syllabus on Quercus for assigned problems; the ballistic pendulum, needing both conservation laws in sequence, is a standard exam question.

Core idea map

Impulse changes momentum; with no external force, total momentum holds. Energy may or may not. Problems combining a collision with a swing or slide need momentum for the collision and energy for the rest.

Study routine. For every collision problem, write down which quantities are conserved before solving. Half the marks are lost by applying energy conservation to an inelastic collision.

Deep practice extensions

Use these after the main practice set. They push on definitions and edge cases rather than routine calculation.

Extension 1 · Always conserved

Which quantity is conserved in every collision?

Guide

Momentum, provided the net external force is negligible. Kinetic energy only in elastic collisions.

Extension 2 · Reducing force

How does extending collision time help?

Guide

The impulse is fixed by the momentum change, so a longer time means a smaller average force.

Extension 3 · Maximum loss

Which collision loses the most kinetic energy?

Guide

Perfectly inelastic, where the bodies stick. Momentum conservation sets a floor below which the kinetic energy cannot fall.

Extension 4 · Internal forces

Why can internal forces not change total momentum?

Guide

By the third law they occur in equal and opposite pairs, so they cancel when summed over the whole system.

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