Newton's laws and momentum: VCE Physics Unit 2 Year 11

What this dot point is asking

VCAA wants you to apply Newton's three laws and the concept of momentum to motion problems, draw free-body diagrams, and use conservation of momentum in collision problems.

Newton's three laws

First law (inertia)

An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by a net external force.

Second law

$F_{\text{net}} = m a$. The net force on an object equals its mass times its acceleration. Force and acceleration are vectors in the same direction.

Third law

For every action, there is an equal and opposite reaction. If A exerts a force on B, then B exerts an equal-magnitude, opposite-direction force on A.

The forces in the third law act on different bodies; they do not cancel.

Forces commonly encountered

• Gravity. $W = mg$ where $g = 9.8$ m s$^{-2}$. Always downward.
• Normal force. Perpendicular to a surface, from the surface on the object.
• Friction. Parallel to a surface, opposing relative motion. $f \leq \mu N$ for static; $f = \mu_k N$ for kinetic.
• Tension. Along a rope or string. Always pulling.
• Applied force. Pushed or pulled by something external.

Free-body diagrams

Draw the object as a dot or simple shape. Show every force acting on the object as a vector arrow originating from the dot. Label each force.

Apply Newton's second law: sum forces vectorially (component by component), set equal to $ma$.

For an object on a surface with friction:

• Normal force balances gravity perpendicular to surface.
• Net force parallel to surface = applied minus friction = $ma$.

Momentum and impulse

Momentum $\vec{p} = m \vec{v}$. Vector quantity, in the direction of velocity.

Impulse $\vec{J} = \vec{F} \Delta t = \Delta \vec{p}$. The impulse on an object equals the change in its momentum.

The impulse-momentum theorem follows from Newton's second law: $F = ma = m \Delta v / \Delta t = \Delta p / \Delta t$, so $F \Delta t = \Delta p$.

Force-time graph

The area under a force-time graph equals the impulse (= change in momentum).

For non-constant forces, integrate (or estimate area under the curve graphically).

Conservation of momentum

In an isolated system (no external forces), total momentum is conserved.

For a collision between objects 1 and 2:

where $u$ = before-collision velocity, $v$ = after-collision velocity.

Elastic vs inelastic collisions

Elastic collision

Both momentum and kinetic energy are conserved. Examples: collisions between hard balls (close to elastic in practice).

Inelastic collision

Momentum conserved; kinetic energy not (some lost to heat, sound, deformation).

Perfectly inelastic

The two objects stick together after collision. Maximum kinetic energy loss for given initial conditions.

Worked example: elastic collision

A 2 kg ball at 5 m/s collides elastically with a stationary 3 kg ball. Find final velocities.

Conservation of momentum: $2(5) + 3(0) = 2 v_1 + 3 v_2$, so $10 = 2 v_1 + 3 v_2$.

Conservation of KE: $\frac{1}{2}(2)(5)^2 = \frac{1}{2}(2) v_1^2 + \frac{1}{2}(3) v_2^2$, so $25 = v_1^2 + 1.5 v_2^2$.

From momentum: $v_2 = (10 - 2 v_1)/3$. Substitute and solve.

Result: $v_1 = -1$ m/s (ball reverses direction); $v_2 = 4$ m/s.

Check KE: $\frac{1}{2}(2)(1)^2 + \frac{1}{2}(3)(16) = 1 + 24 = 25$ J. Conserved.

Applications

Car safety

Crumple zones extend the time of a collision, reducing the force (impulse-momentum theorem: $F = \Delta p / \Delta t$). Larger $\Delta t$ means smaller $F$.

Rocket propulsion

Reaction mass expelled backwards gives forward momentum to the rocket (Newton's third law / conservation of momentum).

Sports

Follow-through in tennis or golf extends contact time to deliver more impulse.

Examples in context

Example 1. Melbourne Port container crane snatch load. A Patrick Terminals ship-to-shore crane at the Port of Melbourne lifts a $30$ tonne container vertically. If the cable accelerates the container from rest to $1.5$ m s$^{-1}$ in $1.0$ s, Newton's second law gives tension $T - mg = ma$, so $T = m(g + a) = 30\,000 \times (9.8 + 1.5) = 3.39 \times 10^5$ N. Compared with the static weight of $2.94 \times 10^5$ N, this is a $15$% dynamic-load increase that drives the cable rating (minimum breaking load of around $10$ times working load). The impulse during the $1.0$ s acceleration is $\Delta p = m\Delta v = 30\,000 \times 1.5 = 4.5 \times 10^4$ N s.

Example 2. AFL kick-after-mark at the MCG. An AFL ball ($450$ g) is kicked from rest to a launch speed of $25$ m s$^{-1}$ in a contact time of $0.012$ s. Newton's second law in impulse form gives $F\Delta t = m\Delta v$, so $F = 0.45 \times 25 / 0.012 = 938$ N, comparable to the player's body weight. After the kick, the ball's momentum is $p = mv = 0.45 \times 25 = 11.25$ kg m s$^{-1}$. Conservation of momentum in player-ball interaction would predict a tiny recoil of the kicker's body (mass $90$ kg) of order $11.25/90 \approx 0.13$ m s$^{-1}$, absorbed mostly by the planted foot via friction with the turf.

Try this

Q1. State Newton's second law and define impulse. [2 marks]

• Cue. $\vec{F}_{\rm net} = m\vec{a}$ for constant mass; impulse $\vec{J} = \vec{F}\Delta t = \Delta \vec{p}$.

Q2. A $1200$ kg car travelling at $20$ m s$^{-1}$ brakes to rest in $5.0$ s. Calculate (a) the deceleration, (b) the average braking force, and (c) the impulse delivered. [5 marks]

• Cue. (a) $a = -4.0$ m s$^{-2}$. (b) $F = ma = -4800$ N. (c) $\Delta p = -1200 \times 20 = -2.4 \times 10^4$ N s.

Q3. Refer to the Port of Melbourne crane example. (a) Calculate the tension during acceleration. (b) Determine the impulse delivered. (c) Explain why dynamic loading affects cable rating. [2+2+2 marks]

• Cue. (a) $3.39 \times 10^5$ N. (b) $4.5 \times 10^4$ N s. (c) Cables must handle peak transient load greater than static weight.