Edexcel IAL Physics: Further Mechanics — Revision Notes

Condensed for the final weeks. For the full explanation, use the Further Mechanics study guide.

Momentum in two dimensions

Momentum is a vector, so it is conserved independently in each perpendicular direction. Resolve into x and y, apply conservation separately, then recombine.

p = mv        impulse = Ft = delta-p

Elastic collisions conserve kinetic energy; inelastic do not. Momentum is conserved in both.

For a perfectly elastic collision, the relative speed of approach equals the relative speed of separation — a useful shortcut that avoids simultaneous equations.

Ek = p²/2m is a useful form of the kinetic energy equation when momentum, rather than velocity, is known directly — handy for checking whether a collision is elastic without first converting every momentum back to a velocity.

CORE PRACTICAL 9 investigates the relationship between force and change of momentum. CORE PRACTICAL 10 uses ICT (e.g. video analysis software) to analyse collisions between small spheres such as ball bearings on a table top — the standard practical basis for two-dimensional momentum questions.

Worked example: a 0.15 kg ball moving at 8.0 m/s collides head-on with a
stationary 0.25 kg ball. After the collision the first ball rebounds at
2.0 m/s. Find the velocity of the second ball, and state whether the
collision is elastic.

Momentum:    (0.15 x 8.0) = (0.15 x -2.0) + (0.25 x v)
             1.2 = -0.30 + 0.25v  ->  v = 6.0 m/s

KE before:   0.5 x 0.15 x 8.0^2 = 4.8 J
KE after:    0.5 x 0.15 x 2.0^2 + 0.5 x 0.25 x 6.0^2 = 0.3 + 4.5 = 4.8 J
             Kinetic energy conserved -> ELASTIC

Circular motion

omega = theta / t = 2 pi / T = 2 pi f       rad s^-1
v = r omega
a = v^2 / r = r omega^2
F = m v^2 / r = m r omega^2

Angles must be in radians.

Centripetal force is not a new force. It is the resultant of the real forces already present — tension, gravity, friction, normal contact, or magnetic force. Adding it as an extra arrow on a free-body diagram is the standard error.

Situation	Provided by
Ball on a string	Tension
Car on a flat bend	Friction
Planet in orbit	Gravitational attraction
Charged particle in a magnetic field	Magnetic force BQv
Car on a banked track	Horizontal component of the normal force

Why there is acceleration at constant speed: velocity is a vector, and its direction changes continuously, so the velocity changes even though the speed does not.

The force is always perpendicular to the velocity, so it does no work and kinetic energy is constant.

Vertical circles

top:     T + mg = mv^2/r    ->   T = mv^2/r - mg
bottom:  T - mg = mv^2/r    ->   T = mv^2/r + mg

Tension is greatest at the bottom and least at the top. Subtracting the two boxed equations gives a difference of 2mg only when the same speed v is substituted into both — a common simplification, not the real situation. In an actual vertical circle the object is faster at the bottom than at the top, and if energy is conserved, v_bottom^2 exceeds v_top^2 by 4gr (from ½mv_bottom^2 = ½mv_top^2 + mg(2r)). Substituting this back shows the true difference between the bottom and top tensions is 6mg, not 2mg.

Minimum speed at the top is when T = 0 and gravity alone supplies the centripetal force:

mg = mv^2/r    ->    v_min = sqrt(gr)

Below this the object leaves the circular path.

Exam traps

• Drawing centripetal force as an additional force rather than identifying its source.
• Using degrees instead of radians.
• Saying there is no acceleration because speed is constant.
• Claiming centripetal force does work.
• Forgetting that at the top of a vertical circle both tension and weight act towards the centre.
• Failing to resolve momentum into components in a two-dimensional collision.
• Converting every momentum back to a velocity before checking kinetic energy, when Ek = p²/2m gives the answer directly from momentum and mass alone.
• Confusing what each core practical investigates: Core Practical 9 is force and momentum change; Core Practical 10 is two-dimensional collision analysis by video.

Self-test

1. How is momentum conserved in a two-dimensional collision?
2. Give the two expressions for centripetal acceleration.
3. Why is an object in uniform circular motion accelerating?
4. Why does the centripetal force do no work?
5. Derive the minimum speed at the top of a vertical circle.
6. Write the kinetic energy equation in terms of momentum, and state when this form is most useful.
7. A 0.20 kg ball moving at 5.0 m/s collides with a stationary 0.30 kg ball; after the collision the first ball moves at 1.0 m/s in the same direction. Find the second ball’s velocity and state whether the collision is elastic.

Answers: 1. It is conserved independently in each of two perpendicular directions, so resolve into components and apply conservation to each separately. 2. a = v²/r and a = rω². 3. Velocity is a vector whose direction is continuously changing, so the velocity changes even at constant speed. 4. It acts perpendicular to the direction of motion, and work requires a component of force along the displacement. 5. At minimum speed the tension is zero, so weight alone provides the centripetal force: mg = mv²/r, giving v = √(gr). 6. Ek = p²/2m; most useful when momentum, rather than velocity, is already known. 7. Momentum: (0.20 × 5.0) = (0.20 × 1.0) + (0.30 × v) → 1.0 = 0.20 + 0.30v → v = 2.67 m/s. KE before = ½(0.20)(5.0)² = 2.5 J; KE after = ½(0.20)(1.0)² + ½(0.30)(2.67)² ≈ 0.1 + 1.07 = 1.17 J — kinetic energy is not conserved, so the collision is inelastic.