Title: InkMaths / Further Mechanics 1: impulse, energy and collisions
On every note of this deck, unless a note says otherwise:
Footer: Continue on InkMaths · © 2026 InkMaths. All rights reserved.

Front: STANDARD RESULT — LEARN IT
Bouncing ball, dropped from height h:
height after one bounce? total distance until it stops?\(\text{height after one bounce?} \text{ total distance until it stops?}\)height after one bounce? total distance until it stops?
Back: e^2h, h(1 + e^2)1 - e^2\(e^{2}h, \frac{h(1 + e^{2})}{1 - e^{2}}\)e^2h, ((h(1 + e^2))/(1 - e^2))

Front: STANDARD RESULT — LEARN IT
Work and power, θ between force and displacement:
work = ?, P = ?\(\text{work} = ?, P = ?\)work = ?, P = ?
Back: Fd cos θ, Fv\(Fd \cos θ, Fv\)Fd cos θ, Fv

Front: DEFINE
Define: conservation of momentum.
Back: In a collision the internal forces are equal and opposite for the same time, so the impulses cancel and the total momentum before equals the total after. It holds whatever happens to the energy.

Front: STANDARD RESULT — LEARN IT
Impulse:
I = ?\(\text{I} = ?\)I = ?
Back: mv - mu, equal to Ft for a constant force. A vector, in N s\(m\text{v} - m\text{u}, \text{ equal to Ft for a constant force. A vector, in N s}\)m v - m u, equal to Ft for a constant force. A vector, in N s

Front: SPOT THE ERROR
Say what is wrong with this claim:
A ball that bounces back off a wall at the speed it arrived has received no impulse.
Back: Momentum is a vector. Reversing it removes the original momentum and supplies the same again the other way, so the impulse is the mass times the sum of the two speeds, not their difference.

Front: CHECK YOURSELF
Why is momentum not conserved when a ball bounces off the floor?
Back: The floor is attached to the Earth, so the ball alone is not an isolated system and receives an external impulse. Only the law of restitution applies.

Front: STANDARD RESULT — LEARN IT
Hooke's law:
T = ?\(T = ?\)T = ?
Back: λ xl, x the extension and l the natural length\(\frac{λ x}{l}, \text{ x the extension and l the natural length}\)((λ x)/l), x the extension and l the natural length

Front: RECALL
State how impulse works in two dimensions.
Back: The i components and the j components each obey I = mv − mu on their own. Take magnitudes only at the end: a change of direction at constant speed still needs an impulse.

Front: RECALL
State what happens to momentum and kinetic energy in an impact.
Back: Momentum is conserved when no outside impulse acts, which covers two spheres meeting but not a bounce off a wall; kinetic energy is lost unless e = 1. Find the loss by adding ½mv² for both spheres before and after and subtracting.

Front: RECALL
State the rule for oblique impact on a smooth surface.
Back: A smooth surface leaves the parallel component untouched and multiplies the perpendicular one by −e. So unless e = 1 the outgoing path hugs the surface more closely than the incoming one, and the angle to the surface shrinks.

Front: DEFINE
Define: work-energy principle.
Back: The total work done by all the forces equals the change in kinetic energy. The normal reaction never contributes, since it acts at right angles to the motion.

Front: STANDARD RESULT — LEARN IT
Elastic energy:
EPE = ?\(\text{EPE} = ?\)EPE = ?
Back: λ x^22l, the triangle under the Hooke line\(\frac{λ x^{2}}{2l}, \text{ the triangle under the Hooke line}\)((λ x^2)/2l), the triangle under the Hooke line

Front: STANDARD RESULT — LEARN IT
Newton's law of restitution:
e = ?, range?\(e = ?, \text{range}?\)e = ?, range?
Back: separation speedapproach speed, 0 ≤ e ≤ 1\(\frac{\text{separation speed}}{\text{approach speed}}, 0 \le e \le 1\)(separation speed/approach speed), 0 ≤ e ≤ 1