10 questions · 1 mark each · Correct answer highlighted with explanation
10 questions · 3 marks each · Full model answers included
The work-energy theorem states that the work done by the net force on a body equals the change in its kinetic energy.
Using $v^2 = u^2 + 2as \Rightarrow as = \dfrac{v^2-u^2}{2}$. Multiplying both sides by mass $m$:
$mas = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$
Since $F = ma$, the left side is $Fs = W$ (work done). Therefore: $W = KE_f - KE_i$
Conservative forces do work that depends only on the initial and final positions, not on the path taken — the work done in a closed path is zero. Example: gravitational force, spring force.
Non-conservative forces do work that depends on the actual path taken, and the work done over a closed path is non-zero — mechanical energy is not conserved when they act. Example: friction, air resistance.
The restoring force in a spring stretched by $y$ is $F = -ky$ (Hooke's law); the applied force needed to stretch it further is $F_{applied} = ky$.
Small work done to stretch it by $dy$: $dW = ky\,dy$
Total work (= stored PE) to stretch from 0 to $x$: $U = \displaystyle\int_0^x ky\,dy = \left[\dfrac{ky^2}{2}\right]_0^x = \dfrac{1}{2}kx^2$
The law states that the total mechanical energy (kinetic + potential) of a system remains constant, provided only conservative forces do work on it.
$KE_i + PE_i = KE_f + PE_f$
This holds specifically when non-conservative forces like friction or air resistance are absent (or negligible), since such forces dissipate mechanical energy as heat, sound, or other non-mechanical forms, causing total mechanical energy to decrease.
Elastic collision: Both momentum and kinetic energy are conserved. The bodies separate after collision without any permanent deformation or heat loss.
Inelastic collision: Momentum is conserved, but kinetic energy is not — some KE converts into heat, sound, or deformation energy. In a perfectly inelastic collision, the colliding bodies stick together and move with a common velocity, losing the maximum possible kinetic energy consistent with momentum conservation.
Kinetic energy: $KE = \dfrac{1}{2}mv^2$. Momentum: $p = mv \Rightarrow v = \dfrac{p}{m}$
Substituting: $KE = \dfrac{1}{2}m\left(\dfrac{p}{m}\right)^2 = \dfrac{p^2}{2m}$
This shows KE is directly proportional to the square of momentum for a given mass.
Power is the rate at which work is done, or the rate at which energy is transferred: $P = \dfrac{dW}{dt}$.
Since $dW = F\,ds\cos\theta$ (work done for small displacement $ds$), for force and velocity in the same direction:
$P = \dfrac{F\,ds}{dt} = F\dfrac{ds}{dt} = Fv$
More generally, for force and velocity vectors, $P = \vec{F}\cdot\vec{v}$.
In uniform circular motion, the centripetal force always acts radially inward, towards the centre of the circle, while the instantaneous velocity (and hence the direction of displacement) is always tangential to the circle.
Since the centripetal force is always perpendicular to the direction of motion ($\theta = 90°$ at every instant), and $W = Fs\cos90° = 0$, the work done by the centripetal force is always zero — consistent with the fact that speed (and hence kinetic energy) remains constant in uniform circular motion.
Efficiency is the ratio of useful output energy (or work) to the total input energy, usually expressed as a percentage: $\eta = \dfrac{\text{Output energy}}{\text{Input energy}} \times 100\%$
Efficiency of any real machine is always less than 100% because some input energy is inevitably lost to friction, heat, sound, or other dissipative effects during operation — by the law of conservation of energy, total energy is conserved, but not all of it converts to the intended useful output.
Consider a body of mass $m$ falling from height $h$. At height $h$ (start): $KE=0$, $PE=mgh$, Total $E = mgh$.
At height $(h-x)$, after falling distance $x$: velocity $v^2 = 2gx$ (from $v^2=u^2+2as$), so $KE = \frac{1}{2}mv^2 = \frac{1}{2}m(2gx) = mgx$, and $PE = mg(h-x)$.
Total energy = $mgx + mg(h-x) = mgx + mgh - mgx = mgh$
Since total energy remains $mgh$ at every point during the fall, mechanical energy is conserved.
10 questions · 5 marks each · Detailed step-by-step solutions
$W = Fs\cos\theta = 20 \times 10 \times \cos60°$
$\cos60° = 0.5$
$W = 20 \times 10 \times 0.5 = \mathbf{100\text{ J}}$
Work done by gravity as the body falls: $W = mgh = 2 \times 10 \times 10 = 200\text{ J}$
By the work-energy theorem, this work equals the gain in kinetic energy (initial KE = 0, since dropped from rest):
$W = \dfrac{1}{2}mv^2 \Rightarrow 200 = \dfrac{1}{2}(2)v^2 = v^2$
$v = \sqrt{200} \approx 14.14\text{ m/s}$
Work done against gravity (per minute) = $mgh = 600 \times 10 \times 25 = 150000\text{ J}$
Kinetic energy imparted (per minute) = $\frac{1}{2}mv^2 = \frac{1}{2}(600)(50)^2 = \frac{1}{2}(600)(2500) = 750000\text{ J}$
Total work per minute = $150000 + 750000 = 900000\text{ J}$
Power = Total work / time = $\dfrac{900000}{60} = \mathbf{15000\text{ W} = 15\text{ kW}}$
(a) Common velocity (perfectly inelastic collision, momentum conserved):
$m_1u_1 = (m_1+m_2)v \Rightarrow (0.2)(5) = (0.5)v$
$1 = 0.5v \Rightarrow v = 2\text{ m/s}$
(b) KE lost: Initial KE = $\frac{1}{2}(0.2)(5)^2 = \frac{1}{2}(0.2)(25) = 2.5\text{ J}$
Final KE = $\frac{1}{2}(0.5)(2)^2 = \frac{1}{2}(0.5)(4) = 1\text{ J}$
KE lost = 2.5 − 1 = 1.5 J
$x = 10\text{ cm} = 0.1\text{ m}$
Elastic PE: $U = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.1)^2 = \frac{1}{2}(200)(0.01) = 1\text{ J}$
Launch speed (energy conservation): $\frac{1}{2}mv^2 = U \Rightarrow \frac{1}{2}(0.5)v^2 = 1$
$0.25v^2 = 1 \Rightarrow v^2 = 4 \Rightarrow v = \mathbf{2\text{ m/s}}$
Using the work-energy theorem: Work done by braking force = change in kinetic energy
$-Fs = 0 - \frac{1}{2}mu^2$
$F(50) = \frac{1}{2}(1000)(20)^2 = \frac{1}{2}(1000)(400) = 200000\text{ J}$
$F = \dfrac{200000}{50} = \mathbf{4000\text{ N}}$
Maximum height: $H = \dfrac{u^2}{2g} = \dfrac{400}{20} = 20\text{ m}$; half of this is $10\text{ m}$.
Total mechanical energy (conserved throughout, equals initial KE): $E = \frac{1}{2}mu^2 = \frac{1}{2}(1)(400) = 200\text{ J}$
PE at half height: $PE = mgh = (1)(10)(10) = 100\text{ J}$
KE at half height (by conservation): $KE = E - PE = 200 - 100 = \mathbf{100\text{ J}}$; and $\mathbf{PE = 100\text{ J}}$ — exactly equal at the halfway point, as expected.
Mass of water ejected per second: $\dfrac{dm}{dt} = \rho A v$ (density × area × speed — volume flow rate times density).
Kinetic energy imparted per second (power) = rate of KE delivery = $\dfrac{1}{2}\left(\dfrac{dm}{dt}\right)v^2$
$P = \dfrac{1}{2}(\rho A v)v^2 = \dfrac{1}{2}\rho A v^3$
Since $\rho$ and $A$ are constants for a given pipe and fluid, $P \propto v^3$ — power required increases very steeply (as the cube) with the ejection speed.
Using the work-energy theorem, the work done by the resistive force equals the loss in kinetic energy (final KE = 0, bullet stops).
$F \times s = \frac{1}{2}mu^2$
$F \times 0.1 = \frac{1}{2}(0.01)(200)^2 = \frac{1}{2}(0.01)(40000) = 200\text{ J}$
$F = \dfrac{200}{0.1} = \mathbf{2000\text{ N}}$
Speed at lowest point (energy conservation, PE converts fully to KE): $mgh = \frac{1}{2}mv^2 \Rightarrow v^2 = 2gh = 2(10)(0.8) = 16 \Rightarrow v = 4\text{ m/s}$
Condition to complete a vertical circle (minimum speed at the top, for a rigid rod, is when gravity alone cannot exceed the centripetal requirement — for a rod, minimum speed at top can be zero, but for a full swing from height $h=0.8\text{ m}$ with radius $r=0.8\text{ m}$, this height exactly equals the radius, meaning the bob would only just reach the same height as the pivot, not go over the top):
Since the bob is released from a height equal to the radius (0.8 m), it does not have enough energy to reach the top of the circle (height $2r = 1.6\text{ m}$ needed) — it cannot complete a full vertical circle; it would only swing up to the height it was released from (on the opposite side, ignoring losses), not continue over the top.