Paper 3 — Mechanics

Time allowed: 75 minutes Total marks: 50 Topics covered: All 5 mechanics topics

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Instructions

Answer all questions. Calculators are permitted. Take $g = 9.8$ m/s$^2$ unless otherwise stated. Show all working — marks are awarded for method as well as final answer.

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Questions

Q1 [10 marks] — Kinematics

A particle moves in a straight line so that its velocity $v$ m/s at time $t$ seconds ($t \geq 0$) is given by $v = 6t - t^2 - 5$.

(a) Find the times at which the particle is instantaneously at rest. [3 marks]

(b) Calculate the total distance travelled by the particle from $t = 0$ to $t = 7$. [5 marks]

(c) A student calculates the displacement over $[0, 7]$ by evaluating $\int_0^7 v\,dt$ and obtains a positive answer. The student then claims this integral equals the total distance. Calculate the percentage error in the student's answer. [2 marks]

Q2 [10 marks] — Forces and Newton's Laws

A block of mass $8$ kg rests on a rough horizontal surface. The coefficient of friction between the block and the surface is $\mu = 0.4$. A horizontal force $P$ is applied to the block.

(a) Find the range of values of $P$ for which the block remains in equilibrium. [2 marks]

(b) When $P = 20$ N, find the magnitude and direction of the frictional force acting on the block. [2 marks]

(c) A student, upon seeing the value $\mu = 0.4$, immediately writes $F = \mu R = 0.4 \times 78.4 = 31.36$ N for the frictional force, regardless of the applied force $P$. Explain why this is incorrect for $P = 20$ N, and calculate the percentage by which the student overestimates the friction. [3 marks]

(d) The force $P$ is now applied at an angle of $30°$ above the horizontal. Find the maximum value of $P$ for which the block remains in equilibrium, and explain why this maximum is greater than the answer in part (a). [3 marks]

Q3 [10 marks] — Moments

A force of $50$ N acts at one end $B$ of a uniform rod $AB$ of length $3$ m. The rod is hinged at end $A$ and held at an angle of $40°$ to the horizontal. The force acts vertically downwards.

(a) Find the moment of the $50$ N force about the hinge $A$. [3 marks]

(b) A student calculates the moment as $50 \times 3 = 150$ Nm. Explain the error and calculate the percentage overestimate. [4 marks]

(c) The force at $B$ is now replaced by a force of $50$ N acting perpendicular to the rod (not vertically). Find the new moment about $A$ and explain why it is larger than the answer in part (a). [3 marks]

Q4 [10 marks] — Energy and Work

A car of mass $800$ kg travels on a level road. The engine works at constant power $40$ kW. The resistance to motion is a constant $200$ N.

(a) Show that the acceleration of the car is given by $a = \frac{P}{mv} - \frac{R}{m}$, where $P$ is the power, $v$ is the speed, and $R$ is the resistance. [2 marks]

(b) Find the maximum speed of the car. [2 marks]

(c) Find the acceleration when the speed is $10$ m/s, and when the speed is $100$ m/s. [2 marks]

(d) A student claims that "since the power is constant, the acceleration is constant." Use your answers from part (c) to refute this claim. [2 marks]

(e) Find the time taken for the car to accelerate from $5$ m/s to $15$ m/s, giving your answer in terms of an integral that you need not evaluate. [2 marks]

Q5 [10 marks] — Momentum

Two particles $A$ (mass $4$ kg) and $B$ (mass $6$ kg) move towards each other along the same straight line. $A$ has speed $5$ m/s and $B$ has speed $3$ m/s. After the collision, $A$ moves in the opposite direction with speed $2$ m/s.

(a) Taking the direction of $A$'s initial motion as positive, apply conservation of momentum to find the velocity of $B$ after the collision. [3 marks]

(b) Find the coefficient of restitution for the collision. [3 marks]

(c) A student defines positive as the direction of $B$'s initial motion and obtains a different numerical value for $v_B$. Show that the physical velocity is the same regardless of the sign convention. [2 marks]

(d) Determine whether the collision is elastic, inelastic, or perfectly inelastic, and calculate the kinetic energy lost. [2 marks]

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Solutions

Q1

(a) $v = 0 \implies 6t - t^2 - 5 = 0 \implies t^2 - 6t + 5 = 0 \implies (t-1)(t-5) = 0$.

The particle is at rest at $t = 1$ s and $t = 5$ s.

(b) First, determine the sign of $v$ in each interval.

For $0 \lt{} t \lt{} 1$: test $t = 0.5$, $v = 3 - 0.25 - 5 = -2.25 \lt{} 0$ (moving in negative direction).

For $1 \lt{} t \lt{} 5$: test $t = 3$, $v = 18 - 9 - 5 = 4 \gt{} 0$ (moving in positive direction).

For $t \gt{} 5$: test $t = 6$, $v = 36 - 36 - 5 = -5 \lt{} 0$ (moving in negative direction).

The particle reverses direction at $t = 1$ and $t = 5$. Total distance requires integrating $\lvert v \rvert$, which means splitting at the turning points and taking the magnitude of each segment.

$s(t) = \int v\,dt = \int (6t - t^2 - 5)\,dt = 3t^2 - \frac{t^3}{3} - 5t + C$

With $s(0) = 0$: $C = 0$, so $s(t) = 3t^2 - \frac{t^3}{3} - 5t$.

$s(1) = 3 - \frac{1}{3} - 5 = -\frac{7}{3}$ m.

$s(5) = 75 - \frac{125}{3} - 25 = 50 - \frac{125}{3} = \frac{150 - 125}{3} = \frac{25}{3}$ m.

$s(7) = 147 - \frac{343}{3} - 35 = 112 - \frac{343}{3} = \frac{336 - 343}{3} = -\frac{7}{3}$ m.

$\text{Distance} = \lvert s(1) - s(0) \rvert + \lvert s(5) - s(1) \rvert + \lvert s(7) - s(5) \rvert$

$= \left\lvert -\frac{7}{3} \right\rvert + \left\lvert \frac{25}{3} - \left(-\frac{7}{3}\right) \right\rvert + \left\lvert -\frac{7}{3} - \frac{25}{3} \right\rvert$

$= \frac{7}{3} + \frac{32}{3} + \frac{32}{3} = \frac{71}{3} \approx 23.67 \text{ m}$

(c) The student's displacement answer:

$\text{Displacement} = s(7) - s(0) = -\frac{7}{3} \approx -2.33 \text{ m}$

The student claims the distance is $2.33$ m (taking the magnitude). Actual distance is $\frac{71}{3} \approx 23.67$ m.

$\text{Percentage error} = \frac◆LB◆\lvert \frac{71}{3} - \frac{7}{3} \rvert◆RB◆◆LB◆\frac{71}{3}◆RB◆ \times 100\% = \frac{64}{71} \times 100\% \approx 90.1\%$

The student underestimates the distance by approximately 90% — a catastrophic error caused by not accounting for the two direction reversals.

Q2

(a) The block remains in equilibrium as long as the applied force does not exceed the maximum static friction.

Normal reaction: $R = mg = 8 \times 9.8 = 78.4$ N.

Maximum friction: $F_{\max} = \mu R = 0.4 \times 78.4 = 31.36$ N.

For equilibrium, the friction must balance $P$: $F = P$.

Since $F \leq F_{\max} = 31.36$ N, we need $P \leq 31.36$ N.

The block remains in equilibrium for $0 \leq P \leq 31.36$ N.

(b) When $P = 20$ N (which is less than $31.36$ N), the block does not move. The frictional force adjusts to exactly balance the applied force:

$F = P = 20 \text{ N}$

The frictional force acts in the direction opposite to $P$ (i.e., opposing the tendency to move).

(c) The student writes $F = 31.36$ N, but the actual friction is only $20$ N. The student has assumed the block is on the point of sliding, but $20 \lt{} 31.36$ N, so the block is not even close to sliding. The friction adjusts to match the applied force.

$\text{Percentage overestimate} = \frac{31.36 - 20}{20} \times 100\% = 56.8\%$

(d) Resolving perpendicular to the surface:

$R + P\sin 30° = mg \implies R = 78.4 - 0.5P$

Resolving horizontally, at limiting equilibrium:

$P\cos 30° = \mu R = 0.4(78.4 - 0.5P)$

$0.866P = 31.36 - 0.2P$

$1.066P = 31.36 \implies P = \frac{31.36}{1.066} \approx 29.42 \text{ N}$

This is less than $31.36$ N, not greater. Applying the force at an angle above the horizontal reduces the normal reaction ($R = 78.4 - 0.5P \lt{} 78.4$), which in turn reduces the maximum friction. Although the horizontal component of $P$ is only $P\cos 30° \approx 0.866P$, the reduction in $R$ means the maximum available horizontal force is reduced overall.

Q3

(a) The moment of a force about a point equals the force multiplied by the perpendicular distance from the point to the line of action of the force.

The force acts vertically downwards at $B$. The perpendicular distance from $A$ to the vertical line through $B$ is the horizontal distance from $A$ to $B$:

$d = AB \times \cos 40° = 3\cos 40° \approx 2.298 \text{ m}$

$\text{Moment} = F \times d = 50 \times 3\cos 40° = 150\cos 40° \approx 114.9 \text{ Nm}$

(b) The student used the distance $AB = 3$ m instead of the perpendicular distance $3\cos 40° \approx 2.298$ m. The moment is $F \times d_{\perp}$, not $F \times d_{\text{along rod}}$.

$\text{Student's answer} = 150 \text{ Nm}$

$\text{Correct answer} = 150\cos 40° \approx 114.9 \text{ Nm}$

$\text{Percentage overestimate} = \frac◆LB◆150 - 150\cos 40°◆RB◆◆LB◆150\cos 40°◆RB◆ \times 100\% = \frac◆LB◆1 - \cos 40°◆RB◆◆LB◆\cos 40°◆RB◆ \times 100\% = \left(\frac◆LB◆1◆RB◆◆LB◆\cos 40°◆RB◆ - 1\right) \times 100\% \approx 30.5\%$

(c) If the $50$ N force acts perpendicular to the rod at $B$, the perpendicular distance from $A$ to the line of action is simply the length of the rod:

$\text{Moment} = 50 \times 3 = 150 \text{ Nm}$

This is larger because the perpendicular distance equals the full length of the rod ($3$ m), whereas in part (a) the perpendicular distance was only $3\cos 40° \approx 2.298$ m. A force applied perpendicular to a rod always produces the maximum possible moment for a given force magnitude and application point.

Q4

(a) The driving force at speed $v$ is $F = \frac{P}{v}$ (from $P = Fv$).

Net force $= F - R = \frac{P}{v} - R$.

By Newton's Second Law: $ma = \frac{P}{v} - R$.

$a = \frac{P}{mv} - \frac{R}{m} \quad \blacksquare$

(b) At maximum speed, $a = 0$:

$\frac◆LB◆P◆RB◆◆LB◆v_{\max}◆RB◆ = R \implies v_{\max} = \frac{P}{R} = \frac{40000}{200} = 200 \text{ m/s}$

(c) At $v = 10$ m/s:

$a = \frac◆LB◆40000◆RB◆◆LB◆800 \times 10◆RB◆ - \frac{200}{800} = 5 - 0.25 = 4.75 \text{ m/s}^2$

At $v = 100$ m/s:

$a = \frac◆LB◆40000◆RB◆◆LB◆800 \times 100◆RB◆ - \frac{200}{800} = 0.5 - 0.25 = 0.25 \text{ m/s}^2$

(d) The acceleration at $10$ m/s is $4.75$ m/s$^2$ and at $100$ m/s is $0.25$ m/s$^2$. The acceleration decreases by a factor of 19 as the speed increases by a factor of 10. Constant power does not imply constant acceleration; in fact, the acceleration decreases hyperbolically with speed.

(e) From $a = \frac{dv}{dt} = \frac{P}{mv} - \frac{R}{m}$:

$dt = \frac◆LB◆dv◆RB◆◆LB◆\frac{P}{mv} - \frac{R}{m}◆RB◆ = \frac{mv\,dv}{P - Rv}$

$t = \int_{5}^{15} \frac{mv}{P - Rv}\,dv = 800\int_{5}^{15} \frac{v}{40000 - 200v}\,dv$

Let $u = 40000 - 200v$, $du = -200\,dv$, $dv = -\frac{du}{200}$, $v = \frac{40000 - u}{200}$:

$t = 800\int \frac{(40000 - u)/200}{u} \cdot \left(-\frac{du}{200}\right) = 800 \times \frac{-1}{40000}\int \frac{40000 - u}{u}\,du$

$= -\frac{1}{50}\int \left(\frac{40000}{u} - 1\right)du = -\frac{1}{50}[40000\ln u - u]$

Evaluating from $v = 5$ ($u = 39000$) to $v = 15$ ($u = 37000$):

$t = -\frac{1}{50}\left[(40000\ln 37000 - 37000) - (40000\ln 39000 - 39000)\right]$

$= \frac{1}{50}\left[40000\ln\!\left(\frac{39000}{37000}\right) + 2000\right]$

$= 800\ln\!\left(\frac{39}{37}\right) + 40 \approx 800(0.05263) + 40 \approx 42.1 + 40 = 82.1 \text{ s}$

Q5

(a) Positive direction = $A$'s initial motion (to the right).

Initial momenta: $p_A = 4 \times 5 = 20$ kg m/s, $p_B = 6 \times (-3) = -18$ kg m/s.

Total initial momentum $= 20 + (-18) = 2$ kg m/s.

After collision: $p_A' = 4 \times (-2) = -8$ kg m/s.

By conservation: $2 = -8 + 6v_B \implies 6v_B = 10 \implies v_B = \frac{5}{3} \approx 1.67$ m/s.

$B$ moves in the positive direction (the same direction as $A$'s initial motion) at $\frac{5}{3}$ m/s.

(b) Coefficient of restitution:

$e = \frac◆LB◆\text{relative speed of separation}◆RB◆◆LB◆\text{relative speed of approach}◆RB◆$

Relative speed of approach $= 5 - (-3) = 8$ m/s.

Relative speed of separation $= v_B - v_A = \frac{5}{3} - (-2) = \frac{5}{3} + 2 = \frac{11}{3}$ m/s.

$e = \frac{11/3}{8} = \frac{11}{24} \approx 0.458$

(c) With positive = $B$'s initial motion (to the left):

$u_A = -5$ m/s, $u_B = 3$ m/s.

Total initial momentum $= 4(-5) + 6(3) = -20 + 18 = -2$ kg m/s.

After collision: $v_A = 2$ m/s (moves in $B$'s initial direction).

Conservation: $-2 = 4(2) + 6v_B \implies -2 = 8 + 6v_B \implies 6v_B = -10 \implies v_B = -\frac{5}{3}$ m/s.

The negative sign means $B$ moves in the opposite direction to the defined positive, i.e., in $A$'s initial direction. This is the same physical velocity as $\frac{5}{3}$ m/s in $A$'s initial direction, confirming the result is convention-independent.

(d) Since $e = \frac{11}{24} \approx 0.458$ and $0 \lt{} e \lt{} 1$, the collision is inelastic.

$\text{KE}_{\text{before}} = \frac{1}{2}(4)(25) + \frac{1}{2}(6)(9) = 50 + 27 = 77 \text{ J}$

$\text{KE}_{\text{after}} = \frac{1}{2}(4)(4) + \frac{1}{2}(6)\!\left(\frac{25}{9}\right) = 8 + \frac{75}{9} = 8 + \frac{25}{3} = \frac{49}{3} \approx 16.33 \text{ J}$

$\Delta\text{KE} = 77 - \frac{49}{3} = \frac{231 - 49}{3} = \frac{182}{3} \approx 60.67 \text{ J lost}$

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Marking Guide

Question	Topic	Marks	Key Skills Tested
Q1	Kinematics	10	Displacement vs distance, direction changes from $v = 0$, splitting integrals, percentage error
Q2	Forces and Newton's Laws	10	Static friction inequality $F \leq \mu R$, non-limiting friction, angled force and normal reaction
Q3	Moments	10	Perpendicular distance vs distance to pivot, trigonometric moments, percentage error analysis
Q4	Energy and Work	10	$P = Fv$ derivation, maximum speed, decreasing acceleration at constant power, integration for time
Q5	Momentum	10	Sign convention consistency, conservation of momentum, coefficient of restitution, energy classification
Total		50