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A-Level Maths – Mechanics

SUVAT, forces, Newton’s laws, connected particles, moments, projectiles, friction.

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11 questions  ·  71 marks  ·  answers included  ·  A-Level (Edexcel/AQA/OCR)
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This worksheet covers:

  • Constant-acceleration (suvat) equations
  • Forces and Newton’s laws
  • Connected particles
  • Moments
  • Projectiles
  • Friction

Take $g = 9.8\,\text{m s}^{-2}$ throughout and give answers to 3 significant figures unless exact. Calculator allowed; allow about 80 minutes.

Questions

Section A – Skills

Short questions.
1

A particle moves in a straight line with constant acceleration $2.5\,\text{m s}^{-2}$. Its initial velocity is $4\,\text{m s}^{-1}$.(a) Find its velocity after $6$ s.(b) Find the distance travelled in the first $6$ s.

Show answer
(a) $v = 4 + 2.5(6) = 19\,\text{m s}^{-1}$(b) $s = 4(6) + \frac{1}{2}(2.5)(36) = 69$ m
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2

A ball is projected vertically upwards from ground level with speed $14\,\text{m s}^{-1}$.(a) Find the maximum height reached.(b) Find the time taken for the ball to return to the ground.(c) State one modelling assumption you have made.

Show answer
(a) $0 = 14^2 – 2(9.8)h \Rightarrow h = 10$ m(b) $0 = 14t – 4.9t^2 \Rightarrow t = 2.86$ s ($20/7$)(c) Air resistance negligible / ball is a particle / $g$ constant
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3

A car travelling at $30\,\text{m s}^{-1}$ brakes with constant deceleration and comes to rest in $75$ m.(a) Find the deceleration.(b) Find the time taken to stop.

Show answer
(a) $0 = 900 – 2a(75) \Rightarrow a = 6\,\text{m s}^{-2}$ (deceleration)(b) $t = 30/6 = 5$ s
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4

A lift of mass $400$ kg carries a person of mass $60$ kg.(a) The lift accelerates upwards at $1.5\,\text{m s}^{-2}$. Find the normal reaction force exerted by the floor on the person.(b) The lift then decelerates at $2\,\text{m s}^{-2}$ while still moving upwards. Find the new normal reaction.

Show answer
(a) $R – 60g = 60(1.5) \Rightarrow R = 678$ N(b) $60g – R = 60(2) \Rightarrow R = 468$ N
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Section B – Exam-style

Show full method, including a clear force diagram description.
5

Two particles $P$ and $Q$, of masses $3$ kg and $5$ kg respectively, are connected by a light inextensible string passing over a smooth fixed pulley. The system is released from rest with the string taut.(a) Find the acceleration of the system.(b) Find the tension in the string.(c) State how you have used the fact that the pulley is smooth.

Show answer
(a) $5g – 3g = 8a \Rightarrow a = \dfrac{g}{4} = 2.45\,\text{m s}^{-2}$(b) $T – 3g = 3a \Rightarrow T = 36.75 \approx 36.8$ N(c) Tension is the same on both sides of the pulley
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6

A block $A$ of mass $4$ kg rests on a rough horizontal table. It is connected by a light inextensible string over a smooth pulley at the edge of the table to a block $B$ of mass $2$ kg hanging freely. The coefficient of friction between $A$ and the table is $0.3$. The system is released from rest.(a) Find the acceleration of the system.(b) Find the tension in the string.(c) Find the speed of $B$ after it has fallen $0.5$ m.

Show answer
(a) $B$: $2g – T = 2a$; $A$: $T – 0.3(4g) = 4a$; adding: $2g – 1.2g = 6a$, $a = 1.31\,\text{m s}^{-2}$ ($0.8g/6$)(b) $T = 2(g – a) = 17.0$ N(c) $v^2 = 2(1.307)(0.5)$, $v = 1.14\,\text{m s}^{-1}$
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7

A uniform beam $AB$ of length $4$ m and mass $20$ kg rests horizontally on two supports, one at $A$ and the other at $C$ where $AC = 3$ m. A child of mass $30$ kg stands on the beam at a point $3.5$ m from $A$.(a) Find the magnitude of the reaction at $C$.(b) Find the magnitude of the reaction at $A$.(c) Find the greatest distance from $A$ at which the child can stand without the beam tipping.

Show answer
(a) Moments about $A$: $3R_C = 20g(2) + 30g(3.5) = 145g$, $R_C = 474$ N(b) $R_A = 50g – R_C = 16.3$ N(c) Tipping: $R_A = 0$; moments about $C$: $30g(d – 3) = 20g(1) \Rightarrow d = 3\frac{2}{3} = 3.67$ m
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8

A particle is projected from a point on horizontal ground with speed $25\,\text{m s}^{-1}$ at an angle of $40^\circ$ above the horizontal.(a) Find the time of flight.(b) Find the horizontal range.(c) Find the maximum height reached.

Show answer
(a) $0 = 25\sin 40^\circ\, t – 4.9t^2 \Rightarrow t = 3.28$ s(b) $25\cos 40^\circ \times 3.280 = 62.8$ m(c) $\dfrac{(25\sin 40^\circ)^2}{2g} = 13.2$ m
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9

A box of mass $5$ kg is placed on a rough plane inclined at $25^\circ$ to the horizontal. The coefficient of friction between the box and the plane is $0.4$.(a) Find the component of the weight acting down the plane.(b) Find the maximum possible frictional force.(c) Show that the box slides down the plane, and find its acceleration.

Show answer
(a) $5g\sin 25^\circ = 20.7$ N(b) $F_{\max} = 0.4 \times 5g\cos 25^\circ = 17.8$ N(c) $20.7 > 17.8$ so it slides; $5a = 20.71 – 17.76$, $a = 0.589\,\text{m s}^{-2}$
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Section C – Extended

Longer structured questions.
10

A stone is thrown from the top of a vertical cliff $40$ m high with speed $20\,\text{m s}^{-1}$ at an angle of $30^\circ$ above the horizontal. It lands in the sea at the foot of the cliff.(a) Show that the time taken to reach the sea satisfies $4.9t^2 – 10t – 40 = 0$.(b) Find the time taken to reach the sea.(c) Find the horizontal distance from the foot of the cliff to the point where the stone lands.(d) Find the speed of the stone as it hits the sea, and the angle its velocity makes with the horizontal.

Show answer
(a) $u_y = 20\sin 30^\circ = 10$; $-40 = 10t – 4.9t^2$(b) $t = 4.05$ s(c) $20\cos 30^\circ \times 4.054 = 70.2$ m(d) $v_x = 17.32$, $v_y = 10 – 9.8(4.054) = -29.7$; speed $= 34.4\,\text{m s}^{-1}$ at $59.8^\circ$ below the horizontal
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11

A car of mass $1000$ kg tows a trailer of mass $400$ kg along a straight horizontal road using a light rigid tow-bar. The engine produces a driving force of $2500$ N. The resistances to motion on the car and trailer are $300$ N and $200$ N respectively.(a) Find the acceleration of the car and trailer.(b) Find the tension in the tow-bar.(c) The driver now applies the brakes, producing a braking force of $1800$ N on the car only. The resistances are unchanged. Find the deceleration of the car and trailer.(d) Find the magnitude of the force in the tow-bar during braking, and state whether it is a tension or a thrust.

Show answer
(a) $2500 – 500 = 1400a \Rightarrow a = 1.43\,\text{m s}^{-2}$(b) Trailer: $T – 200 = 400(1.4286)$, $T = 771$ N(c) $1800 + 500 = 1400a \Rightarrow a = 1.64\,\text{m s}^{-2}$(d) Trailer: $200 + F = 400(1.6429)$, $F = 457$ N; thrust (tow-bar in compression)
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