Further mechanics (circular motion, SHM) – A-level Physics exam questions with mark scheme

1Core3 marks

The blades of a wind turbine have a length of 42 m. The blades rotate at a constant rate of 15 revolutions per minute.

(a) Calculate the angular speed of the blades. [1]

(b) Calculate the linear speed of a blade tip. [1]

(a) $\omega = 2\pi f = 2\pi \times \frac{15}{60}$ = 1.6 rad s$^{-1}$ (1.57) allow $\pi/2$(1)
(b) $v = \omega r = 1.57 \times 42$ = 66 m s$^{-1}$ (ecf from (a))(1)
(c) $a = \omega^2 r$ or $v^2/r$ = 104 m s$^{-2}$ (103.6; allow 100 to 2 sf) (ecf)(1)
Answer: (a) 1.57 rad s$^{-1}$ (b) 66 m s$^{-1}$ (c) 104 m s$^{-2}$
Examiner tip: Convert revolutions per minute to revolutions per second before multiplying by 2π — leaving it per minute is the most common slip.

2Standard4 marks

A car travels over the top of a hump-backed bridge. The top of the bridge is an arc of a vertical circle of radius 25 m.

(a) Explain why the normal contact force on the car from the road is less than the weight of the car when the car is at the top of the bridge. [2]

(b) Calculate the maximum speed at which the car can travel over the top of the bridge without losing contact with the road. [2]

(a) a resultant (centripetal) force is needed towards the centre of the circle, i.e. downwards(1)
(a) resultant force = weight − normal force $= mv^2/r$, so normal force $= mg - mv^2/r$ which is less than $mg$(1)
(b) contact lost when normal force = 0, so $mg = mv^2/r$ or $v = \sqrt{gr}$(1)
(b) $v = \sqrt{9.81 \times 25}$ = 16 m s$^{-1}$ (15.7)(1)
Answer: (b) 15.7 m s$^{-1}$
Examiner tip: Say which force is bigger and why: the weight must exceed the reaction to leave a downward resultant towards the centre.

3Hard5 marks

A small ball of mass 0.20 kg is attached to a string of length 0.50 m. The ball moves at constant speed in a horizontal circle with the string at 30° to the vertical, as shown.

(a) Show that the tension in the string is about 2.3 N. [2]

(b) Calculate the speed of the ball. [3]

(a) vertical forces balance: $T\cos 30^\circ = mg$(1)
(a) $T = \frac{0.20 \times 9.81}{\cos 30^\circ}$ = 2.27 N (at least 3 sf seen for a 'show that')(1)
(b) radius of circle $r = 0.50 \sin 30^\circ = 0.25$ m(1)
(b) horizontal component of tension provides centripetal force: $T \sin 30^\circ = \frac{mv^2}{r}$ (or $v^2 = gr\tan 30^\circ$)(1)
(b) $v$ = 1.2 m s$^{-1}$ (1.19) (allow use of 2.3 N giving 1.20)(1)
Answer: (a) 2.27 N (b) 1.19 m s$^{-1}$
Examiner tip: The radius of the circle is not the string length — resolve to get r = L sin θ.

4Core4 marks

A mass of 0.40 kg hangs from a spring of stiffness 25 N m$^{-1}$. The mass is pulled down 3.0 cm from its equilibrium position and released so that it oscillates with simple harmonic motion.

(a) Calculate the period of the oscillations. [2]

(b) Calculate the maximum speed of the mass. [2]

(a) $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.40}{25}}$(1)
(a) $T$ = 0.79 s (0.795)(1)
(b) $v_{max} = \omega A$ with $\omega = 2\pi/T$ = 7.9 rad s$^{-1}$ (or $\sqrt{k/m}$)(1)
(b) $v_{max} = 7.91 \times 0.030$ = 0.24 m s$^{-1}$ (ecf from (a))(1)
Answer: (a) 0.79 s (b) 0.24 m s$^{-1}$
Examiner tip: Amplitude must be in metres (0.030 m) before using v = ωA.

5Hard5 marks

The graph shows how the displacement $x$ of an oscillating object varies with time $t$. The object moves with simple harmonic motion.

(a) State one time shown on the graph at which the speed of the object is a maximum. [1]

(b) Calculate the maximum acceleration of the object. [2]

(a) any one of 0.20 s, 0.60 s, 1.0 s, 1.4 s (where $x = 0$)(1)
(b) period read as 0.80 s, so $\omega = 2\pi/0.80$ = 7.85 rad s$^{-1}$(1)
(b) $a_{max} = \omega^2 A = 7.85^2 \times 0.040$ = 2.5 m s$^{-2}$ (2.47) (A must be in m)(1)
(c) $x = A\cos\omega t = 0.040\cos(7.85 \times 0.10)$ (calculator in radians)(1)
(c) $x$ = 0.028 m or 2.8 cm(1)
Answer: (a) e.g. 0.20 s (b) 2.5 m s$^{-2}$ (c) 2.8 cm
Examiner tip: Use x = A cos ωt here because the graph starts at maximum displacement, and put your calculator in radians.

6Standard4 marks

(a) A student wants to make a simple pendulum with a period of 2.0 s. Calculate the length of string needed. [2]

(b) The pendulum is now driven by a periodic force whose frequency can be varied. Describe how the amplitude–frequency graph for the pendulum changes when the damping is increased. [2]

(a) $L = \frac{gT^2}{4\pi^2} = \frac{9.81 \times 2.0^2}{4\pi^2}$(1)
(a) $L$ = 0.99 m (0.994)(1)
(b) peak (maximum) amplitude is lower (accept: amplitude is smaller at all frequencies)(1)
(b) peak is broader / flatter (accept: peak occurs at a slightly lower frequency)(1)
Answer: (a) 0.99 m (b) lower and broader resonance peak
Examiner tip: For damping questions give two separate changes to the graph: the height of the peak and its width.

7Hard7 marks

A ball of mass 0.15 kg is attached to a light string and whirled in a vertical circle of radius 0.80 m. At the top of the circle the speed of the ball is 3.5 m s$^{-1}$. Assume no energy is lost to air resistance.

(a) Calculate the tension in the string when the ball is at the top of the circle. [2]

(b) Calculate the speed of the ball at the bottom of the circle. [2]

(d) Explain why the string is most likely to break when the ball is at the bottom of the circle. [1]

(a) at top: $T + mg = \frac{mv^2}{r}$(1)
(a) $T = 0.15\left(\frac{3.5^2}{0.80} - 9.81\right)$ = 0.83 N (0.825)(1)
(b) loss in GPE = gain in KE: $v_b^2 = v_t^2 + 2g(2r) = 3.5^2 + 2 \times 9.81 \times 1.60$(1)
(b) $v_b$ = 6.6 m s$^{-1}$ (6.61)(1)
(c) at bottom: $T - mg = \frac{mv^2}{r}$(1)
(c) $T = 0.15\left(9.81 + \frac{6.61^2}{0.80}\right)$ = 9.7 N (9.65) (allow 9.6 N from $v$ = 6.6 m s$^{-1}$; ecf from (b))(1)
(d) at the bottom the tension must both support the weight and provide the centripetal force (and speed is greatest), so tension is greatest there(1)
Answer: (a) 0.83 N (b) 6.6 m s$^{-1}$ (c) 9.7 N
Examiner tip: The height change from top to bottom is the diameter (2r), not the radius.

8Hard5 marks

A trolley of mass 0.25 kg is attached between springs on a horizontal track. The combined stiffness of the springs is 40 N m$^{-1}$. The trolley is displaced by 5.0 cm and released, and it oscillates with simple harmonic motion. Friction is negligible.

(a) Calculate the total energy of the oscillating system. [1]

(b) Calculate the speed of the trolley when its displacement is 3.0 cm. [2]

(c) The trolley is replaced by one of mass 1.0 kg, again displaced by 5.0 cm and released. Explain what happens to the maximum speed. [2]

(a) $E = \frac{1}{2}kA^2 = \frac{1}{2} \times 40 \times 0.050^2$ = 0.050 J(1)
(b) $\omega = \sqrt{k/m} = \sqrt{40/0.25}$ = 12.65 rad s$^{-1}$ and $v = \omega\sqrt{A^2 - x^2}$ (or ½kA² − ½kx² = ½mv²)(1)
(b) $v = 12.65\sqrt{0.050^2 - 0.030^2}$ = 0.51 m s$^{-1}$ (allow 0.50 m s$^{-1}$ from $\omega$ = 12.6)(1)
(c) total energy $\frac{1}{2}kA^2$ is unchanged (same $k$ and $A$) / $\omega$ halves because $\omega = \sqrt{k/m}$ and $m$ is 4×(1)
(c) so maximum speed halves (from 0.63 to 0.32 m s$^{-1}$) since $\frac{1}{2}mv_{max}^2$ is constant / $v_{max} = \omega A$(1)
Answer: (a) 0.050 J (b) 0.51 m s$^{-1}$ (c) maximum speed halves
Examiner tip: In 'explain' questions about changing mass, quote the relationship and state the factor (halves) — 'decreases' alone usually loses the second mark.

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