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A2 Forces and MotionCircular MotionOscillations 3.1.3(a)3.1.2(a)(i)
~20 min Difficulty: 6/10
Prior knowledge SUVAT equationsNewton's Second Lawbasic circular motion definitions.
Problem structure
(a) 2, numerical, kinematics (projectile motion). (b) 2, numerical, kinematics (projectile motion). (c) 3, numerical, circular motion (centripetal force). (d) 3, numerical, oscillations (SHM period). (e) 2, numerical, oscillations (SHM acceleration).

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A small robotic probe of mass $m = 0.450\text{ kg}$ is designed to test the surface of a low-gravity asteroid. The probe is launched horizontally from the top of a vertical cliff of height $H = 12.0\text{ m}$ with an initial speed $u$. (a) On this asteroid, the acceleration due to gravity is $g_{ast} = 1.60\text{ m s}^{-2}$. Calculate the time $t$ it takes for the probe to reach the ground, assuming no atmosphere. [2] (b) The probe must land at a horizontal distance (range) of $x = 18.0\text{ m}$ from the base of the cliff. Calculate the required horizontal launch speed $u$. [2] (c) Upon landing, the probe is immediately tethered to a fixed point on the ground and performs a horizontal circular sweep to scan the area. It moves in a circle of radius $r = 5.00\text{ m}$ at a constant speed. The maximum horizontal centripetal force the tether can withstand is $2.25\text{ N}$. Calculate the maximum angular velocity $\omega$ the probe can maintain without the tether snapping. [3] (d) The probe's internal sensor uses a small mass on a spring to detect vibrations. The sensor behaves as a simple harmonic oscillator with a period $T = 0.850\text{ s}$. If the mass of the sensor's internal block is $0.0250\text{ kg}$, calculate the stiffness (spring constant) $k$ of the spring. [3] (e) During a test, the sensor mass is displaced by $15.0\text{ mm}$ from its equilibrium position and released. Calculate the maximum acceleration $a_{max}$ of the sensor mass. [2]
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(a) For vertical motion, we use the equation of motion $s = ut + \frac{1}{2}at^2$. Since the initial vertical velocity is zero: $H = \frac{1}{2}g_{ast}t^2$ $t = \sqrt{\frac{2H}{g_{ast}}} = \sqrt{\frac{2 \times 12.0}{1.60}} = \sqrt{15.0}$ $t = 3.8729... \approx 3.87\text{ s}$ (to 3 SF). Unit check: $\sqrt{\text{m} / (\text{m s}^{-2})} = \sqrt{\text{s}^2} = \text{s}$. Correct. (b) For horizontal motion, speed is constant: $x = u \times t$ $u = \frac{x}{t} = \frac{18.0}{3.8729...}$ $u = 4.647... \approx 4.65\text{ m s}^{-1}$ (to 3 SF). (c) The centripetal force $F$ is provided by the tether: $F = m r \omega^2$ $\omega = \sqrt{\frac{F}{mr}}$ Using $F_{max} = 2.25\text{ N}$, $m = 0.450\text{ kg}$, and $r = 5.00\text{ m}$: $\omega = \sqrt{\frac{2.25}{0.450 \times 5.00}} = \sqrt{\frac{2.25}{2.25}} = 1.00\text{ rad s}^{-1}$. Unit check: $\sqrt{\text{N} / (\text{kg m})} = \sqrt{(\text{kg m s}^{-2}) / (\text{kg m})} = \sqrt{\text{s}^{-2}} = \text{s}^{-1}$. Correct. (d) For a mass-spring system in SHM: $T = 2\pi\sqrt{\frac{m_{sensor}}{k}}$ $T^2 = 4\pi^2 \frac{m_{sensor}}{k}$ $k = \frac{4\pi^2 m_{sensor}}{T^2}$ $k = \frac{4 \times \pi^2 \times 0.0250}{0.850^2} = \frac{0.98696...}{0.7225}$ $k = 1.366... \approx 1.37\text{ N m}^{-1}$ (to 3 SF). (e) Maximum acceleration in SHM is given by $a_{max} = \omega_{shm}^2 A$, where $\omega_{shm} = \frac{2\pi}{T}$. $\omega_{shm} = \frac{2\pi}{0.850} = 7.392... \text{ rad s}^{-1}$ $a_{max} = (7.392...)^2 \times 0.0150$ $a_{max} = 54.64... \times 0.0150 = 0.8196... \approx 0.820\text{ m s}^{-2}$ (to 3 SF). Alternatively: $a_{max} = \frac{k A}{m_{sensor}} = \frac{1.366 \times 0.0150}{0.0250} = 0.8196\text{ m s}^{-2}$.

Marking guidance

(a) M1: Correct use of $s = \frac{1}{2}at^2$ or equivalent. A1: $3.87\text{ s}$ (accept $3.9\text{ s}$ if 2 SF used consistently). (b) M1: $u = \text{distance} / \text{time}$ using answer from (a). A1: $4.65\text{ m s}^{-1}$ (allow ecf from (a)). (c) M1: Correct formula for centripetal force $F = m r \omega^2$ or $F = \frac{mv^2}{r}$ and $v = r\omega$. M1: Correct rearrangement for $\omega$. A1: $1.00\text{ rad s}^{-1}$ (must have units). (d) M1: Correct formula for period of mass-spring system $T = 2\pi\sqrt{m/k}$. M1: Correct rearrangement for $k$. A1: $1.37\text{ N m}^{-1}$ (accept range $1.36-1.38$). (e) M1: Use of $a_{max} = \omega^2 A$ or $F = ma = kx$. A1: $0.820\text{ m s}^{-2}$ (allow $0.82$). Common error: Forgetting to convert $15\text{ mm}$ to $0.015\text{ m}$.