Forces and Motion, Circular Motion, Oscillations problem-solving question

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]

(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) 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 $0.015\text{ m}$.

(a) Hint 1: Vertical and horizontal motions are independent. Use the vertical height and the asteroid's gravity to find the time of flight. (b) Hint 2: The horizontal velocity is constant because there is no air resistance. Use the time from part (a). (c) Hint 3: Centripetal force is $m \omega^2 r$. Make sure to use the mass of the whole probe, not the sensor mass. (d) Hint 4: The period of a mass-spring oscillator depends on the mass and the spring constant. Rearrange the standard formula for $k$. (e) Hint 5: Maximum acceleration occurs at maximum displacement (amplitude). You can use $a = \omega^2 x$ or $F = kx = ma$.