2025 Edexcel IAS-Level Physics (XPH11) 模擬試題連答案詳解

Using the kinematic equation for vertical displacement: \(s_y = u_y t + \frac{1}{2} a_y t^2\).\
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The vertical component of the initial velocity is \(u_y = u \sin\theta = 20 \sin(30^\circ)\).\
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The vertical acceleration is \(a_y = -g = -9.81\text{ m s}^{-2}\).\
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Therefore, the vertical displacement is:\
\(s_y = 20t \sin(30^\circ) - \frac{1}{2}(9.81)t^2 \approx 20t \sin(30^\circ) - 4.9t^2\).\
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This matches option B.

[1] Correct option B: identifying vertical velocity component as \(20 \sin(30^\circ)\) and vertical acceleration term as \(4.9t^2\).

1. Calculate the work done in lifting the mass:\
\(W = mgh = 80\text{ kg} \times 9.81\text{ m s}^{-2} \times 15\text{ m} = 11772\text{ J}\).\
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2. Calculate the useful power output:\
\(P_{\text{out}} = \frac{W}{t} = \frac{11772\text{ J}}{10\text{ s}} = 1177.2\text{ W}\).\
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3. Use the efficiency formula to find the electrical power input:\
\(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{1177.2\text{ W}}{0.60} \approx 1962\text{ W} = 1.96\text{ kW}\).\
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This matches option C.

[1] Correct option C: calculating output power as \(1.18\text{ kW}\) and dividing by \(0.60\) to get \(1.96\text{ kW}\).

Since the block moves at a constant velocity, it is in equilibrium (net force parallel to the slope is zero).\
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The component of weight acting down the slope is \(mg \sin\theta\).\
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The frictional force acts up the slope to oppose the motion. Therefore, the magnitude of the frictional force is equal to the component of weight parallel to the slope:\
\(F_{\text{friction}} = mg \sin\theta\).\
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This matches option A.

[1] Correct option A: recognizing that at constant velocity, forces parallel to the plane balance so friction equals \(mg \sin\theta\).

The beam is uniform, so its weight of \(120\text{ N}\) acts at its centre of gravity, which is at the pivot. Therefore, the weight of the beam has a perpendicular distance of zero from the pivot and produces no moment.\
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The pivot is at the centre, \(1.0\text{ m}\) from either end.\
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An \(80\text{ N}\) weight is \(0.5\text{ m}\) from the left end, which is \(1.0 - 0.5 = 0.5\text{ m}\) to the left of the pivot. This produces an anticlockwise moment of:\
\(\tau_{\text{anticlockwise}} = 80\text{ N} \times 0.5\text{ m} = 40\text{ N m}\).\
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To balance this, a downward force \(F\) is applied at the right-hand end, which is \(1.0\text{ m}\) from the pivot. This produces a clockwise moment:\
\(\tau_{\text{clockwise}} = F \times 1.0\text{ m}\).\
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For rotational equilibrium:\
\(\tau_{\text{clockwise}} = \tau_{\text{anticlockwise}} \implies F \times 1.0 = 40 \implies F = 40\text{ N}\).\
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This matches option B.

[1] Correct option B: identifying pivot-to-load distances and setting clockwise moment equal to anticlockwise moment to solve for force.

First, find the effective spring constant of the parallel combination:\
\(k_p = k + k = 2k\).\
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Next, combine this in series with the third spring of constant \(k\):\
\(\frac{1}{k_{\text{eff}}} = \frac{1}{k_p} + \frac{1}{k} = \frac{1}{2k} + \frac{1}{k} = \frac{3}{2k}\).\
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Inverting this gives:\
\(k_{\text{eff}} = \frac{2}{3}k\).\
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This matches option A.

[1] Correct option A: applying parallel and series spring combinations correctly.

The relationship for the Young Modulus \(E\) is given by:\
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta x / L} = \frac{FL}{A\Delta x}\).\
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Rearranging for extension:\
\(\Delta x = \frac{FL}{AE}\).\
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For the second wire, the length is \(2L\) and the area is \(\frac{1}{2}A\). Since it is the same material, the Young Modulus \(E\) is the same. Under the same force \(F\), the new extension \(\Delta x_2\) is:\
\(\Delta x_2 = \frac{F(2L)}{(\frac{1}{2}A)E} = 4 \left(\frac{FL}{AE}\right) = 4\Delta x\).\
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This matches option D.

[1] Correct option D: showing that extension is proportional to \(L/A\), leading to \(2 / (1/2) = 4\) times the original extension.

At terminal velocity, the upward forces (viscous drag and upthrust) balance the downward force (weight):\
\(F_{\text{drag}} + U = W\).\
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Using Stokes' Law, the viscous drag is:\
\(F_{\text{drag}} = 6\pi \eta r v\).\
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Weight \(W = \rho_{\text{solid}} V g\) and upthrust \(U = \rho_{\text{fluid}} V g\), where \(V = \frac{4}{3}\pi r^3\).\
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Substituting these into the balance equation:\
\(6\pi \eta r v = (\rho_{\text{solid}} - \rho_{\text{fluid}}) \frac{4}{3}\pi r^3 g\).\
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Solving for \(v\):\
\(v = \frac{2g(\rho_{\text{solid}} - \rho_{\text{fluid}})}{9\eta} r^2\).\
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Since \(g\), \(\rho_{\text{solid}}\), \(\rho_{\text{fluid}}\), and \(\eta\) are constants, we have:\
\(v \propto r^2\).\
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This matches option D.

[1] Correct option D: deriving or recalling that terminal velocity of a falling sphere in a viscous fluid is proportional to the square of its radius.

Stiffness is defined by a high Young modulus, which means the material resists elastic deformation. Therefore, the stiffest materials are W and Y.\
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Ductility is the ability of a material to be drawn into a wire, which corresponds to undergoing large plastic deformation before fracturing. Therefore, the most ductile materials are W and X.\
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The material that is both the stiffest and the most ductile is Material W.\
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This matches option A.

[1] Correct option A: identifying high Young modulus with stiffness and large plastic deformation with ductility.

The forces acting on the block parallel to the ramp are the component of gravity pulling it down the slope, \(mg \sin\theta\), and the frictional force resisting motion, \(F\). Since the block is moving at a constant speed, it is in equilibrium, so \(F = mg \sin\theta\). The normal contact force perpendicular to the ramp is \(R = mg \cos\theta\). Since \(F = \mu R\), we have \(mg \sin\theta = \mu mg \cos\theta\). Dividing both sides by \(mg \cos\theta\) gives \(\mu = \tan\theta\).

1 mark: Correctly identifies the relation for equilibrium parallel and perpendicular to the slope, leading to \(\mu = \tan\theta\).

The Young modulus \(E\) is given by \(E = \frac{\text{stress}}{\text{strain}} = \frac{F L}{A \Delta x}\), which rearranges to \(\Delta x = \frac{F L}{A E}\). Since both wires are made of the same metal, they have the same Young modulus \(E\). The cross-sectional area of wire X is \(A_X = \frac{\pi d^2}{4}\), and for wire Y, \(A_Y = \frac{\pi (2d)^2}{4} = 4 A_X\). Using the formula for extension: \(\Delta x_X = \frac{F L}{A_X E}\) and \(\Delta x_Y = \frac{F (2L)}{A_Y E} = \frac{2 F L}{4 A_X E} = 0.5 \frac{F L}{A_X E}\). Thus, \(\Delta x_Y = 0.5 \Delta x_X\), which gives the ratio \(\frac{\Delta x_X}{\Delta x_Y} = 2\).

1 mark: Correctly relates area to the square of diameter, applies the Young modulus formula to both wires, and simplifies to find the ratio is 2.

(a) The vertical component of velocity is calculated using trigonometry:
\(u_y = u \sin\theta = 22 \sin(30^\circ) = 11\text{ m s}^{-1}\).

(b) Taking upwards as positive, we can write the kinematic equation for vertical motion:
\(s = u_y t + \frac{1}{2} a t^2\)
Substituting the values: \(s = -45\text{ m}\), \(u_y = 11\text{ m s}^{-1}\), and \(a = -9.81\text{ m s}^{-2}\):
\(-45 = 11 t - 4.905 t^2\)
Rearranging this into a quadratic equation:
\(4.905 t^2 - 11 t - 45 = 0\)
Solving for \(t\) using the quadratic formula:
\(t = \frac{11 \pm \sqrt{(-11)^2 - 4(4.905)(-45)}}{2(4.905)}\)
\(t = \frac{11 \pm \sqrt{121 + 882.9}}{9.81} = \frac{11 + 31.68}{9.81} \approx 4.35\text{ s}\).

(c) The horizontal speed remains constant and is given by:
\(u_x = 22 \cos(30^\circ) \approx 19.05\text{ m s}^{-1}\).

The horizontal distance travelled is:
\(d = u_x \times t = 19.05 \times 4.35 \approx 82.9\text{ m}\) (or \(83\text{ m}\) to 2 significant figures).

(a) [1.7 marks]
- Use of \(u_y = u \sin\theta\) [1 mark]
- Correct calculation showing \(22 \sin(30^\circ) = 11\text{ m s}^{-1}\) [0.7 marks]

(b) [3 marks]
- Use of \(s = ut + \frac{1}{2}at^2\) with correct sign convention [1 mark]
- Correct substitution of values into the quadratic formula [1 mark]
- Correct final value of time \(t = 4.35\text{ s}\) (accept \(4.4\text{ s}\)) [1 mark]

(c) [3 marks]
- Correct calculation of horizontal velocity component \(19.1\text{ m s}^{-1}\) [1 mark]
- Use of distance = velocity \(\times\) time [1 mark]
- Correct evaluation of distance \(83\text{ m}\) (accept \(82.9\text{ m}\)) [1 mark]

(a) The tension in the wire is equal to the weight of the mass:
\(T = mg = 6.0 \times 9.81 = 58.86\text{ N}\).
Tensile stress is given by:
\(\sigma = \frac{T}{A} = \frac{58.86}{1.2 \times 10^{-7}} = 4.91 \times 10^8\text{ Pa}\).

(b) Using the definition of Young Modulus:
\(E = \frac{\text{Stress}}{\text{Strain}}\)
\(\text{Strain} = \frac{\text{Stress}}{E} = \frac{4.91 \times 10^8}{2.0 \times 10^{11}} = 2.455 \times 10^{-3}\).

Since \(\text{Strain} = \frac{\Delta x}{L}\), the extension is:
\(\Delta x = \text{Strain} \times L = 2.455 \times 10^{-3} \times 2.5 \approx 6.14 \times 10^{-3}\text{ m}\) (or \(6.1\text{ mm}\)).

(c) Assumptions include: The elastic limit of the steel has not been exceeded, the cross-sectional area of the wire remains constant during stretching, or the weight of the wire itself is negligible.

(a) [2.7 marks]
- Use of \(T = mg\) [1 mark]
- Use of \(\sigma = \frac{F}{A}\) [1 mark]
- Correct calculation of stress \(= 4.9 \times 10^8\text{ Pa}\) (accept \(4.91 \times 10^8\text{ Pa}\)) [0.7 marks]

(b) [3 marks]
- Use of relation \(E = \frac{\sigma}{\varepsilon}\) or equivalent [1 mark]
- Correct substitution of values [1 mark]
- Correct extension of \(6.1\text{ mm}\) (or \(6.1 \times 10^{-3}\text{ m}\)) [1 mark]

(c) [2 marks]
- State one reasonable assumption (e.g. limit of proportionality/elastic limit not exceeded, cross-sectional area remains constant, or weight of wire is negligible) [2 marks]

(a) The weight of the rocket is:
\(W = mg = 0.35 \times 9.81 = 3.4335\text{ N}\).
The net upward force is:
\(F_{\text{net}} = T - W = 8.5 - 3.4335 = 5.0665\text{ N}\).
Using Newton's Second Law:
\(a = \frac{F_{\text{net}}}{m} = \frac{5.0665}{0.35} \approx 14.48\text{ m s}^{-2}\), which is approximately \(14\text{ m s}^{-2}\).

(b) Using the equation of motion:
\(v = u + at\)
\(v = 0 + (14.48 \times 1.8) \approx 26.1\text{ m s}^{-1}\) (or \(25.2\text{ m s}^{-1}\) if using the given approximation of \(14\text{ m s}^{-2}\)).

(c) The change in momentum is given by impulse:
\(\Delta p = F_{\text{net}} \Delta t = 5.0665 \times 1.8 \approx 9.12\text{ N s}\) (or \(9.12\text{ kg m s}^{-1}\)).
Alternatively, \(\Delta p = m \Delta v = 0.35 \times 26.1 = 9.14\text{ kg m s}^{-1}\).

(a) [2.7 marks]
- Calculation of the rocket's weight [1 mark]
- Newton's second law expression for net force: \(F_{\text{net}} = T - W\) [1 mark]
- Calculation of acceleration to show it is \(14.5\text{ m s}^{-2}\) (approx \(14\text{ m s}^{-2}\)) [0.7 marks]

(b) [2 marks]
- Use of \(v = u + at\) with \(u = 0\) [1 mark]
- Correct calculation of velocity: \(26\text{ m s}^{-1}\) (allow \(25\text{ m s}^{-1}\) if using \(14\text{ m s}^{-2}\)) [1 mark]

(c) [3 marks]
- Use of \(\Delta p = F \Delta t\) or \(\Delta p = m \Delta v\) [1 mark]
- Correct substitution of values [1 mark]
- Correct change in momentum: \(9.1\text{ kg m s}^{-1}\) (accept range \(9.0\text{ kg m s}^{-1}\) to \(9.2\text{ kg m s}^{-1}\)) [1 mark]

(a) The free-body diagram must display three vertical forces acting on the sphere:
- Weight (\(W\)) acting vertically downwards.
- Upthrust (\(U\)) acting vertically upwards.
- Viscous drag (\(F_{\text{drag}}\)) acting vertically upwards.

(b) The volume of the spherical bead is:
\(V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.5 \times 10^{-3})^3 \approx 1.414 \times 10^{-8}\text{ m}^3\).

Upthrust is equal to the weight of the displaced fluid:
\(U = \rho_{\text{oil}} V g = 920 \times (1.414 \times 10^{-8}) \times 9.81 \approx 1.28 \times 10^{-4}\text{ N}\).

(c) The weight of the bead is:
\(W = \rho_{\text{bead}} V g = 2500 \times (1.414 \times 10^{-8}) \times 9.81 \approx 3.47 \times 10^{-4}\text{ N}\).

At terminal velocity, the upward forces balance the downward force:
\(W = U + F_{\text{drag}}\)
\(F_{\text{drag}} = W - U = 3.47 \times 10^{-4} - 1.28 \times 10^{-4} = 2.19 \times 10^{-4}\text{ N}\).

Using Stokes' Law:
\(F_{\text{drag}} = 6 \pi \eta r v\)
\(2.19 \times 10^{-4} = 6 \pi \eta (1.5 \times 10^{-3}) (0.12)\)
\(\eta = \frac{2.19 \times 10^{-4}}{6 \pi \times 1.5 \times 10^{-3} \times 0.12} \approx 0.0645\text{ Pa s}\) (or \(\text{kg m}^{-1}\text{ s}^{-1}\)).

(a) [2 marks]
- One downward force arrow labelled Weight (or \(W\) or mg) [1 mark]
- Two upward force arrows labelled Upthrust (or \(U\)) and Viscous Drag (or \(F\) or \(D\)) [1 mark]

(b) [2.7 marks]
- Use of sphere volume formula \(V = \frac{4}{3}\pi r^3\) [1 mark]
- Use of Archimedes' Principle \(U = \rho_{\text{oil}} V g\) [1 mark]
- Correct upthrust value \(1.3 \times 10^{-4}\text{ N}\) (accept \(1.28 \times 10^{-4}\text{ N}\)) [0.7 marks]

(c) [3 marks]
- Correct equilibrium expression \(F_{\text{drag}} = W - U\) and calculation of drag [1 mark]
- Use of Stokes' Law formula \(F = 6\pi\eta r v\) [1 mark]
- Correct viscosity calculation \(0.065\text{ Pa s}\) (accept \(0.064\text{ Pa s}\) to \(0.065\text{ Pa s}\)) [1 mark]

(a) The useful work done is the increase in gravitational potential energy:
\(W = mgh = 85 \times 9.81 \times 12 = 10006.2\text{ J} \approx 1.0 \times 10^4\text{ J}\).

(b) The electrical energy input is given by:
\(E_{\text{in}} = V I t = 230 \times 1.5 \times 35 = 12075\text{ J} \approx 1.2 \times 10^4\text{ J}\).

(c) The efficiency of the system is:
\(\text{Efficiency} = \frac{\text{Useful Work Output}}{\text{Total Energy Input}} = \frac{10006.2}{12075} \approx 0.8287\) or \(82.9\%\) (accept \(83\%\)).

(a) [2 marks]
- Use of \(E_p = mgh\) [1 mark]
- Correct work value \(1.0 \times 10^4\text{ J}\) (or \(10000\text{ J}\)) [1 mark]

(b) [2.7 marks]
- Use of \(E = VIt\) [1 mark]
- Correct values substituted [1 mark]
- Correct energy input \(1.2 \times 10^4\text{ J}\) (accept \(12075\text{ J}\)) [0.7 marks]

(c) [3 marks]
- Use of \(\text{Efficiency} = \frac{\text{Useful output}}{\text{Total input}}\) [1 mark]
- Substitution of their values from (a) and (b) [1 mark]
- Correct percentage efficiency \(83\%\) (or fraction \(0.83\)) [1 mark]

(a) The difference between the curves is due to elastic hysteresis. During stretching (loading), work is done to align and slide polymer chains past one another. During contraction (unloading), some of this energy is not recovered as mechanical work and is instead transferred into thermal energy because of internal friction between molecular chains.

(b) The thermal energy transferred is the difference between the mechanical energy put in during loading and the energy recovered during unloading:
\(\text{Thermal Energy} = 4.2\text{ J} - 3.1\text{ J} = 1.1\text{ J}\).

(c) The work done in stretching is represented by the area under the loading curve down to the extension axis. The student can estimate this area by counting the total number of grid squares under the loading curve up to the maximum extension point, and then multiplying this total by the scale energy value of a single square.

(a) [2.7 marks]
- Reference to elastic hysteresis or internal molecular friction [1 mark]
- Explanation that polymer chains slide past each other and work is dissipated [1 mark]
- Identification of energy transformed into thermal energy [0.7 marks]

(b) [2 marks]
- Indication that thermal energy is the difference between the loading and unloading areas [1 mark]
- Correct subtraction resulting in \(1.1\text{ J}\) [1 mark]

(c) [3 marks]
- Identifies that work done is the entire area under the loading curve [1 mark]
- Details the method of counting grid squares under the curve [1 mark]
- Explains how to scale the counted squares to get the total energy [1 mark]

(a) The free-body force diagram must display:
- Upward normal reaction force \(F_A\) at support A (left-hand end, \(x = 0\)).
- Upward normal reaction force \(F_B\) at support B (\(1.0\text{ m}\) from the right end, \(x = 3.0\text{ m}\)).
- Downward weight force \(W = mg\) acting at the midpoint (\(x = 2.0\text{ m}\)).

(b) The plank is in equilibrium, so we can take moments about support A:
\(\sum M_A = 0\)
Clockwise moments = Counter-clockwise moments
\(W \times 2.0 = F_B \times 3.0\)
Since \(W = mg = 18 \times 9.81 = 176.58\text{ N}\):
\(176.58 \times 2.0 = 3.0 \times F_B\)
\(F_B = \frac{353.16}{3.0} \approx 117.7\text{ N}\) (or \(118\text{ N}\), which is \(120\text{ N}\) to 2 significant figures).

(c) If a heavy object is placed directly on top of support B, its entire weight is supported directly by support B. Because the line of action of this weight passes directly through the pivot B, it produces no additional moment about B. Therefore, the upward force exerted by support A remains completely unchanged.

(a) [2.7 marks]
- Downward arrow representing weight located at the center [1 mark]
- Two upward arrows representing forces from A and B correctly located [1 mark]
- Correctly labeled forces (e.g. \(W\), \(R_A\), \(R_B\)) [0.7 marks]

(b) [3 marks]
- Statement or formula representing the Principle of Moments [1 mark]
- Correct substitution of values using A as the pivot [1 mark]
- Correct calculation of the reaction force at B as \(118\text{ N}\) (or \(120\text{ N}\)) [1 mark]

(c) [2 marks]
- State that the force at A remains unchanged [1 mark]
- Explain that placing the object directly over B means B carries all the extra load directly (no extra moment about B) [1 mark]

(a) Newton's third law states that when body A exerts a force on body B, body B exerts an equal and opposite force on body A.

(b) Using the conservation of linear momentum (taking the initial drift direction as positive):
\(p_{\text{initial}} = p_{\text{final}}\)
\((m_{\text{astro}} + m_{\text{kit}}) u = m_{\text{astro}} v_{\text{astro}} + m_{\text{kit}} v_{\text{kit}}\)
\((80 + 2.5) \times 0.50 = 80 \times v_{\text{astro}} + 2.5 \times 8.0\)
\(41.25 = 80 v_{\text{astro}} + 20\)
\(80 v_{\text{astro}} = 21.25\)
\(v_{\text{astro}} = \frac{21.25}{80} \approx 0.266\text{ m s}^{-1}\) (or \(0.27\text{ m s}^{-1}\)).

(c) According to Newton's third law, when the astronaut exerts a forward force to throw the tool kit, the tool kit exerts an equal and opposite backward force on the astronaut. Newton's second law states that this unbalanced backward force causes a backward acceleration (deceleration) of the astronaut, reducing their velocity.

(a) [2 marks]
- Correct formulation of Newton's third law (action-reaction pairs between two distinct bodies) [2 marks]

(b) [3 marks]
- Application of the principle of conservation of momentum [1 mark]
- Correct substitution of values [1 mark]
- Correct final velocity \(0.27\text{ m s}^{-1}\) (accept \(0.266\text{ m s}^{-1}\)) [1 mark]

(c) [2.7 marks]
- Newton's third law application: throwing the kit forward causes a backward force on the astronaut [1 mark]
- Newton's second law link: this force results in acceleration in the opposite direction of motion [1 mark]
- Clear connection showing why the drift speed decreases [0.7 marks]

(a) The gain in kinetic energy equals the loss in gravitational potential energy: \(m g h = \frac{1}{2} m v^2\). Rearranging gives \(v = \sqrt{2 g h}\). Substituting the values: \(v = \sqrt{2 \times 9.81\text{ m s}^{-2} \times 0.85\text{ m}} = 4.08\text{ m s}^{-1}\), which is approximately \(4.1\text{ m s}^{-1}\). (b) Work done against friction is \(W = F \times d = 0.45\text{ N} \times 1.2\text{ m} = 0.54\text{ J}\). The initial kinetic energy at the bottom of the ramp is \(E_{\text{k}} = m g h = 0.35\text{ kg} \times 9.81\text{ m s}^{-2} \times 0.85\text{ m} = 2.92\text{ J}\). The kinetic energy at the table edge is \(E_{\text{k, edge}} = 2.92\text{ J} - 0.54\text{ J} = 2.38\text{ J}\). Thus, the speed is \(v_{\text{edge}} = \sqrt{\frac{2 \times E_{\text{k, edge}}}{m}} = \sqrt{\frac{2 \times 2.38\text{ J}}{0.35\text{ kg}}} = 3.69\text{ m s}^{-1}\) (or \(3.7\text{ m s}^{-1}\)). (c) For the vertical motion, \(s = u_y t + \frac{1}{2} g t^2\). Since \(u_y = 0\), \(0.95\text{ m} = \frac{1}{2} \times 9.81\text{ m s}^{-2} \times t^2\), which gives \(t = 0.440\text{ s}\). For the horizontal motion, the distance is \(x = v_{\text{edge}} \times t = 3.69\text{ m s}^{-1} \times 0.440\text{ s} = 1.62\text{ m}\) (or \(1.6\text{ m}\) to 2 significant figures).

(a) MP1: Use of \(m g h = \frac{1}{2} m v^2\) or \(v = \sqrt{2 g h}\). MP2: Correct calculation showing \(4.08\text{ m s}^{-1}\). (b) MP3: Calculation of work done against friction (\(W = 0.54\text{ J}\)) OR deceleration (\(a = 1.29\text{ m s}^{-2}\)). MP4: Use of conservation of energy or equations of motion to find final kinetic energy (\(2.38\text{ J}\)) or speed squared (\(13.6\text{ m}^2\text{ s}^{-2}\)). MP5: Correct speed \(v = 3.7\text{ m s}^{-1}\) (accept \(3.69\text{ m s}^{-1}\)). (c) MP6: Use of \(s = \frac{1}{2} g t^2\) to find time. MP7: Correct time \(t = 0.44\text{ s}\). MP8: Correct horizontal distance \(x = 1.6\text{ m}\) (accept range \(1.6\text{ m}\) to \(1.63\text{ m}\) depending on rounding, allow ECF from b).