2024 Edexcel IAL Physics (YPH11) Practice Paper with Answers

The horizontal motion has constant velocity, so \(d = vt\) and therefore the time of flight is \(t = \frac{d}{v}\). The vertical motion is under constant acceleration \(g\) from rest, so \(h = \frac{1}{2}gt^2\). Substituting \(t\) into the vertical displacement equation gives \(h = \frac{1}{2}g \left(\frac{d}{v}\right)^2 = \frac{gd^2}{2v^2}\, which corresponds to option A.

1 mark for the correct option A.

The equation of motion starting from rest is \(s = ut + \frac{1}{2}at^2\). Since the car is released from rest, \(u = 0\), which simplifies the equation to \(s = \left(\frac{1}{2}a\right)t^2\). This shows that a graph of \(s\) against \(t^2\) is linear with a gradient equal to \(\frac{1}{2}a\). Therefore, \(k = \frac{1}{2}a\), which means the acceleration is \(a = 2k\).

1 mark for the correct option C.

The initial kinetic energy is \(E_i = \frac{1}{2}mu^2\). By conservation of momentum, \(mu = (m+M)v\), where \(v\) is the common velocity of the embedded pellet and block. Thus, \(v = \frac{m}{m+M}u\). The final kinetic energy is \(E_f = \frac{1}{2}(m+M)v^2 = \frac{1}{2}\frac{m^2 u^2}{m+M}\). The fraction of kinetic energy lost is \(\frac{E_i - E_f}{E_i} = 1 - \frac{E_f}{E_i} = 1 - \frac{m}{m+M} = \frac{M}{m+M}\, which corresponds to option B.

1 mark for the correct option B.

At terminal velocity, the net force is zero, meaning weight is balanced by upthrust and viscous drag: \(W = U + F_{\text{drag}}\). This leads to \(\frac{4}{3}\pi r^3\rho_s g = \frac{4}{3}\pi r^3\rho_f g + 6\pi\eta r v\). Rearranging for velocity gives \(v = \frac{2r^2 g(\rho_s - \rho_f)}{9\eta}\). Since \(g\), densities, and viscosity are constant, terminal velocity is directly proportional to the square of the radius (\(v \propto r^2\)). Doubling the radius increases the terminal velocity by a factor of \(2^2 = 4\), giving \(4v\).

1 mark for the correct option C.

The extension is given by \(\Delta L = \frac{FL}{AE}\), where \(A = \frac{\pi d^2}{4}\). Thus, \(\Delta L = \frac{4FL}{\pi d^2 E}\). Since both wires have the same material (same Young modulus \(E\)) and identical loads (same force \(F\)), the extension is proportional to \(\frac{L}{d^2}\). Given \(L_X = 2L_Y\) and \(d_X = 0.5d_Y\), the ratio of extensions is \(\frac{\Delta L_X}{\Delta L_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2 = 2 \times \left(\frac{1}{0.5}\right)^2 = 2 \times 4 = 8\).

1 mark for the correct option C.

The useful work done in lifting the mass is equal to the gain in gravitational potential energy, \(E_p = mgh\). The useful power output of the motor is the work done divided by the time, \(P_{\text{out}} = \frac{mgh}{t}\). The efficiency is the ratio of useful power output to power input: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{mgh}{Pt}\).

1 mark for the correct option B.

The area under the loading curve represents the work done on the rubber band when stretching it. The area under the unloading curve represents the work done by the rubber band when contracting. The difference between these two areas (the area enclosed by the loop) is the energy that is not recovered mechanically, which is dissipated as thermal energy during the cycle.

1 mark for the correct option C.

The forces acting vertically on the helicopter are the upward lift force \(L\), the downward weight \(Mg\), and the downward air resistance \(R\) (since the motion is upwards). The resultant force acting upwards is \(F = L - Mg - R\). Applying Newton's second law, \(F = Ma\), gives \(L - Mg - R = Ma\).

1 mark for the correct option A.

To find the force \(F\), resolve the forces acting on the box in both the vertical and horizontal directions.

1. Vertical forces: The normal contact force \(N\) acts upwards, the vertical component of the pulling force \(F \sin\theta\) acts upwards, and the weight \(mg\) acts downwards.
Since there is no vertical acceleration:
\(N + F \sin\theta = mg\)
\(N = mg - F \sin\theta\)

2. Horizontal forces: The horizontal component of the pulling force \(F \cos\theta\) acts in the direction of motion, and the frictional force \(f = \mu N\) opposes motion.
Since the box moves at a constant velocity, the horizontal forces are in equilibrium:
\(F \cos\theta = f\)
\(F \cos\theta = \mu(mg - F \sin\theta)\)

3. Rearrange to solve for \(F\):
\(F \cos\theta = \mu mg - \mu F \sin\theta\)
\(F \cos\theta + \mu F \sin\theta = \mu mg\)
\(F(\cos\theta + \mu \sin\theta) = \mu mg\)
\(F = \frac{\mu mg}{\cos\theta + \mu \sin\theta}\)

1 mark: Correct option B selected.

The extension \(\Delta x\) of a wire of length \(L\), cross-sectional area \(A\), and Young modulus \(E\) under a tension \(F\) is given by:
\(\Delta x = \frac{FL}{AE}\)

Since the wires are connected in series, the tension \(F\) is the same in both wires. Since they are made of the same material, the Young modulus \(E\) is also the same.

The cross-sectional area \(A\) is related to the diameter \(d\) by \(A = \frac{\pi d^2}{4}\). Thus, \(A \propto d^2\).

Therefore, the extension \(\Delta x\) is proportional to \:\frac{L}{d^2}\):
\(\Delta x \propto \frac{L}{d^2}\)

For wire X:
\(\Delta x_X \propto \frac{L}{d^2}\)

For wire Y:
\(\Delta x_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\)

Taking the ratio of their extensions:
\(\frac{\Delta x_X}{\Delta x_Y} = \frac{L/d^2}{L/(2d^2)} = 2\)

1 mark: Correct option C selected.

(a) Mechanical energy is a scalar quantity. Since air resistance is negligible, the total mechanical energy of the system is conserved. The work done by gravity depends only on the change in vertical height, and the initial elastic energy depends only on the compression of the spring. Therefore, the final kinetic energy (and thus the speed) depends only on the initial total mechanical energy and is independent of the direction of launch.

(b) Initial elastic potential energy:
\(E_e = \frac{1}{2} k x^2 = 0.5 \times 240 \text{ N m}^{-1} \times (0.050 \text{ m})^2 = 0.30 \text{ J}\)

Initial gravitational potential energy:
\(E_p = m g h = 0.045 \text{ kg} \times 9.81 \text{ m s}^{-2} \times 1.2 \text{ m} = 0.530 \text{ J}\)

Total initial energy:
\(E_{\text{total}} = 0.30 \text{ J} + 0.530 \text{ J} = 0.830 \text{ J}\)

Since mechanical energy is conserved:
\(E_k = \frac{1}{2} m v^2 = 0.830 \text{ J}\)

\(v = \sqrt{\frac{2 \times 0.830 \text{ J}}{0.045 \text{ kg}}} = 6.07 \text{ m s}^{-1} \approx 6.1 \text{ m s}^{-1}\)

(a) [2 marks]
- Statement that energy is a scalar quantity / conservation of energy depends only on initial and final heights, not the path taken [1]
- Statement that work done against air resistance is zero, so mechanical energy is conserved [1]

(b) [5.7 marks]
- Use of \(E_e = \frac{1}{2} k x^2\) to find initial elastic potential energy = 0.30 J [1.7]
- Use of \(E_p = m g h\) to find gravitational potential energy = 0.530 J [1]
- Addition of energies to find total initial energy = 0.830 J [1]
- Equating total initial energy to final kinetic energy \(E_k = \frac{1}{2} m v^2\) [1]
- Correct calculation of final speed with correct units: \(6.1 \text{ m s}^{-1}\) (Accept \(6.07 \text{ m s}^{-1}\)) [1]

(a) Using the Young modulus definition:
\(E = \frac{\text{stress}}{\text{strain}} = \frac{F L}{A \Delta x}\)

Rearranging to solve for extension \(\Delta x\):
\(\Delta x = \frac{F L}{A E} = \frac{75 \text{ N} \times 2.4 \text{ m}}{4.5 \times 10^{-7} \text{ m}^2 \times 2.0 \times 10^{11} \text{ Pa}} = \frac{180}{90000} = 2.0 \times 10^{-3} \text{ m}\) (or \(2.0 \text{ mm}\))

(b) The elastic strain energy stored is given by:
\(E_{\text{el}} = \frac{1}{2} F \Delta x = 0.5 \times 75 \text{ N} \times 2.0 \times 10^{-3} \text{ m} = 0.075 \text{ J}\)

(a) [3.7 marks]
- Use of the Young modulus formula rearranged for extension: \(\Delta x = \frac{F L}{A E}\) [1.7]
- Correct substitution of values [1]
- Correct calculation of extension with units: \(2.0 \times 10^{-3} \text{ m}\) or \(2.0 \text{ mm}\) [1]

(b) [4.0 marks]
- Use of \(E_{\text{el}} = \frac{1}{2} F \Delta x\) [2]
- Substitution of their extension value [1]
- Correct calculation of elastic strain energy with units: \(0.075 \text{ J}\) [1]

(a) According to the conservation of linear momentum:
\(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f\)

Since glider 2 is initially stationary (\(v_2 = 0\)):
\(0.35 \text{ kg} \times 1.8 \text{ m s}^{-1} + 0 = (0.35 \text{ kg} + 0.15 \text{ kg}) v_f\)
\(0.63 \text{ kg m s}^{-1} = 0.50 \text{ kg} \times v_f\)
\(v_f = \frac{0.63}{0.50} = 1.26 \text{ m s}^{-1}\)

(b) Initial kinetic energy of the system:
\(E_{ki} = \frac{1}{2} m_1 v_1^2 = 0.5 \times 0.35 \text{ kg} \times (1.8 \text{ m s}^{-1})^2 = 0.567 \text{ J}\)

Final kinetic energy of the system:
\(E_{kf} = \frac{1}{2} (m_1 + m_2) v_f^2 = 0.5 \times 0.50 \text{ kg} \times (1.26 \text{ m s}^{-1})^2 = 0.3969 \text{ J}\)

Loss in kinetic energy:
\(\Delta E_k = E_{ki} - E_{kf} = 0.567 \text{ J} - 0.397 \text{ J} = 0.170 \text{ J}\) (or \(0.17 \text{ J}\))

(a) [3.0 marks]
- Statement of conservation of momentum or use of \(m_1 v_1 = (m_1 + m_2) v_f\) [1]
- Calculation of total initial momentum as \(0.63 \text{ kg m s}^{-1}\) [1]
- Correct calculation of final speed with units: \(1.26 \text{ m s}^{-1}\) (Accept \(1.3 \text{ m s}^{-1}\)) [1]

(b) [4.7 marks]
- Use of \(E_k = \frac{1}{2} m v^2\) to find initial kinetic energy = 0.567 J [1.7]
- Calculation of final kinetic energy = 0.397 J [1]
- Subtraction of energies to find the loss [1]
- Correct calculation of energy loss with units: \(0.17 \text{ J}\) (Accept \(0.170 \text{ J}\)) [1]

(a) The volume of the sphere is:
\(V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.2 \times 10^{-3} \text{ m})^3 = 7.238 \times 10^{-9} \text{ m}^3\)

According to Archimedes' principle, the upthrust is equal to the weight of the fluid displaced:
\(U = \rho_{\text{fluid}} V g = 960 \text{ kg m}^{-3} \times 7.238 \times 10^{-9} \text{ m}^3 \times 9.81 \text{ m s}^{-2} = 6.817 \times 10^{-5} \text{ N} \approx 6.8 \times 10^{-5} \text{ N}\)

(b) The weight of the copper bead is:
\(W = \rho_{\text{copper}} V g = 8900 \text{ kg m}^{-3} \times 7.238 \times 10^{-9} \text{ m}^3 \times 9.81 \text{ m s}^{-2} = 6.319 \times 10^{-4} \text{ N}\)

At terminal velocity, the upward forces equal the downward forces:
\(W = U + F_{\text{drag}}\)
\(F_{\text{drag}} = W - U = 6.319 \times 10^{-4} \text{ N} - 6.817 \times 10^{-5} \text{ N} = 5.637 \times 10^{-4} \text{ N}\)

Using Stokes' Law:
\(F_{\text{drag}} = 6 \pi \eta r v\)

Rearranging to solve for terminal velocity \(v\):
\(v = \frac{F_{\text{drag}}}{6 \pi \eta r} = \frac{5.637 \times 10^{-4} \text{ N}}{6 \pi \times 0.95 \text{ Pa s} \times 1.2 \times 10^{-3} \text{ m}} = 0.0262 \text{ m s}^{-1}\)

(a) [2.7 marks]
- Use of \(V = \frac{4}{3} \pi r^3\) to find volume = \(7.2 \times 10^{-9} \text{ m}^3\) [1]
- Use of \(U = \rho_{\text{fluid}} V g\) [1]
- Clear substitution and calculation resulting in \(6.8 \times 10^{-5} \text{ N}\) [0.7]

(b) [5.0 marks]
- Use of \(W = \rho V g\) to calculate the bead weight = \(6.3 \times 10^{-4} \text{ N}\) [1]
- Use of equilibrium equation \(W = U + F_{\text{drag}}\) to find drag force = \(5.6 \times 10^{-4} \text{ N}\) [1]
- Identification of Stokes' Law \(F = 6 \pi \eta r v\) [1]
- Rearranging for terminal velocity \(v\) [1]
- Correct calculation of terminal velocity with units: \(0.026 \text{ m s}^{-1}\) (Accept \(2.6 \text{ cm s}^{-1}\)) [1]

(a) The three forces acting on the shelf are:
1. The weight of the shelf \(W = m g\) acting vertically downwards from the geometric center (midpoint, 0.60 m from A).
2. The tension \(T\) in the cable acting at end B, pulling at an angle of 35° above the horizontal.
3. The reaction force \(R\) from the hinge at end A, which has both horizontal and vertical components to keep the shelf in equilibrium.

(b) Take moments about the hinge (A) to eliminate the force \(R\):
Clockwise moment about A:
\(\tau_{\text{clockwise}} = W \times 0.60 \text{ m} = (4.5 \text{ kg} \times 9.81 \text{ m s}^{-2}) \times 0.60 \text{ m} = 26.49 \text{ N m}\)

Anticlockwise moment about A:
\(\tau_{\text{anticlockwise}} = (T \sin 35^\circ) \times 1.2 \text{ m} = 0.6883 T \text{ m}\)

For rotational equilibrium:
\(\tau_{\text{clockwise}} = \tau_{\text{anticlockwise}}\)
\(26.49 \text{ N m} = 0.6883 T \text{ m}\)

\(T = \frac{26.49}{0.6883} = 38.5 \text{ N} \approx 39 \text{ N}\)

(a) [3.0 marks]
- Weight shown acting vertically downwards at the midpoint [1]
- Tension shown acting at the end B pointing up and left at 35° [1]
- Hinge force shown at end A with components directed upwards and rightwards [1]

(b) [4.7 marks]
- Use of moment formula \(\text{moment} = F d\) [1]
- Taking moments about A with weight acting at 0.60 m [1]
- Calculating the moment of the tension force using perpendicular component \(T \sin 35^\circ \times 1.2\) [1]
- Equating clockwise and anticlockwise moments [0.7]
- Correct calculation of tension with units: \(39 \text{ N}\) (Accept \(38.5 \text{ N}\)) [1]

(a) First, find the total volume of the wooden block:
\(V = A \times h = 0.012 \text{ m}^2 \times 0.25 \text{ m} = 0.0030 \text{ m}^3\)

Calculate the mass using density:
\(m = \rho_{\text{wood}} \times V = 680 \text{ kg m}^{-3} \times 0.0030 \text{ m}^3 = 2.04 \text{ kg}\)

(b) Since the block is floating, the upthrust \(U\) is equal to its weight \(W\):
\(U = W = m g = 2.04 \text{ kg} \times 9.81 \text{ m s}^{-2} = 20.01 \text{ N}\)

According to Archimedes' principle:
\(U = \rho_{\text{water}} \times V_{\text{submerged}} \times g\)

Equating the expressions for upthrust:
\(\rho_{\text{water}} \times V_{\text{submerged}} \times g = m g\)
\(V_{\text{submerged}} = \frac{m}{\rho_{\text{water}}} = \frac{2.04 \text{ kg}}{1000 \text{ kg m}^{-3}} = 2.04 \times 10^{-3} \text{ m}^3\)

The submerged depth \(d\) is given by:
\(d = \frac{V_{\text{submerged}}}{A} = \frac{2.04 \times 10^{-3} \text{ m}^3}{0.012 \text{ m}^2} = 0.17 \text{ m}\)

(a) [2.7 marks]
- Use of \(V = A h\) to find block volume = \(0.0030 \text{ m}^3\) [1]
- Use of \(m = \rho V\) [1]
- Correct calculation of mass: \(2.04 \text{ kg}\) [0.7]

(b) [5.0 marks]
- Statement that upthrust equals weight of the floating object [1]
- Use of Archimedes' Principle formula: \(U = \rho_{\text{water}} V_{\text{submerged}} g\) [1]
- Equating upthrust to weight of the block to find submerged volume = \(2.04 \times 10^{-3} \text{ m}^3\) [1]
- Use of \(d = \frac{V_{\text{submerged}}}{A}\) [1]
- Correct calculation of submerged depth with units: \(0.17 \text{ m}\) [1]

(a) Since the crate is being lifted at a constant speed, the force exerted by the motor is equal to the weight of the crate:
\(F = m g = 85 \text{ kg} \times 9.81 \text{ m s}^{-2} = 833.85 \text{ N}\)

Useful power output:
\(P_{\text{out}} = F v = 833.85 \text{ N} \times 1.5 \text{ m s}^{-1} = 1250.8 \text{ W} \approx 1250 \text{ W}\)

(b) The electrical power input is related to useful power output by the efficiency:
\(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%\)
\(0.72 = \frac{1250.8 \text{ W}}{P_{\text{in}}}\)
\(P_{\text{in}} = \frac{1250.8}{0.72} = 1737.2 \text{ W} \approx 1740 \text{ W}\)

Electrical energy supplied to the motor in 15 seconds:
\(E = P_{\text{in}} \times t = 1737.2 \text{ W} \times 15 \text{ s} = 26058 \text{ J} \approx 2.6 \times 10^4 \text{ J}\) (or \(26 \text{ kJ}\))

(a) [3.0 marks]
- Calculation of weight as force \(F = 834 \text{ N}\) [1]
- Use of \(P = F v\) [1]
- Correct calculation of useful power output with units: \(1250 \text{ W}\) [1]

(b) [4.7 marks]
- Use of \(\text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\) [1.7]
- Correct calculation of input power: \(1740 \text{ W}\) (Accept \(1737 \text{ W}\)) [1]
- Use of \(E = P_{\text{in}} \times t\) [1]
- Correct calculation of electrical energy with units: \(2.6 \times 10^4 \text{ J}\) or \(26 \text{ kJ}\) [1]

(a)
- Limit of proportionality: The point beyond which stress is no longer directly proportional to strain (Hooke's Law is no longer obeyed).
- Ultimate tensile strength: The maximum stress that a material can withstand before breaking under tension.

(b) The Young modulus is the gradient of the linear region of the stress-strain graph:
\(E = \frac{\text{stress}}{\text{strain}} = \frac{1.2 \times 10^8 \text{ Pa}}{9.6 \times 10^{-4}} = 1.25 \times 10^{11} \text{ Pa}\)

At the limit of proportionality:
\(\text{stress} = 1.2 \times 10^8 \text{ Pa}\)
\(F = 150 \text{ N}\)

Using stress definition:
\(\text{stress} = \frac{F}{A}\)
\(A = \frac{F}{\text{stress}} = \frac{150 \text{ N}}{1.2 \times 10^8 \text{ Pa}} = 1.25 \times 10^{-6} \text{ m}^2\)

(a) [3.0 marks]
- Limit of proportionality defined as the point beyond which stress and strain are no longer directly proportional [1.5]
- Ultimate tensile strength defined as the maximum tensile stress a material can withstand [1.5]

(b) [4.7 marks]
- Use of gradient of stress-strain graph to find Young modulus \(E = \frac{\text{stress}}{\text{strain}}\) [1.7]
- Correct calculation of Young modulus: \(1.25 \times 10^{11} \text{ Pa}\) [1]
- Use of \(\text{stress} = \frac{F}{A}\) [1]
- Correct calculation of cross-sectional area: \(1.25 \times 10^{-6} \text{ m}^2\) [1]

**(a)(i)** Tensile stress is the tensile force applied per unit cross-sectional area: \(\sigma = \frac{F}{A}\).

**(a)(ii)** Tensile strain is the extension per unit original length: \(\varepsilon = \frac{\Delta L}{L}\).

**(b)** First, calculate the cross-sectional area \(A\) of the string:
\[A = \frac{\pi d^2}{4} = \frac{\pi (0.85 \times 10^{-3}\text{ m})^2}{4} = 5.67 \times 10^{-7}\text{ m}^2\]

Using the Young modulus formula:
\[E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta L}\]

Rearranging for extension \(\Delta L\):
\[\Delta L = \frac{F L}{A E} = \frac{85\text{ N} \times 1.20\text{ m}}{5.67 \times 10^{-7}\text{ m}^2 \times 2.4 \times 10^9\text{ Pa}} \approx 0.075\text{ m}\text{ (or } 7.5 \times 10^{-2}\text{ m)}\]

**(c)** The strain energy \(E_{\text{str}}\) stored in the elastic region is given by:
\[E_{\text{str}} = \frac{1}{2} F \Delta L\]

\[E_{\text{str}} = 0.5 \times 85\text{ N} \times 0.0749\text{ m} \approx 3.2\text{ J}\text{ (or } 3.18\text{ J)}\]

**(d)** If the diameter is halved, the cross-sectional area \(A\) decreases to \(\frac{1}{4}\) of its original value (since \(A \propto d^2\)).
Since extension \(\Delta L = \frac{FL}{AE}\), the extension \(\Delta L\) will increase by a factor of 4.
Since strain energy is \(E_{\text{str}} = \frac{1}{2} F \Delta L\) and the tension \(F\) is kept constant, the strain energy stored will increase by a factor of 4 (quadruple).

**(a)(i)**
- Force divided by cross-sectional area (1)

**(a)(ii)**
- Extension divided by original length (1)

**(b)**
- Use of \(A = \frac{\pi d^2}{4}\) to obtain \(5.67 \times 10^{-7}\text{ m}^2\) (1)
- Use of \(E = \frac{F L}{A \Delta L}\) or equivalent rearranged (1)
- Correct calculation of \(\Delta L = 0.075\text{ m}\) or \(7.5 \times 10^{-2}\text{ m}\) (1)

**(c)**
- Use of \(E_{\text{str}} = \frac{1}{2} F \Delta L\) (1)
- Correct calculation of energy = \(3.2\text{ J}\) (accept \(3.18\text{ J}\) to \(3.2\text{ J}\) depending on rounding) (1)

**(d)**
- States that halving the diameter reduces the area by 4, which increases the extension by 4, thus quadrupling (increasing by a factor of 4) the strain energy (1)