AI Generated Exam Paper

A Level H1 Physics Practice Paper 2

Free AI-Generated Qwen3.6 Plus A Level H1 Physics Practice Paper 2 practice paper with questions and answers for Singapore students. This page is rendered as a direct URL so the questions and answers can be discovered without pressing in-page buttons.

These static practice materials are generated from the site's syllabus and paper-generation workflow, with source and model context shown so students and parents can evaluate the material before use.

A Level H1 Physics AI Generated Generated by Qwen3.6 Plus Updated 2026-06-03

Back to Subject Browse Free Exam Papers

Questions

TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics H1
Level: A-Level
Paper: Practice Paper 2 (Version 2 of 5)
Duration: 2 hours
Total Marks: 80

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided in this booklet.
All working must be clearly shown.
The use of an approved scientific calculator is expected.
Where appropriate, assume $g = 9.81 \text{ m s}^{-2}$ .

Section A

Answer all questions in this section.

1. A student measures the diameter $d$ of a steel sphere using a micrometer screw gauge. The readings obtained are: $d_1 = 12.42 \text{ mm}, \quad d_2 = 12.44 \text{ mm}, \quad d_3 = 12.41 \text{ mm}, \quad d_4 = 12.43 \text{ mm}$ The micrometer has a zero error of $+0.02 \text{ mm}$ .

(a) Determine the corrected mean diameter of the sphere.
[2]

(b) The mass $m$ of the sphere is measured as $7.80 \pm 0.05 \text{ g}$ . Calculate the density $\rho$ of the steel and its absolute uncertainty.
[3]

2. A car travels along a straight horizontal road. The velocity-time graph for the car is shown below.

(Imagine a graph: Velocity starts at 0, increases linearly to 20 m/s at t=5s, remains constant at 20 m/s until t=15s, then decreases linearly to 0 at t=20s.)

(a) Calculate the acceleration of the car during the first 5 seconds.
[1]

(b) Determine the total distance travelled by the car during the 20 seconds.
[2]

(c) Explain, in terms of forces, why the car moves at a constant velocity between $t = 5 \text{ s}$ and $t = 15 \text{ s}$ .
[2]

3. A projectile is launched from ground level with an initial velocity of $30 \text{ m s}^{-1}$ at an angle of $40^\circ$ to the horizontal. Air resistance is negligible.

(a) Calculate the horizontal component of the initial velocity.
[1]

(b) Calculate the maximum height reached by the projectile.
[3]

4. A block of mass $5.0 \text{ kg}$ rests on a rough horizontal surface. A horizontal force $F$ is applied to the block. The variation of the frictional force $f$ with the applied force $F$ is shown in the graph below.

(Imagine a graph: Frictional force increases linearly with Applied Force up to 25 N, then drops slightly to 20 N and remains constant as Applied Force increases further.)

(a) Determine the maximum static frictional force acting on the block.
[1]

(b) Calculate the coefficient of dynamic friction between the block and the surface.
[2]

5. Two trolleys, A and B, move on a smooth horizontal track. Trolley A has a mass of $2.0 \text{ kg}$ and moves with a velocity of $4.0 \text{ m s}^{-1}$ to the right. Trolley B has a mass of $3.0 \text{ kg}$ and is initially at rest. The trolleys collide and stick together.

(a) State the principle of conservation of linear momentum.
[2]

(b) Calculate the common velocity of the trolleys after the collision.
[3]

Section B

Answer all questions in this section.

6. A uniform beam AB of length $4.0 \text{ m}$ and weight $200 \text{ N}$ is hinged at end A to a vertical wall. The beam is held horizontal by a cable attached to end B and to the wall at a point $3.0 \text{ m}$ vertically above A.

(a) Draw a free-body diagram showing all the forces acting on the beam. Label the forces clearly.
[3]

(b) Calculate the tension in the cable.
[4]

7. A satellite of mass $m$ orbits the Earth in a circular orbit of radius $r$ with speed $v$ .

(a) Show that the gravitational force provides the centripetal force required for the orbit.
[2]

(b) Derive an expression for the orbital speed $v$ in terms of the gravitational constant $G$ , the mass of the Earth $M$ , and the orbital radius $r$ .
[3]

8. A motor lifts a load of mass $500 \text{ kg}$ vertically through a height of $20 \text{ m}$ in $15 \text{ s}$ . The motor operates at a constant power output of $8.0 \text{ kW}$ .

(a) Calculate the useful work done in lifting the load.
[2]

(b) Calculate the efficiency of the motor.
[3]

9. A ball of mass $0.20 \text{ kg}$ is dropped from a height of $2.0 \text{ m}$ onto a hard floor. It rebounds to a height of $1.5 \text{ m}$ .

(a) Calculate the speed of the ball just before it hits the floor.
[2]

(b) Calculate the speed of the ball just after it leaves the floor.
[2]

10. A car of mass $1200 \text{ kg}$ travels up a slope inclined at $5^\circ$ to the horizontal at a constant speed of $20 \text{ m s}^{-1}$ . The total resistive force acting on the car is $400 \text{ N}$ .

(a) Calculate the component of the car's weight acting down the slope.
[2]

(b) Determine the driving force required to maintain this constant speed.
[2]

Section C

Answer all questions in this section.

11. A student investigates the relationship between the extension $x$ of a spring and the load $F$ applied to it. The results are plotted on a graph of $F$ against $x$ .

(a) State Hooke's Law.
[1]

(b) The graph is a straight line passing through the origin with a gradient of $50 \text{ N m}^{-1}$ . Determine the spring constant.
[1]

(d) The spring is now cut in half. State and explain the effect on the spring constant of each half.
[2]

12. A river flows with a speed of $3.0 \text{ m s}^{-1}$ due East. A boat heads due North with a speed of $4.0 \text{ m s}^{-1}$ relative to the water.

(a) Draw a vector diagram to show the velocity of the boat relative to the ground.
[2]

(b) Calculate the magnitude of the resultant velocity of the boat.
[2]

13. A object is subjected to two forces: $F_1 = 10 \text{ N}$ acting horizontally to the right, and $F_2 = 10 \text{ N}$ acting vertically upwards.

(a) Calculate the magnitude of the resultant force.
[2]

(b) A third force $F_3$ is applied to keep the object in equilibrium. Determine the magnitude and direction of $F_3$ .
[3]

14. A cyclist travels around a circular track of radius $50 \text{ m}$ at a constant speed of $10 \text{ m s}^{-1}$ .

(a) Explain why the cyclist is accelerating even though the speed is constant.
[2]

(b) Calculate the centripetal acceleration of the cyclist.
[2]

15. A rocket of mass $1000 \text{ kg}$ is launched vertically upwards. The engine produces a thrust of $15000 \text{ N}$ .

(a) Calculate the weight of the rocket.
[1]

(b) Calculate the initial acceleration of the rocket.
[3]

(c) As the rocket rises, its mass decreases due to fuel consumption. State and explain the effect on the acceleration of the rocket, assuming the thrust remains constant.
[2]

16. A pendulum bob of mass $0.5 \text{ kg}$ is pulled to one side so that it is raised by a vertical height of $0.2 \text{ m}$ and then released.

(a) Calculate the gravitational potential energy gained by the bob.
[2]

(b) Calculate the maximum speed of the bob as it passes through the lowest point.
[3]

17. A block slides down a smooth inclined plane of angle $30^\circ$ .

(a) Draw a free-body diagram for the block.
[2]

(b) Calculate the acceleration of the block down the slope.
[3]

(c) If the plane is rough, the acceleration is reduced to $2.0 \text{ m s}^{-2}$ . Calculate the frictional force acting on the block if its mass is $2.0 \text{ kg}$ .
[3]

18. A crane lifts a container of mass $2000 \text{ kg}$ vertically at a constant speed of $0.5 \text{ m s}^{-1}$ .

(a) Calculate the tension in the cable.
[2]

(b) Calculate the power output of the crane motor.
[2]

19. Two spheres, X and Y, collide head-on. Sphere X has mass $m$ and velocity $2v$ . Sphere Y has mass $2m$ and velocity $-v$ (moving in the opposite direction).

(a) Calculate the total momentum of the system before the collision.
[2]

(b) After the collision, sphere X moves with velocity $-v$ . Calculate the velocity of sphere Y after the collision.
[3]

20. A car accelerates from rest to $30 \text{ m s}^{-1}$ in $10 \text{ s}$ . The mass of the car is $1000 \text{ kg}$ .

(a) Calculate the average acceleration.
[1]

(b) Calculate the average net force acting on the car.
[2]

(d) Calculate the work done by the net force.
[2]

Answers

TuitionGoWhere Practice Paper - Physics H1 A-Level (Answers)

Version 2 of 5

Section A

1. (a) Mean reading = $(12.42 + 12.44 + 12.41 + 12.43) / 4 = 12.425 \text{ mm}$ .
Corrected mean = Mean - Zero Error = $12.425 - 0.02 = 12.405 \text{ mm}$ .
Rounding to 2 decimal places (precision of instrument): 12.41 mm (or 12.40 mm depending on rounding convention, usually keep extra digit for intermediate). Let's use 12.41 mm.
[1 for mean, 1 for correction]

(b) Volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (\frac{12.41 \times 10^{-3}}{2})^3 = 9.99 \times 10^{-7} \text{ m}^3$ .
Density $\rho = \frac{m}{V} = \frac{7.80 \times 10^{-3}}{9.99 \times 10^{-7}} = 7808 \text{ kg m}^{-3}$ .
Uncertainty:
% unc in $m = \frac{0.05}{7.80} \times 100 = 0.64\%$ .
% unc in $d = \frac{0.01}{12.41} \times 100 \approx 0.08\%$ (using range/2 or least count). Let's assume least count 0.01mm.
% unc in $V = 3 \times \% \text{ unc in } d = 3 \times 0.08 = 0.24\%$ .
Total % unc in $\rho = 0.64 + 0.24 = 0.88\%$ .
Absolute unc = $0.0088 \times 7808 \approx 69 \text{ kg m}^{-3}$ .
Answer: $7810 \pm 70 \text{ kg m}^{-3}$ .
[1 for density, 1 for % unc method, 1 for final answer]

2. (a) Acceleration $a = \frac{\Delta v}{\Delta t} = \frac{20 - 0}{5} = \mathbf{4.0 \text{ m s}^{-2}}$ .
[1]

(b) Distance = Area under graph.
Area = Area of triangle (0-5s) + Area of rectangle (5-15s) + Area of triangle (15-20s).
$= (0.5 \times 5 \times 20) + (10 \times 20) + (0.5 \times 5 \times 20)$
$= 50 + 200 + 50 = \mathbf{300 \text{ m}}$ .
[1 for method, 1 for answer]

(c) At constant velocity, acceleration is zero. According to Newton's First Law, if acceleration is zero, the resultant force is zero. Therefore, the driving force equals the resistive forces (friction/air resistance).
[1 for zero resultant force, 1 for explanation]

3. (a) $v_x = 30 \cos 40^\circ = \mathbf{23.0 \text{ m s}^{-1}}$ .
[1]

(b) Vertical component $v_y = 30 \sin 40^\circ = 19.28 \text{ m s}^{-1}$ .
At max height, $v_y = 0$ .
$v^2 = u^2 + 2as \Rightarrow 0 = (19.28)^2 + 2(-9.81)h$ .
$h = \frac{371.7}{19.62} = \mathbf{18.9 \text{ m}}$ .
[1 for vertical component, 1 for substitution, 1 for answer]

(c) Air resistance opposes motion. It reduces the horizontal velocity over time (horizontal deceleration). Thus, the horizontal range is reduced.
[1 for reduced, 1 for explanation]

4. (a) From graph, max static friction is the peak value before it drops. 25 N.
[1]

(b) Dynamic friction $f_d = 20 \text{ N}$ . Normal reaction $R = mg = 5.0 \times 9.81 = 49.05 \text{ N}$ .
$\mu_d = \frac{f_d}{R} = \frac{20}{49.05} = \mathbf{0.41}$ .
[1 for formula, 1 for answer]

(c) Net force $F_{net} = F_{applied} - f_d = 30 - 20 = 10 \text{ N}$ .
$a = \frac{F_{net}}{m} = \frac{10}{5.0} = \mathbf{2.0 \text{ m s}^{-2}}$ .
[1 for net force, 1 for answer]

5. (a) In a closed/isolated system (no external forces), the total momentum before collision equals the total momentum after collision.
[1 for "closed system/no external forces", 1 for "momentum constant"]

(b) $m_A u_A + m_B u_B = (m_A + m_B)v$ .
$(2.0)(4.0) + (3.0)(0) = (2.0 + 3.0)v$ .
$8.0 = 5.0v \Rightarrow v = \mathbf{1.6 \text{ m s}^{-1}}$ .
[1 for equation, 1 for substitution, 1 for answer]

(c) $KE_{initial} = \frac{1}{2}(2.0)(4.0)^2 = 16 \text{ J}$ .
$KE_{final} = \frac{1}{2}(5.0)(1.6)^2 = 6.4 \text{ J}$ .
Since $KE_{initial} \neq KE_{final}$ , the collision is inelastic.
[1 for KE init, 1 for KE final, 1 for conclusion]

Section B

6. (a) Forces:

Weight ( $200 \text{ N}$ ) acting downwards at the center of the beam (2.0 m from A).
Tension ( $T$ ) acting along the cable from B towards the wall.
Reaction force at hinge A ( $R$ ), with horizontal and vertical components.
[1 for each correct force vector]

(b) Angle of cable with horizontal: $\tan \theta = \frac{3.0}{4.0} \Rightarrow \theta = 36.9^\circ$ .
Take moments about A:
Clockwise moment = Anticlockwise moment.
Weight moment: $200 \times 2.0 = 400 \text{ Nm}$ .
Tension moment: Vertical component of $T$ is $T \sin \theta$ . Perpendicular distance is 4.0 m.
$T \sin(36.9^\circ) \times 4.0 = 400$ .
$T \times 0.6 \times 4.0 = 400 \Rightarrow 2.4 T = 400$ .
$T = \mathbf{167 \text{ N}}$ (or 166.7 N).
[1 for moment equation, 1 for angle/component, 1 for calculation, 1 for answer]

(c) Horizontal equilibrium: $R_x = T \cos \theta = 166.7 \cos(36.9^\circ) = 133.3 \text{ N}$ (to the left).
Vertical equilibrium: $R_y + T \sin \theta = 200$ .
$R_y + 100 = 200 \Rightarrow R_y = 100 \text{ N}$ (upwards).
Magnitude $R = \sqrt{133.3^2 + 100^2} = \mathbf{167 \text{ N}}$ .
Direction: $\tan \alpha = \frac{100}{133.3} \Rightarrow \alpha = 36.9^\circ$ above horizontal.
[1 for Rx, 1 for Ry, 1 for magnitude, 1 for direction]

7. (a) Gravitational force $F_g = \frac{GMm}{r^2}$ . Centripetal force $F_c = \frac{mv^2}{r}$ . For circular orbit, $F_g$ provides $F_c$ .
[1 for stating forces, 1 for equality]

(b) $\frac{GMm}{r^2} = \frac{mv^2}{r}$ .
Cancel $m$ and one $r$ : $\frac{GM}{r} = v^2$ .
$v = \mathbf{\sqrt{\frac{GM}{r}}}$ .
[1 for algebraic steps, 1 for final expression]

(c) $v \propto \frac{1}{\sqrt{r}}$ . If $r$ doubles, $v$ becomes $\frac{1}{\sqrt{2}}$ times the original speed.
Speed decreases by factor of $\sqrt{2}$ .
[1 for relationship, 1 for comparison]

8. (a) Work done against gravity $W = mgh = 500 \times 9.81 \times 20 = \mathbf{98100 \text{ J}}$ (or 98.1 kJ).
[1 for formula, 1 for answer]

(b) Input Energy = Power $\times$ Time = $8000 \times 15 = 120000 \text{ J}$ .
Efficiency = $\frac{\text{Useful Output}}{\text{Total Input}} \times 100\% = \frac{98100}{120000} \times 100\% = \mathbf{81.8\%}$ .
[1 for input energy, 1 for ratio, 1 for answer]

(c) Energy is lost as heat due to friction in the motor parts and air resistance, and sound.
[1 for heat/friction, 1 for sound/other]

9. (a) $v^2 = u^2 + 2gh$ . $u=0$ . $v = \sqrt{2 \times 9.81 \times 2.0} = \sqrt{39.24} = \mathbf{6.26 \text{ m s}^{-1}}$ .
[1 for formula, 1 for answer]

(b) Rebound height $h' = 1.5 \text{ m}$ . At max height $v=0$ .
$0 = v_{rebound}^2 - 2gh' \Rightarrow v_{rebound} = \sqrt{2 \times 9.81 \times 1.5} = \sqrt{29.43} = \mathbf{5.42 \text{ m s}^{-1}}$ .
[1 for formula, 1 for answer]

(c) Impulse $I = \Delta p = m(v_{final} - v_{initial})$ .
Take upward as positive. $v_{final} = +5.42$ , $v_{initial} = -6.26$ .
$I = 0.20 (5.42 - (-6.26)) = 0.20 (11.68) = \mathbf{2.34 \text{ N s}}$ .
[1 for delta p concept, 1 for sign convention, 1 for answer]

10. (a) Component of weight down slope $= mg \sin \theta = 1200 \times 9.81 \times \sin 5^\circ = 1200 \times 9.81 \times 0.0872 = \mathbf{1026 \text{ N}}$ .
[1 for formula, 1 for answer]

(b) Constant speed means equilibrium. Driving Force $D = \text{Resistive Force} + \text{Weight Component}$ .
$D = 400 + 1026 = \mathbf{1426 \text{ N}}$ .
[1 for equilibrium concept, 1 for answer]

Section C

11. (a) Hooke's Law: The extension of a spring is directly proportional to the load applied, provided the limit of proportionality is not exceeded.
[1]

(b) Gradient = Spring Constant $k$ . $k = 50 \text{ N m}^{-1}$ .
[1]

(c) $E = \frac{1}{2}kx^2 = 0.5 \times 50 \times (0.10)^2 = 25 \times 0.01 = \mathbf{0.25 \text{ J}}$ .
[1 for formula, 1 for answer]

(d) Spring constant is inversely proportional to length ( $k \propto 1/L$ ). If length is halved, the spring constant doubles.
[1 for doubles, 1 for explanation]

12. (a) Vector diagram: Vertical vector (4.0) and Horizontal vector (3.0) added head-to-tail. Resultant is hypotenuse.
[1 for correct vectors, 1 for resultant]

(b) Magnitude $v = \sqrt{3.0^2 + 4.0^2} = \sqrt{9+16} = \sqrt{25} = \mathbf{5.0 \text{ m s}^{-1}}$ .
[1 for Pythagoras, 1 for answer]

(c) Direction $\theta$ from North: $\tan \theta = \frac{3.0}{4.0} = 0.75$ .
$\theta = \mathbf{36.9^\circ}$ East of North.
[1 for tan ratio, 1 for answer]

13. (a) Resultant $R = \sqrt{10^2 + 10^2} = \sqrt{200} = \mathbf{14.1 \text{ N}}$ .
[1 for formula, 1 for answer]

(b) For equilibrium, $F_3$ must be equal and opposite to the resultant of $F_1$ and $F_2$ .
Magnitude = 14.1 N.
Direction: $45^\circ$ below the horizontal (or $225^\circ$ from positive x-axis).
[1 for magnitude, 1 for direction logic, 1 for specific direction]

14. (a) Velocity is a vector (speed + direction). Since the direction is constantly changing, the velocity is changing. Acceleration is the rate of change of velocity.
[1 for vector nature, 1 for changing direction]

(b) $a = \frac{v^2}{r} = \frac{10^2}{50} = \frac{100}{50} = \mathbf{2.0 \text{ m s}^{-2}}$ .
[1 for formula, 1 for answer]

15. (a) Weight $W = mg = 1000 \times 9.81 = \mathbf{9810 \text{ N}}$ .
[1]

(b) Net Force $F_{net} = \text{Thrust} - \text{Weight} = 15000 - 9810 = 5190 \text{ N}$ .
$a = \frac{F_{net}}{m} = \frac{5190}{1000} = \mathbf{5.19 \text{ m s}^{-2}}$ .
[1 for net force, 1 for formula, 1 for answer]

(c) Mass decreases. Since $F_{net}$ is constant (Thrust constant, Weight decreases slightly but Thrust >> Weight, actually Net Force increases as Weight drops), but primarily $a = F/m$ . As $m$ decreases, $a$ increases.
[1 for increases, 1 for explanation]

16. (a) $GPE = mgh = 0.5 \times 9.81 \times 0.2 = \mathbf{0.981 \text{ J}}$ .
[1 for formula, 1 for answer]

(b) Conservation of energy: $GPE_{lost} = KE_{gained}$ .
$0.981 = \frac{1}{2}mv^2 = 0.5 \times 0.5 \times v^2 = 0.25 v^2$ .
$v^2 = \frac{0.981}{0.25} = 3.924$ .
$v = \sqrt{3.924} = \mathbf{1.98 \text{ m s}^{-1}}$ .
[1 for conservation principle, 1 for substitution, 1 for answer]

(c) Energy is dissipated as heat and sound due to air resistance and friction at the pivot.
[1 for heat/air resistance, 1 for sound/pivot friction]

17. (a) Diagram: Weight (down), Normal Reaction (perpendicular to slope). No friction (smooth).
[1 for Weight, 1 for Normal]

(b) Component of weight down slope $= mg \sin 30^\circ$ .
$ma = mg \sin 30^\circ \Rightarrow a = g \sin 30^\circ = 9.81 \times 0.5 = \mathbf{4.91 \text{ m s}^{-2}}$ .
[1 for component, 1 for Newton's 2nd law, 1 for answer]

(c) $F_{net} = ma = 2.0 \times 2.0 = 4.0 \text{ N}$ .
Driving force (weight component) $= mg \sin 30^\circ = 2.0 \times 9.81 \times 0.5 = 9.81 \text{ N}$ .
$F_{net} = \text{Weight Component} - \text{Friction}$ .
$4.0 = 9.81 - f \Rightarrow f = 9.81 - 4.0 = \mathbf{5.81 \text{ N}}$ .
[1 for net force, 1 for weight component, 1 for friction]

18. (a) Constant speed means equilibrium. Tension $T = \text{Weight} = mg = 2000 \times 9.81 = \mathbf{19620 \text{ N}}$ .
[1 for equilibrium, 1 for answer]

(b) Power $P = Fv = 19620 \times 0.5 = \mathbf{9810 \text{ W}}$ (or 9.81 kW).
[1 for formula, 1 for answer]

(c) Tension is 19620 N (same). Since speed is constant, acceleration is zero, so resultant force is zero. $T = mg$ still holds.
[1 for same value, 1 for explanation]

19. (a) Total Momentum $P = m_X u_X + m_Y u_Y = m(2v) + 2m(-v) = 2mv - 2mv = \mathbf{0}$ .
[1 for substitution, 1 for answer]

(b) Conservation of momentum: $P_{initial} = P_{final}$ .
$0 = m_X v_X' + m_Y v_Y'$ .
$0 = m(-v) + 2m(v_Y')$ .
$mv = 2m v_Y' \Rightarrow v_Y' = \mathbf{0.5v}$ (in the original direction of X).
[1 for equation, 1 for substitution, 1 for answer]

(c) $KE_{initial} = \frac{1}{2}m(2v)^2 + \frac{1}{2}(2m)(-v)^2 = 2mv^2 + mv^2 = 3mv^2$ .
$KE_{final} = \frac{1}{2}m(-v)^2 + \frac{1}{2}(2m)(0.5v)^2 = 0.5mv^2 + 0.25mv^2 = 0.75mv^2$ ?
Wait. Let's recheck.
$KE_{final} = \frac{1}{2}m v^2 + \frac{1}{2}(2m)(\frac{v}{2})^2 = \frac{1}{2}mv^2 + m(\frac{v^2}{4}) = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$ .
$KE_{initial} = 3mv^2$ .
They are NOT equal. The question asks to "Show that kinetic energy is conserved". My calculation shows it is NOT.
Let's re-read the prompt. "Sphere X moves with velocity -v".
If $v_X' = -v$ , then $v_Y' = 0.5v$ .
$KE_i = 3mv^2$ . $KE_f = 0.75mv^2$ .
Kinetic energy is NOT conserved. The collision is inelastic.
Correction for Answer Key: The question likely implies an elastic collision scenario or I must state it is NOT conserved. Given the phrasing "Show that...", usually implies it IS. Did I make a mistake?
Momentum: $2mv - 2mv = 0$ . Final: $-mv + 2m(v_Y') = 0 \rightarrow v_Y' = v/2$ .
KE Initial: $0.5(m)(4v^2) + 0.5(2m)(v^2) = 2mv^2 + mv^2 = 3mv^2$ .
KE Final: $0.5(m)(v^2) + 0.5(2m)(v^2/4) = 0.5mv^2 + 0.25mv^2 = 0.75mv^2$ .
Energy is lost.
Answer: Calculate both. State they are unequal. Conclude KE is not conserved. (If the question strictly says "Show it is conserved", the question premise might be flawed for these specific numbers, but in an exam, you show the working and state the fact).
[1 for KE init, 1 for KE final, 1 for conclusion "Not Conserved"]

20. (a) $a = \frac{30 - 0}{10} = \mathbf{3.0 \text{ m s}^{-2}}$ .
[1]

(b) $F = ma = 1000 \times 3.0 = \mathbf{3000 \text{ N}}$ .
[1 for formula, 1 for answer]

(d) Work Done $W = Fs = 3000 \times 150 = \mathbf{450,000 \text{ J}}$ (or 450 kJ).
Alternatively, $W = \Delta KE = \frac{1}{2}(1000)(30^2) = 500 \times 900 = 450,000 \text{ J}$ .
[1 for formula, 1 for answer]