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A Level H1 Physics Practice Paper 3

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TuitionGoWhere Exam Practice (AI) - Physics H1 A-Level

Subject: Physics H1
Level: A-Level
Paper: Practice Paper (Version 3 of 5)
Duration: 1 hour 30 minutes
Total Marks: 60

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Answer all questions.
Write your answers in the spaces provided.
You are advised to spend approximately 5 minutes reading the paper and 5 minutes checking your answers.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use an approved scientific calculator where appropriate.
Assume $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.

Section A: Structured Questions

Answer all questions in this section.

1. A student is investigating the motion of a trolley on a horizontal track. The trolley has a mass of $0.85 \text{ kg}$ .

(a) Define the term linear momentum.
[1]

(b) The trolley moves with a velocity of $1.2 \text{ m s}^{-1}$ to the right. Calculate the magnitude of its momentum.
[2]

Answer: __________________________ $\text{kg m s}^{-1}$

(c) The student states that because the trolley is moving at a constant velocity, the net force acting on it is zero. Explain, using Newton’s First Law of Motion, whether this statement is correct.
[2]

........................................................................................................................................
........................................................................................................................................
........................................................................................................................................

2. A uniform plank AB of length $4.0 \text{ m}$ and weight $120 \text{ N}$ rests horizontally on two supports. Support X is at end A, and support Y is $1.0 \text{ m}$ from end B. A student of weight $500 \text{ N}$ stands on the plank at a distance $x$ from end A.

(a) On the diagram below (representing the plank), draw and label arrows to represent all the forces acting on the plank.
[3]

(Diagram space: Draw a horizontal line representing the plank. Mark positions A, B, X, and Y.)

(b) Calculate the maximum distance $x$ from end A that the student can stand before the plank begins to tip over support Y.
[4]

Answer: __________________________ $\text{m}$

3. Two ice skaters, Skater P (mass $60 \text{ kg}$ ) and Skater Q (mass $80 \text{ kg}$ ), are initially at rest on a frictionless ice surface. They push against each other and move apart. After the push, Skater P moves with a velocity of $2.5 \text{ m s}^{-1}$ to the left.

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

(b) Calculate the velocity of Skater Q immediately after the push.
[3]

Answer: __________________________ $\text{m s}^{-1}$ (direction: _______________)

Answer: __________________________ $\text{J}$

4. A ball is thrown vertically upwards with an initial speed of $15 \text{ m s}^{-1}$ . Air resistance is negligible.

(a) Calculate the maximum height reached by the ball.
[3]

Answer: __________________________ $\text{m}$

(b) Sketch a graph of the vertical velocity $v$ against time $t$ for the motion of the ball from the instant it is thrown until it returns to the starting height. Take upward velocity as positive.
[2]

(Graph space)

(c) Explain the shape of the graph if air resistance were not negligible. Specifically, compare the time taken to reach maximum height with the time taken to fall back down.
[2]

5. A car of mass $1200 \text{ kg}$ travels up a slope inclined at $5.0^\circ$ to the horizontal at a constant speed of $20 \text{ m s}^{-1}$ . The total resistive force (air resistance and friction) acting on the car is $400 \text{ N}$ .

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

Answer: __________________________ $\text{N}$

(b) Determine the power developed by the car’s engine to maintain this constant speed.
[3]

Answer: __________________________ $\text{W}$

Section B: Data and Context Questions

Answer all questions in this section.

6. A student performs an experiment to determine the acceleration due to gravity, $g$ , using a free-fall method. A steel ball is dropped from rest, and the time $t$ taken to fall a distance $h$ is recorded. The results are shown below.

$h$ (m)	$t$ (s)	$t^2$ ( $\text{s}^2$ )
0.50	0.32	0.102
1.00	0.45	0.203
1.50	0.55	0.303
2.00	0.64	0.410
2.50	0.71	0.504

(a) The equation of motion is $h = \frac{1}{2}gt^2$ . Explain why a graph of $h$ against $t^2$ should be a straight line through the origin.
[2]

(b) Plot a graph of $h$ (y-axis) against $t^2$ (x-axis) on the grid provided. Draw the line of best fit.
[4]

(Grid space: X-axis 0 to 0.6 $\text{s}^2$ , Y-axis 0 to 3.0 m)

Answer: Gradient = __________________________ $\text{m s}^{-2}$

(d) Use your gradient to calculate the value of $g$ .
[2]

Answer: $g$ = __________________________ $\text{m s}^{-2}$

7. A toy car of mass $0.20 \text{ kg}$ is released from rest at the top of a curved track. The top of the track is $0.80 \text{ m}$ above the bottom. The car leaves the track horizontally at the bottom and lands on the floor $1.2 \text{ m}$ away horizontally. The vertical drop from the end of the track to the floor is $0.45 \text{ m}$ .

(a) Calculate the theoretical speed of the car at the bottom of the track, assuming conservation of energy and no resistance.
[3]

Answer: __________________________ $\text{m s}^{-1}$

(b) Calculate the actual speed of the car as it leaves the track, based on the projectile motion data provided.
[4]

Answer: __________________________ $\text{m s}^{-1}$

........................................................................................................................................

8. A spring obeys Hooke’s Law. A force-extension graph for the spring is shown below.

(Graph description: A straight line starting from origin (0,0) passing through point (0.10 m, 20 N))

(a) Determine the spring constant $k$ of the spring.
[2]

Answer: __________________________ $\text{N m}^{-1}$

(b) Calculate the elastic potential energy stored in the spring when it is extended by $0.10 \text{ m}$ .
[2]

Answer: __________________________ $\text{J}$

(c) The spring is now used to launch a $0.05 \text{ kg}$ projectile vertically upwards. Assuming all elastic potential energy is converted to gravitational potential energy, calculate the maximum height reached by the projectile from its launch position.
[3]

Answer: __________________________ $\text{m}$

9. Two trolleys, A and B, move along a straight horizontal track. Trolley A (mass $2.0 \text{ kg}$ ) moves to the right with velocity $3.0 \text{ m s}^{-1}$ . Trolley B (mass $1.0 \text{ kg}$ ) is stationary. They collide and stick together.

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

Answer: __________________________ $\text{m s}^{-1}$

(b) Show that this collision is inelastic by comparing the kinetic energy before and after the collision.
[4]

10. A block of mass $5.0 \text{ kg}$ is pulled along a rough horizontal surface by a horizontal force of $30 \text{ N}$ . The block accelerates at $2.0 \text{ m s}^{-2}$ .

(a) Calculate the net force acting on the block.
[2]

Answer: __________________________ $\text{N}$

(b) Calculate the magnitude of the frictional force acting on the block.
[2]

Answer: __________________________ $\text{N}$

Section C: Extended Response

Answer the question in this section.

11. A skydiver jumps from a stationary helicopter.

(a) Describe and explain the variation in the skydiver’s acceleration from the moment he jumps until he reaches terminal velocity. Refer to the forces acting on him.
[4]

(b) The skydiver opens his parachute. Explain, in terms of forces, why his speed decreases rapidly after opening the parachute.
[3]

(c) Eventually, the skydiver reaches a new, lower terminal velocity. State the relationship between the air resistance and the weight of the skydiver at this stage.
[1]

........................................................................................................................................

12. A crane lifts a load of mass $500 \text{ kg}$ vertically upwards from rest. The load accelerates uniformly at $0.50 \text{ m s}^{-2}$ for $4.0 \text{ s}$ .

(a) Calculate the tension in the cable during this acceleration phase.
[4]

Answer: __________________________ $\text{N}$

(b) Calculate the work done by the tension in the cable during these $4.0 \text{ s}$ .
[4]

Answer: __________________________ $\text{J}$

Answer: __________________________ $\text{W}$

13. A ball of mass $0.15 \text{ kg}$ strikes a vertical wall horizontally with a speed of $12 \text{ m s}^{-1}$ and rebounds horizontally with a speed of $10 \text{ m s}^{-1}$ . The contact time with the wall is $0.02 \text{ s}$ .

(a) Calculate the change in momentum of the ball. Indicate the direction.
[3]

Answer: __________________________ $\text{kg m s}^{-1}$ (Direction: _______________)

(b) Calculate the average force exerted by the wall on the ball.
[2]

Answer: __________________________ $\text{N}$

14. A uniform ladder of weight $W$ and length $L$ rests against a smooth vertical wall and on a rough horizontal ground. The ladder makes an angle of $60^\circ$ with the ground.

(a) Explain why the force exerted by the wall on the ladder is horizontal.
[1]

........................................................................................................................................

(b) Draw a free-body diagram showing all forces acting on the ladder.
[3]

(Space for diagram)

(c) By taking moments about the base of the ladder, derive an expression for the normal reaction force from the wall ( $R_w$ ) in terms of $W$ .
[4]

Answer: $R_w$ = __________________________

15. A car travels around a circular bend of radius $50 \text{ m}$ on a flat horizontal road. The coefficient of static friction between the tires and the road is $0.80$ .

(a) Identify the force that provides the centripetal acceleration.
[1]

........................................................................................................................................

(b) Calculate the maximum speed at which the car can travel around the bend without skidding.
[4]

Answer: __________________________ $\text{m s}^{-1}$

16. A projectile is fired from ground level with an initial velocity of $20 \text{ m s}^{-1}$ at an angle of $30^\circ$ to the horizontal.

(a) Calculate the horizontal and vertical components of the initial velocity.
[2]

Answer: $v_x$ = __________ $\text{m s}^{-1}$ , $v_y$ = __________ $\text{m s}^{-1}$

(b) Calculate the time of flight.
[3]

Answer: __________________________ $\text{s}$

Answer: __________________________ $\text{m}$

17. A block of mass $2.0 \text{ kg}$ slides down a smooth inclined plane at an angle of $30^\circ$ to the horizontal.

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

Answer: __________________________ $\text{m s}^{-2}$

(b) If the plane is rough and the block slides down at a constant velocity, calculate the magnitude of the frictional force.
[2]

Answer: __________________________ $\text{N}$

18. Two forces, $F_1 = 10 \text{ N}$ acting horizontally to the right, and $F_2 = 10 \text{ N}$ acting vertically upwards, act on a particle.

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

Answer: __________________________ $\text{N}$

(b) Determine the direction of the resultant force relative to the horizontal.
[2]

Answer: __________________________ degrees

19. A rocket of mass $1000 \text{ kg}$ is launched vertically. The engine produces a thrust of $15,000 \text{ N}$ .

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

Answer: __________________________ $\text{m s}^{-2}$

(b) As the rocket rises, its mass decreases. Explain the effect of this on its acceleration, assuming thrust remains constant.
[2]

20. A pendulum bob of mass $0.5 \text{ kg}$ is pulled to one side so that it is $0.20 \text{ m}$ higher than its lowest point. It is released from rest.

(a) Calculate the speed of the bob at the lowest point of its swing.
[3]

Answer: __________________________ $\text{m s}^{-1}$

(b) State one assumption made in this calculation.
[1]

........................................................................................................................................

End of Paper

Answers

TuitionGoWhere Exam Practice (AI) - Physics H1 A-Level

Answer Key and Marking Scheme
Paper: Practice Paper (Version 3 of 5)
Topic: Mechanics

Section A: Structured Questions

1.
(a) Linear momentum is the product of mass and velocity. [B1]
(b) $p = mv = 0.85 \times 1.2$ [M1]
$p = 1.02 \text{ kg m s}^{-1}$ [A1]
(c) Newton’s First Law states that an object remains at rest or in uniform motion unless acted upon by a net external force. [B1]
Since the velocity is constant (zero acceleration), the net force must be zero. [B1]

2.
(a) Forces:

Weight of plank ( $120 \text{ N}$ ) acting downwards at the center ( $2.0 \text{ m}$ from A). [B1]
Weight of student ( $500 \text{ N}$ ) acting downwards at distance $x$ . [B1]
Reaction force at X ( $R_X$ ) acting upwards at A. [B1]
Reaction force at Y ( $R_Y$ ) acting upwards at $3.0 \text{ m}$ from A. [B1]
(Note: Accept arrows labeled clearly. Max 3 marks if only 3 forces shown but correct.)
(b) At the point of tipping, the plank loses contact with support X, so $R_X = 0$ . [B1]
Take moments about support Y (position $3.0 \text{ m}$ from A).
Clockwise moment = Anti-clockwise moment.
Weight of plank acts at $2.0 \text{ m}$ from A. Distance from Y = $3.0 - 2.0 = 1.0 \text{ m}$ .
Moment of plank = $120 \times 1.0 = 120 \text{ N m}$ . [M1]
Student is at distance $x$ from A. Distance from Y = $3.0 - x$ .
Moment of student = $500 \times (3.0 - x)$ . [M1]
Equilibrium: $500(3.0 - x) = 120$
$1500 - 500x = 120$
$500x = 1380$
$x = 2.76 \text{ m}$ [A1]

3.
(a) In a closed/isolated system (no external forces), the total linear momentum before interaction equals the total linear momentum after interaction. [B1] [B1]
(b) Total initial momentum = $0$ .
Let right be positive.
$p_{\text{initial}} = p_{\text{final}}$
$0 = m_P v_P + m_Q v_Q$
$0 = 60(-2.5) + 80(v_Q)$ [M1]
$150 = 80 v_Q$
$v_Q = 1.875 \text{ m s}^{-1}$ [A1]
Direction: To the right. [B1]
(c) $KE_P = \frac{1}{2}(60)(2.5)^2 = 187.5 \text{ J}$ [M1]
$KE_Q = \frac{1}{2}(80)(1.875)^2 = 140.625 \text{ J}$ [M1]
Total $KE = 187.5 + 140.625 = 328.125 \text{ J}$ (approx $328 \text{ J}$ ) [A1]

4.
(a) $v^2 = u^2 + 2as$
At max height, $v = 0$ . $u = 15$ , $a = -9.81$ .
$0 = 15^2 + 2(-9.81)s$ [M1]
$19.62 s = 225$
$s = 11.47 \text{ m}$ (approx $11.5 \text{ m}$ ) [A1]
(b) Graph: Straight line with negative gradient. [B1]
Starts at positive $v$ ( $+15$ ), crosses t-axis (v=0), ends at negative $v$ ( $-15$ ). [B1]
(c) Air resistance opposes motion.
On the way up, air resistance acts downwards (with gravity), so deceleration $> g$ . Time to top is shorter. [B1]
On the way down, air resistance acts upwards (against gravity), so acceleration $< g$ . Time to fall is longer. [B1]

5.
(a) Component of weight down slope = $mg \sin \theta$
$= 1200 \times 9.81 \times \sin(5.0^\circ)$ [M1]
$= 1025 \text{ N}$ (approx $1030 \text{ N}$ ) [A1]
(b) Since speed is constant, driving force $F_D$ balances resistive forces.
$F_D = \text{Weight component} + \text{Resistive force}$
$F_D = 1025 + 400 = 1425 \text{ N}$ [M1]
Power $P = F_D v$
$P = 1425 \times 20$ [M1]
$P = 28,500 \text{ W}$ (or $28.5 \text{ kW}$ ) [A1]

Section B: Data and Context Questions

6.
(a) $h = \frac{1}{2}gt^2$ is in the form $y = mx$ where $y=h$ , $x=t^2$ , and $m = \frac{1}{2}g$ . [B1]
Since there is no constant term ( $c=0$ ), the line passes through the origin. [B1]
(b) Plot points correctly. [B1]
Line of best fit is a straight line through the origin. [B1]
Axes labeled with units. [B1]
Scale appropriate. [B1]
(c) Gradient = $\frac{\Delta h}{\Delta t^2}$ . Using points from line (e.g., $(0.5, 2.5)$ ):
Gradient $\approx \frac{2.5}{0.5} = 5.0 \text{ m s}^{-2}$ (Accept range $4.8 - 5.2$ ). [M1] [A1]
(d) Gradient $= \frac{1}{2}g \Rightarrow g = 2 \times \text{Gradient}$ .
$g = 2 \times 5.0 = 10.0 \text{ m s}^{-2}$ (Accept range $9.6 - 10.4$ ). [M1] [A1]

7.
(a) Loss in GPE = Gain in KE
$mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.80}$ [M1]
$v = \sqrt{15.696} = 3.96 \text{ m s}^{-1}$ [A1]
(b) Vertical motion: $s = ut + \frac{1}{2}at^2$ . $u_y = 0$ , $s = 0.45$ , $a = 9.81$ .
$0.45 = 0 + \frac{1}{2}(9.81)t^2$
$t^2 = \frac{0.90}{9.81} \Rightarrow t = 0.303 \text{ s}$ [M1]
Horizontal motion: $s_x = v_x t$ .
$1.2 = v_x (0.303)$
$v_x = \frac{1.2}{0.303} = 3.96 \text{ m s}^{-1}$ [M1] [A1]
(Note: If student uses $g=10$ , answers will vary slightly but method marks apply.)
(c) Energy lost due to friction/resistance on the track. [B1]

8.
(a) $k = \frac{F}{x} = \frac{20}{0.10} = 200 \text{ N m}^{-1}$ . [M1] [A1]
(b) $EPE = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.10)^2$ [M1]
$EPE = 1.0 \text{ J}$ . [A1]
(c) $EPE = GPE$
$1.0 = mgh$
$1.0 = 0.05 \times 9.81 \times h$ [M1]
$h = \frac{1.0}{0.4905} = 2.04 \text{ m}$ . [A1]

9.
(a) Conservation of momentum:
$m_A u_A + m_B u_B = (m_A + m_B)v$
$2.0(3.0) + 1.0(0) = (2.0 + 1.0)v$ [M1]
$6.0 = 3.0v$
$v = 2.0 \text{ m s}^{-1}$ . [A1] Direction: Right. [B1]
(b) $KE_{\text{initial}} = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \text{ J}$ . [M1]
$KE_{\text{final}} = \frac{1}{2}(3.0)(2.0)^2 = 6.0 \text{ J}$ . [M1]
$KE_{\text{initial}} \neq KE_{\text{final}}$ ( $9.0 \neq 6.0$ ), so kinetic energy is not conserved. [B1]
Therefore, the collision is inelastic. [B1]

10.
(a) $F_{\text{net}} = ma = 5.0 \times 2.0 = 10 \text{ N}$ . [M1] [A1]
(b) $F_{\text{net}} = F_{\text{pull}} - F_{\text{friction}}$
$10 = 30 - F_{\text{friction}}$ [M1]
$F_{\text{friction}} = 20 \text{ N}$ . [A1]
(c) The block will decelerate (slow down) due to the frictional force acting opposite to motion. [B1]
It will eventually come to rest. [B1]

Section C: Extended Response

11.
(a) Initially, air resistance is zero, so net force is weight ( $W$ ). Acceleration is $g$ . [B1]
As speed increases, air resistance ( $R$ ) increases. Net force $= W - R$ . [B1]
Since net force decreases, acceleration decreases. [B1]
When $R = W$ , net force is zero, acceleration is zero, and terminal velocity is reached. [B1]
(b) Opening parachute greatly increases surface area, causing a large increase in air resistance. [B1]
Air resistance becomes much greater than weight ( $R > W$ ). [B1]
This creates a large net upward force, causing rapid deceleration (upward acceleration). [B1]
(c) Air resistance equals weight ( $R = W$ ). [B1]

12.
(a) Forces: Tension $T$ (up), Weight $mg$ (down).
$F_{\text{net}} = T - mg = ma$ [M1]
$T = m(g + a)$
$T = 500(9.81 + 0.50)$ [M1]
$T = 500(10.31) = 5155 \text{ N}$ . [A1] (Accept $5150$ or $5160$ )
(b) Distance moved $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(0.50)(4.0)^2 = 4.0 \text{ m}$ . [M1]
Work Done $= T \times s$ [M1]
$W = 5155 \times 4.0 = 20,620 \text{ J}$ . [A1] (Accept $20,600$ )
(c) Average Power $= \frac{\text{Work}}{\text{Time}}$ [M1]
$P = \frac{20620}{4.0} = 5155 \text{ W}$ . [A1]

13.
(a) Take direction towards wall as positive.
$u = +12 \text{ m s}^{-1}$ , $v = -10 \text{ m s}^{-1}$ .
$\Delta p = m(v - u) = 0.15(-10 - 12)$ [M1]
$\Delta p = 0.15(-22) = -3.3 \text{ kg m s}^{-1}$ . [A1]
Magnitude: $3.3 \text{ kg m s}^{-1}$ . Direction: Away from wall. [B1]
(b) $F_{\text{avg}} = \frac{\Delta p}{\Delta t}$ [M1]
$F = \frac{3.3}{0.02} = 165 \text{ N}$ . [A1]
(c) $KE_{\text{initial}} = \frac{1}{2}(0.15)(12)^2 = 10.8 \text{ J}$ .
$KE_{\text{final}} = \frac{1}{2}(0.15)(10)^2 = 7.5 \text{ J}$ .
KE is lost to sound, heat, and deformation of the ball/wall. [B1] [B1]

14.
(a) The wall is smooth, so there is no friction. The reaction force must be perpendicular (normal) to the surface. [B1]
(b) Diagram:

Weight $W$ down from center. [B1]
Normal reaction from ground $R_g$ up from base. [B1]
Friction $F$ at base towards wall. [B1]
Normal reaction from wall $R_w$ horizontal away from wall at top.
(c) Take moments about base.
Clockwise moment: $R_w \times (L \sin 60^\circ)$ . [M1]
Anti-clockwise moment: $W \times (\frac{L}{2} \cos 60^\circ)$ . [M1]
Equilibrium: $R_w L \sin 60^\circ = W \frac{L}{2} \cos 60^\circ$ [M1]
$R_w = \frac{W \cos 60^\circ}{2 \sin 60^\circ} = \frac{W}{2 \tan 60^\circ}$
$R_w = \frac{W}{2\sqrt{3}}$ or $0.289 W$ . [A1]

15.
(a) Frictional force between tires and road. [B1]
(b) Centripetal force $F_c = \frac{mv^2}{r}$ .
Max friction $F_f = \mu R = \mu mg$ .
$\mu mg = \frac{mv^2}{r}$ [M1]
$v^2 = \mu gr$
$v = \sqrt{0.80 \times 9.81 \times 50}$ [M1]
$v = \sqrt{392.4} = 19.8 \text{ m s}^{-1}$ . [A1]
(c) Banking allows the normal reaction force to have a horizontal component. [B1]
This component helps provide the centripetal force, reducing reliance on friction. [B1]

16.
(a) $v_x = 20 \cos 30^\circ = 17.32 \text{ m s}^{-1}$ . [B1]
$v_y = 20 \sin 30^\circ = 10.0 \text{ m s}^{-1}$ . [B1]
(b) Time to max height: $v_y = u_y + at \Rightarrow 0 = 10 - 9.81 t \Rightarrow t = 1.02 \text{ s}$ .
Total time $= 2 \times 1.02 = 2.04 \text{ s}$ . [M1] [A1] (Allow 2.0 s if $g=10$ )
(c) Range $= v_x \times t = 17.32 \times 2.04 = 35.3 \text{ m}$ . [M1] [A1]

17.
(a) Force down slope $= mg \sin 30^\circ$ .
$ma = mg \sin 30^\circ \Rightarrow a = g \sin 30^\circ$ . [M1]
$a = 9.81 \times 0.5 = 4.905 \text{ m s}^{-2}$ . [A1]
(b) Constant velocity means zero acceleration, so net force is zero.
Friction $= \text{Component of weight down slope}$ .
$F_f = mg \sin 30^\circ = 2.0 \times 9.81 \times 0.5 = 9.81 \text{ N}$ . [M1] [A1]

18.
(a) Resultant $R = \sqrt{10^2 + 10^2} = \sqrt{200} = 14.1 \text{ N}$ . [M1] [A1]
(b) $\tan \theta = \frac{10}{10} = 1$ .
$\theta = 45^\circ$ . [M1] [A1]

19.
(a) $F_{\text{net}} = \text{Thrust} - \text{Weight} = 15000 - (1000 \times 9.81) = 5190 \text{ N}$ . [M1]
$a = \frac{F_{\text{net}}}{m} = \frac{5190}{1000} = 5.19 \text{ m s}^{-2}$ . [A1]
(b) As mass $m$ decreases, and Thrust is constant, the net force increases (since weight decreases). [B1]
Since $a = F/m$ , and $F$ increases while $m$ decreases, acceleration increases. [B1]

20.
(a) $mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$ . [M1]
$v = \sqrt{2 \times 9.81 \times 0.20} = \sqrt{3.924} = 1.98 \text{ m s}^{-1}$ . [A1]
(b) Air resistance is negligible / No energy loss to heat/sound. [B1]