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A Level H1 Physics Practice Paper 4

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TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics H1 (8867)
Level: A-Level
Paper: Practice Paper – Mechanics (Version 4 of 5)
Duration: 1 hour 30 minutes
Total Marks: 60

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.
You may lose marks if you do not show your working or if you do not use appropriate units.
Take the acceleration of free fall $g = 9.81 \text{ m s}^{-2}$ .

Section A: Structured Questions

Answer all questions in this section.

1. 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=10s, remains constant until t=30s, then decreases linearly to 0 at t=40s.)

(a) Calculate the acceleration of the car during the first 10 seconds.
[2]

(b) Calculate the total distance travelled by the car during the 40 seconds.
[2]

2. A steel ball of mass $0.50 \text{ kg}$ is dropped from rest from a height of $20 \text{ m}$ . Air resistance is negligible.

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

(b) The ball hits the ground and rebounds vertically with a speed of $15 \text{ m s}^{-1}$ . Calculate the change in momentum of the ball during the impact. State the direction of this change.
[3]

(c) The contact time with the ground is $0.10 \text{ s}$ . Calculate the average resultant force acting on the ball during the impact.
[2]

3. 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 $C$ , $3.0 \text{ m}$ vertically above $A$ .

(a) Draw a free-body diagram showing all the forces acting on the beam $AB$ . Label the forces clearly.
[2]

(b) Calculate the tension in the cable.
[3]

4. A block of mass $2.0 \text{ kg}$ rests on a rough horizontal surface. A horizontal force $F$ is applied to the block. The coefficient of static friction between the block and the surface is $0.40$ , and the coefficient of dynamic friction is $0.30$ .

(a) Calculate the minimum force $F$ required to just start the block moving.
[2]

(b) Once the block is moving, the force $F$ is maintained at $10 \text{ N}$ . Calculate the acceleration of the block.
[3]

5. Define the principle of conservation of linear momentum.
[2]

Section B: Data-Based and Application Questions

Answer all questions in this section.

6. A student investigates the motion of a trolley rolling down a slope. The trolley starts from rest. The student measures the distance $s$ travelled and the time $t$ taken. The data is recorded below.

$t / \text{s}$	$s / \text{m}$
0.50	0.12
1.00	0.48
1.50	1.08
2.00	1.92
2.50	3.00

(a) Plot a graph of $s$ against $t^2$ on the grid provided below. Draw the line of best fit.
[3]

(Grid space for plotting: X-axis $t^2 / \text{s}^2$ from 0 to 7.0; Y-axis $s / \text{m}$ from 0 to 3.5)

(b) Determine the gradient of your graph.
[1]

(d) Suggest one reason why the acceleration calculated might be less than $g \sin \theta$ (where $\theta$ is the angle of the slope).
[1]

7. A crane lifts a load of mass $500 \text{ kg}$ vertically upwards at a constant speed of $2.0 \text{ m s}^{-1}$ .

(a) Calculate the power developed by the crane motor.
[2]

(b) The crane motor has an efficiency of $60\%$ . Calculate the input power required by the motor.
[2]

8. Two ice skaters, Skater A (mass $60 \text{ kg}$ ) and Skater B (mass $80 \text{ kg}$ ), are initially at rest on a frictionless ice rink. They push against each other and move apart. Skater A moves with a velocity of $3.0 \text{ m s}^{-1}$ to the left.

(a) Calculate the velocity of Skater B.
[3]

(b) Calculate the total kinetic energy of the system after the push.
[2]

9. A projectile is launched from ground level with an initial velocity of $40 \text{ m s}^{-1}$ at an angle of $30^\circ$ 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]

10. A spring obeys Hooke's Law. When a force of $10 \text{ N}$ is applied, the extension is $5.0 \text{ cm}$ .

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

(b) Calculate the elastic potential energy stored in the spring when the extension is $5.0 \text{ cm}$ .
[2]

(c) Sketch the force-extension graph for this spring up to an extension of $10 \text{ cm}$ . Label the axes with units.
[2]

Section C: Extended Response and Synthesis

Answer all questions in this section.

11. A car of mass $1200 \text{ kg}$ travels around a circular bend of radius $50 \text{ m}$ on a flat horizontal road. The maximum frictional force between the tires and the road is $8000 \text{ N}$ .

(a) Explain what provides the centripetal force in this situation.
[1]

(b) Calculate the maximum speed at which the car can travel around the bend without skidding.
[3]

(c) If the road is banked at an angle $\theta$ , explain how this allows the car to travel at higher speeds safely without relying solely on friction.
[3]

12. A rocket of mass $5000 \text{ kg}$ is launched vertically from the Earth's surface. The engines produce a constant upward thrust of $80,000 \text{ N}$ . Assume the mass of the rocket remains constant for the first 10 seconds and air resistance is negligible.

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

(b) Calculate the velocity of the rocket after 10 seconds.
[2]

(c) In reality, the mass of the rocket decreases as fuel is burned. Explain qualitatively how this affects the acceleration of the rocket, assuming the thrust remains constant.
[2]

13. A pendulum bob of mass $0.20 \text{ kg}$ is pulled to one side so that it is raised vertically by $0.10 \text{ m}$ from its lowest point. It is released from rest.

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

(b) Calculate the maximum speed of the bob as it passes through the lowest point.
[2]

(c) In a real experiment, the bob does not return to the original height. Explain the energy transformations that occur during one complete swing.
[3]

14. Two trolleys, $X$ and $Y$ , move on a straight horizontal track. Trolley $X$ has mass $2.0 \text{ kg}$ and moves at $4.0 \text{ m s}^{-1}$ . Trolley $Y$ has mass $3.0 \text{ kg}$ and is stationary. They collide and stick together.

(a) Show that the common velocity after the collision is $1.6 \text{ m s}^{-1}$ .
[2]

(b) Calculate the loss in kinetic energy during the collision.
[3]

15. A block of mass $5.0 \text{ kg}$ is pushed up a rough inclined plane at a constant speed by a force $P$ acting parallel to the plane. The plane is inclined at $30^\circ$ to the horizontal. The frictional force acting on the block is $10 \text{ N}$ .

(a) Draw a free-body diagram for the block, showing the weight, normal reaction, friction, and force $P$ .
[2]

(b) Calculate the magnitude of the force $P$ .
[3]

16. A satellite orbits the Earth in a circular orbit.

(a) State the force that provides the centripetal acceleration for the satellite.
[1]

(b) Explain why the satellite is considered to be in a state of "free fall" even though it does not get closer to the Earth.
[3]

17. A ball of mass $0.10 \text{ kg}$ is thrown vertically upwards with an initial speed of $20 \text{ m s}^{-1}$ . Air resistance is significant.

(a) On the axes below, sketch the velocity-time graph for the upward journey until the ball reaches maximum height. Compare it with the graph for no air resistance.
[3]

(Grid space: Y-axis Velocity, X-axis Time. Show two curves)

(b) Explain why the time taken to reach maximum height is less than it would be in a vacuum.
[2]

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

(a) Explain why the wall exerts only a horizontal force on the ladder.
[1]

(b) By taking moments about the base of the ladder, derive an expression for the horizontal force exerted by the wall in terms of $W$ .
[3]

19. A car engine delivers a power of $50 \text{ kW}$ to the wheels. The car has a mass of $1000 \text{ kg}$ .

(a) Calculate the driving force exerted by the engine when the car is travelling at a constant speed of $25 \text{ m s}^{-1}$ .
[2]

(b) If the total resistive force at this speed is equal to the driving force, calculate the resistive force.
[1]

(c) The driver accelerates. Explain why the acceleration decreases as the speed increases, assuming constant engine power.
[3]

20. Impulse and Momentum.

(a) Define impulse.
[1]

(b) A golf club strikes a stationary ball of mass $0.045 \text{ kg}$ . The contact time is $0.50 \text{ ms}$ . The ball leaves the club with a speed of $50 \text{ m s}^{-1}$ . Calculate the average force exerted by the club on the ball.
[3]

(c) Explain, using the concept of impulse, why modern cars are designed with "crumple zones" that deform during a collision.
[3]

End of Paper

Answers

TuitionGoWhere Practice Paper - Physics H1 A-Level

Answer Key and Marking Scheme (Version 4)

Subject: Physics H1 (8867)
Topic: Mechanics

Section A: Structured Questions

1. Kinematics Graphs (a) Acceleration $a = \frac{\Delta v}{\Delta t}$
$a = \frac{20 - 0}{10 - 0} = 2.0 \text{ m s}^{-2}$
[M1] for substitution, [A1] for answer.

(b) Distance = Area under graph.
Area = Area of triangle (0-10s) + Area of rectangle (10-30s) + Area of triangle (30-40s)
$= (\frac{1}{2} \times 10 \times 20) + (20 \times 20) + (\frac{1}{2} \times 10 \times 20)$
$= 100 + 400 + 100 = 600 \text{ m}$
[M1] for correct area method, [A1] for answer.

(c) Velocity is constant, so 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).
[B1] for constant velocity/zero acceleration, [B1] for balanced forces.

2. Free Fall and Momentum (a) Using conservation of energy or equations of motion:
$v^2 = u^2 + 2as$
$v^2 = 0 + 2(9.81)(20)$
$v = \sqrt{392.4} = 19.8 \text{ m s}^{-1}$
[M1] for correct equation/substitution, [A1] for answer.

(b) Take upward as positive.
Initial velocity $u = -19.8 \text{ m s}^{-1}$ (downward)
Final velocity $v = +15 \text{ m s}^{-1}$ (upward)
Change in momentum $\Delta p = m(v - u)$
$\Delta p = 0.50 (15 - (-19.8)) = 0.50 (34.8) = 17.4 \text{ N s}$ (or kg m s⁻¹)
Direction: Upwards.
[M1] for correct vector subtraction, [A1] for magnitude, [B1] for direction.

(c) Average Force $F = \frac{\Delta p}{\Delta t}$
$F = \frac{17.4}{0.10} = 174 \text{ N}$
[M1] for formula, [A1] for answer.

3. Equilibrium of Beam (a) Diagram must show:

Weight ( $200 \text{ N}$ ) acting downwards at the center ( $2.0 \text{ m}$ from A).
Tension ( $T$ ) acting at B towards C.
Reaction force at hinge A (can be shown as vertical/horizontal components or resultant).
[B1] for correct weight position/direction, [B1] for correct tension direction.

(b) Take moments about A.
Clockwise moment = Anticlockwise moment.
Weight moment: $200 \times 2.0 = 400 \text{ N m}$ .
Tension acts at angle. Angle $\theta$ with horizontal: $\tan \theta = \frac{3}{4} \Rightarrow \theta = 36.9^\circ$ .
Vertical component of Tension $T_y = T \sin \theta$ .
Moment of Tension: $(T \sin 36.9^\circ) \times 4.0$ .
$400 = T (0.6) (4.0)$
$400 = 2.4 T$
$T = \frac{400}{2.4} = 167 \text{ N}$
[M1] for moment equation, [M1] for resolving tension or using perpendicular distance, [A1] for answer.

(c) If C is moved higher, the angle $\theta$ increases.
$\sin \theta$ increases.
Since $T \sin \theta = \text{constant}$ (to balance weight moment), $T$ decreases.
[B1] for angle increases, [B1] for tension decreases.

4. Friction (a) Normal reaction $R = mg = 2.0 \times 9.81 = 19.62 \text{ N}$ .
Max static friction $F_{max} = \mu_s R = 0.40 \times 19.62 = 7.85 \text{ N}$ .
Minimum force $F = 7.85 \text{ N}$ (or $7.8 \text{ N}$ ).
[M1] for calculation of friction, [A1] for answer.

(b) Dynamic friction $F_d = \mu_d R = 0.30 \times 19.62 = 5.89 \text{ N}$ .
Resultant force $F_{res} = F_{applied} - F_d = 10 - 5.89 = 4.11 \text{ N}$ .
$F_{res} = ma \Rightarrow 4.11 = 2.0 a$ .
$a = 2.05 \text{ m s}^{-2}$ (or $2.1 \text{ m s}^{-2}$ ).
[M1] for resultant force, [M1] for Newton's 2nd law, [A1] for answer.

5. Conservation of Momentum "In a closed system (or isolated system) [B1], the total linear momentum remains constant (or is conserved) provided no external forces act [B1]."

Section B: Data-Based and Application Questions

6. Trolley Experiment (a) Graph:

Axes labeled $t^2 / \text{s}^2$ and $s / \text{m}$ .
Points plotted correctly: (0.25, 0.12), (1.0, 0.48), (2.25, 1.08), (4.0, 1.92), (6.25, 3.00).
Straight line of best fit through origin.
[B1] for axes, [B1] for points, [B1] for line.

(b) Gradient calculation:
Using points $(0,0)$ and $(6.25, 3.00)$ .
Gradient $= \frac{3.00 - 0}{6.25 - 0} = 0.48$ .
[B1] for value in range $0.47 - 0.49$ .

(c) Equation of motion: $s = ut + \frac{1}{2}at^2$ . Since $u=0$ , $s = \frac{1}{2}at^2$ .
Gradient $= \frac{1}{2}a$ .
$a = 2 \times \text{Gradient} = 2 \times 0.48 = 0.96 \text{ m s}^{-2}$ .
[M1] for relation, [A1] for answer.

(d) Friction between trolley and track / Air resistance.
[B1] for valid reason.

7. Crane Power (a) Force required = Weight $= mg = 500 \times 9.81 = 4905 \text{ N}$ .
Power $P = Fv = 4905 \times 2.0 = 9810 \text{ W}$ (or $9.8 \text{ kW}$ ).
[M1] for force, [A1] for power.

(b) Efficiency $= \frac{P_{out}}{P_{in}} \times 100\%$ .
$0.60 = \frac{9810}{P_{in}}$ .
$P_{in} = \frac{9810}{0.60} = 16350 \text{ W}$ (or $16.4 \text{ kW}$ ).
[M1] for substitution, [A1] for answer.

(c) Speed is constant, so kinetic energy ( $\frac{1}{2}mv^2$ ) is constant.
Work done by the crane increases the gravitational potential energy of the load, not its kinetic energy.
[B1] for KE constant due to constant speed, [B1] for work converting to GPE.

8. Ice Skaters (a) Conservation of momentum: $P_{initial} = P_{final}$ .
$0 = m_A v_A + m_B v_B$ .
$0 = 60(-3.0) + 80(v_B)$ . (Taking left as negative)
$180 = 80 v_B$ .
$v_B = \frac{180}{80} = 2.25 \text{ m s}^{-1}$ to the right.
[M1] for equation, [M1] for substitution, [A1] for answer + direction.

(b) $KE_{total} = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2$ .
$KE = \frac{1}{2}(60)(3.0)^2 + \frac{1}{2}(80)(2.25)^2$ .
$KE = 270 + 202.5 = 472.5 \text{ J}$ .
[M1] for sum of KEs, [A1] for answer.

9. Projectile Motion (a) $v_x = v \cos \theta = 40 \cos 30^\circ = 34.6 \text{ m s}^{-1}$ .
[B1] for answer.

(b) Vertical component $u_y = 40 \sin 30^\circ = 20 \text{ m s}^{-1}$ .
At max height, $v_y = 0$ .
$v_y^2 = u_y^2 + 2as$ .
$0 = 20^2 + 2(-9.81)h$ .
$h = \frac{400}{19.62} = 20.4 \text{ m}$ .
[M1] for vertical component, [M1] for equation, [A1] for answer.

(c) Time to max height: $v = u + at \Rightarrow 0 = 20 - 9.81t \Rightarrow t = 2.04 \text{ s}$ .
Total time of flight $= 2 \times t = 4.08 \text{ s}$ .
[M1] for time to peak, [A1] for total time.

10. Springs (a) $F = kx \Rightarrow 10 = k(0.05)$ .
$k = \frac{10}{0.05} = 200 \text{ N m}^{-1}$ .
[M1] for substitution, [A1] for answer.

(b) $EPE = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.05)^2$ .
$EPE = 100 \times 0.0025 = 0.25 \text{ J}$ .
[M1] for formula, [A1] for answer.

(c) Graph: Straight line through origin.
X-axis: Extension (m or cm), Y-axis: Force (N).
Slope is constant.
[B1] for straight line through origin, [B1] for labeled axes.

Section C: Extended Response and Synthesis

11. Circular Motion (a) Friction between the tires and the road.
[B1] for friction.

(b) Centripetal force $F_c = \frac{mv^2}{r}$ .
Max friction provides max centripetal force: $8000 = \frac{1200 v^2}{50}$ .
$v^2 = \frac{8000 \times 50}{1200} = 333.33$ .
$v = \sqrt{333.33} = 18.3 \text{ m s}^{-1}$ .
[M1] for equation, [M1] for substitution, [A1] for answer.

(c) Banking allows the normal reaction force to have a horizontal component.
This horizontal component contributes to the centripetal force.
Thus, less friction is required, or higher speeds can be sustained without relying entirely on friction.
[B1] for normal force component, [B1] for contributes to centripetal force, [B1] for reduced reliance on friction/higher speed.

12. Rocket Dynamics (a) Weight $W = mg = 5000 \times 9.81 = 49050 \text{ N}$ .
Resultant Force $F_{res} = \text{Thrust} - W = 80000 - 49050 = 30950 \text{ N}$ .
$F_{res} = ma \Rightarrow 30950 = 5000 a$ .
$a = 6.19 \text{ m s}^{-2}$ .
[M1] for weight, [M1] for resultant force, [A1] for acceleration.

(b) $v = u + at$ .
$v = 0 + (6.19)(10) = 61.9 \text{ m s}^{-1}$ .
[M1] for equation, [A1] for answer.

(c) As mass $m$ decreases, and Thrust is constant (and Weight decreases), the resultant force increases (or stays high while mass drops).
Since $a = \frac{F}{m}$ , if $m$ decreases, $a$ increases.
[B1] for mass decreases, [B1] for acceleration increases.

13. Pendulum Energy (a) $GPE = mgh = 0.20 \times 9.81 \times 0.10 = 0.196 \text{ J}$ .
[M1] for formula, [A1] for answer.

(b) Conservation of energy: $GPE_{lost} = KE_{gained}$ .
$0.196 = \frac{1}{2}mv^2$ .
$0.196 = 0.5(0.20)v^2 = 0.1v^2$ .
$v^2 = 1.96 \Rightarrow v = 1.4 \text{ m s}^{-1}$ .
[M1] for equivalence, [A1] for answer.

(c) GPE converts to KE as it falls.
KE converts back to GPE as it rises.
Some energy is lost to air resistance/heat at the pivot, so total mechanical energy decreases, and it doesn't reach original height.
[B1] for GPE to KE, [B1] for KE to GPE, [B1] for loss to heat/friction.

14. Inelastic Collision (a) Momentum before: $2.0(4.0) + 3.0(0) = 8.0 \text{ N s}$ .
Momentum after: $(2.0 + 3.0)v = 5.0v$ .
$5.0v = 8.0 \Rightarrow v = 1.6 \text{ m s}^{-1}$ .
[M1] for conservation equation, [A1] for correct result.

(b) $KE_{initial} = \frac{1}{2}(2.0)(4.0)^2 = 16 \text{ J}$ .
$KE_{final} = \frac{1}{2}(5.0)(1.6)^2 = 2.5(2.56) = 6.4 \text{ J}$ .
Loss $= 16 - 6.4 = 9.6 \text{ J}$ .
[M1] for initial KE, [M1] for final KE, [A1] for difference.

(c) Inelastic collision.
Because kinetic energy is not conserved (lost to heat/sound/deformation).
[B1] for inelastic, [B1] for KE not conserved.

15. Inclined Plane (a) Diagram:

Weight vertically down.
Normal reaction perpendicular to slope.
Friction parallel to slope, downwards (opposing motion).
Force P parallel to slope, upwards.
[B1] for correct directions of P and Friction, [B1] for Weight and Normal.

(b) Resolve forces parallel to slope.
Since speed is constant, forces are balanced.
$P = F_{friction} + W \sin \theta$ .
$W \sin 30^\circ = (5.0 \times 9.81) \times 0.5 = 24.525 \text{ N}$ .
$P = 10 + 24.525 = 34.5 \text{ N}$ .
[M1] for resolution of weight, [M1] for balance equation, [A1] for answer.

16. Satellites (a) Gravitational force (between Earth and satellite).
[B1] for gravity.

(b) The satellite has a tangential velocity.
Gravity acts perpendicular to this velocity, changing the direction but not the speed.
It falls towards Earth, but the Earth's surface curves away at the same rate.
[B1] for tangential velocity, [B1] for force perpendicular/changing direction, [B1] for curvature match.

(c) Orbital speed $v = \sqrt{\frac{GM}{r}}$ .
If $r$ increases, $v$ decreases.
[B1] for decreases, [B1] for inverse relationship with square root of radius.

17. Air Resistance Graph (a) Graph:

Starts at $20 \text{ m s}^{-1}$ .
Curve with decreasing gradient (concave up towards time axis).
Reaches $v=0$ at a time $t < t_{vacuum}$ .
Vacuum graph is a straight line from $20$ to $0$ .
[B1] for correct shape (curve), [B1] for starting/ending points, [B1] for comparison line.

(b) Air resistance acts downwards (same direction as weight) during upward motion.
Resultant downward force is greater than weight alone.
Deceleration is greater than $g$ , so it stops in less time.
[B1] for air resistance adds to weight/force, [B1] for greater deceleration.

18. Ladder Equilibrium (a) Wall is smooth, so there is no friction.
Therefore, the reaction force is purely normal (perpendicular) to the wall, i.e., horizontal.
[B1] for smooth/no friction.

(b) Moments about base (A).
Clockwise: Weight $W$ acts at $L/2$ . Horizontal distance from A is $\frac{L}{2} \cos 60^\circ$ .
Moment $= W (\frac{L}{2} \cos 60^\circ)$ .
Anticlockwise: Wall force $R_w$ acts at top. Vertical distance from A is $L \sin 60^\circ$ .
Moment $= R_w (L \sin 60^\circ)$ .
Equating: $R_w L \sin 60^\circ = W \frac{L}{2} \cos 60^\circ$ .
$R_w = \frac{W \cos 60^\circ}{2 \sin 60^\circ} = \frac{W}{2 \tan 60^\circ}$ .
$R_w = \frac{W}{2\sqrt{3}}$ or $0.289 W$ .
[M1] for moment of weight, [M1] for moment of wall force, [A1] for expression.

(c) Friction at base $\le \mu \times \text{Normal Reaction at base}$ .
Or: The horizontal force from the wall must be balanced by static friction at the floor, and this friction must not exceed its maximum limit.
[B1] for friction condition.

19. Car Power (a) $P = Fv \Rightarrow 50000 = F(25)$ .
$F = \frac{50000}{25} = 2000 \text{ N}$ .
[M1] for formula, [A1] for answer.

(b) At constant speed, Driving Force = Resistive Force.
Resistive Force $= 2000 \text{ N}$ .
[B1] for answer.

(c) $P = Fv$ . If $P$ is constant, as $v$ increases, Driving Force $F$ decreases ( $F = P/v$ ).
Resistive forces (air resistance) increase with speed.
Resultant Force $= F_{drive} - F_{resist}$ .
Since $F_{drive}$ drops and $F_{resist}$ rises, Resultant Force decreases.
Since $a = F_{res}/m$ , acceleration decreases.
[B1] for F drive decreases, [B1] for resistive force increases/resultant drops, [B1] for link to acceleration.

20. Impulse (a) Change in momentum (or Force $\times$ time).
[B1] for definition.

(b) Change in momentum $\Delta p = m(v - u) = 0.045(50 - 0) = 2.25 \text{ N s}$ .
Impulse $= F_{avg} \Delta t$ .
$2.25 = F_{avg} (0.50 \times 10^{-3})$ .
$F_{avg} = \frac{2.25}{0.0005} = 4500 \text{ N}$ .
[M1] for momentum change, [M1] for impulse equation, [A1] for answer.

(c) Crumple zones increase the time of contact ( $\Delta t$ ) during a collision.
For a fixed change in momentum ( $\Delta p$ ), increasing $\Delta t$ reduces the average force ( $F = \Delta p / \Delta t$ ).
Lower force means less injury to passengers.
[B1] for increases time, [B1] for reduces force, [B1] for safety implication.

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