A Level H1 Physics Practice Paper 4

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:

1. Weight ($200 \text{ N}$) acting downwards at the center ($2.0 \text{ m}$ from A).
2. Tension ($T$) acting at B towards C.
3. 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.

(c) Chemical potential energy from the skaters' muscles / Internal energy.
[B1] for chemical/internal energy.

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.

(c) Work done $W = F \times d$.
$W = 34.5 \times 2.0 = 69 \text{ J}$.
[M1] for formula, [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.