A Level H2 Physics Practice Paper 5

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TuitionGoWhere Practice Paper - Physics H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics H2
Level: A-Level
Paper: Practice Paper – Version 5 (Mechanics Focus)
Duration: 2 Hours
Total Marks: 80

Name: __________________________
Class: __________________________
Date: __________________________

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Instructions to Candidates

1. Write your name, class, and date in the spaces provided.
2. Answer all questions.
3. Write your answers in the spaces provided in this booklet.
4. You may lose marks if you do not show your working or if you do not use appropriate units.
5. The use of an approved scientific calculator is expected, where appropriate.
6. At the end of the examination, fasten all your work securely together.
7. The number of marks is given in brackets [ ] at the end of each question or part question.

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Section A: Structured Questions

Answer all questions in this section.

1. A student is investigating the motion of a trolley down an inclined plane. The student measures the time $t$ taken for the trolley to travel a distance $s$ from rest. The student plots a graph of $s$ against $t^2$.

(a) State the relationship between $s$ and $t$ for an object accelerating uniformly from rest. [1]

(b) Explain how the acceleration $a$ of the trolley can be determined from the gradient of the graph of $s$ against $t^2$. [2]

(c) The graph obtained is a straight line that does not pass through the origin, having a positive intercept on the $s$-axis. Suggest a systematic error that could cause this result. [1]

2. Define the term linear momentum. [1]

3. State the Principle of Conservation of Linear Momentum. [2]

4. A ball of mass 0.15 kg is dropped from a height of 1.2 m onto a hard horizontal surface. It rebounds to a height of 0.80 m. Assume air resistance is negligible.

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

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

(c) Determine the magnitude of the change in momentum of the ball during the impact. [2]

5. A car of mass 1200 kg travels along a straight horizontal road. The engine provides a constant driving force of 2500 N. The total resistive force acting on the car is proportional to the square of its speed $v$, given by $F_R = kv^2$.

(a) When the car reaches a constant speed of 30 m s⁻¹, calculate the value of the constant $k$. [2]

(b) Calculate the initial acceleration of the car when it starts from rest. [2]

(c) Explain, in terms of forces, why the acceleration of the car decreases as its speed increases. [2]

6. A satellite orbits the Earth in a circular orbit of radius $r$.

(a) Show that the orbital speed $v$ of the satellite is given by $v = \sqrt{\frac{GM}{r}}$, where $M$ is the mass of the Earth and $G$ is the gravitational constant. [3]

(b) Two satellites, A and B, orbit the Earth. Satellite A has an orbital radius $r$, and Satellite B has an orbital radius $4r$. Determine the ratio of the orbital period of Satellite B to the orbital period of Satellite A. [3]

7. A block of mass 2.0 kg is placed 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 horizontal force $F$ required to just start the block moving. [2]

(b) Once the block is moving, the force $F$ is maintained at the value calculated in (a). Calculate the acceleration of the block. [3]

8. A projectile is launched from ground level with an initial velocity of 20 m s⁻¹ at an angle of 30° to the horizontal.

(a) Calculate the maximum height reached by the projectile. [2]

(b) Calculate the horizontal range of the projectile. [2]

(c) State the velocity of the projectile at its maximum height. [1]

9. A uniform beam of length 4.0 m and weight 50 N is hinged at one end to a vertical wall. The beam is held horizontal by a cable attached to the other end of the beam and to the wall at a point 3.0 m above the hinge.

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

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

10. A spring obeys Hooke’s Law. When a load of 10 N is applied, the extension is 4.0 cm.

(a) Calculate the spring constant $k$. [1]

(b) Calculate the elastic potential energy stored in the spring when the extension is 4.0 cm. [2]

(c) The load is increased to 20 N. Calculate the additional work done in extending the spring from 4.0 cm to the new extension. [2]

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Section B: Data-Based and Contextual Questions

Answer all questions in this section.

11. A student performs an experiment to determine the acceleration due to gravity $g$ using a simple pendulum. The student measures the time $T$ for 10 oscillations for different lengths $L$ of the pendulum. The data is recorded below.

Length $L$ (m)	Time for 10 oscillations (s)	Period $T$ (s)	$T^2$ (s²)
0.20	8.9	0.89	0.79
0.40	12.7	1.27	1.61
0.60	15.5	1.55	2.40
0.80	17.9	1.79	3.20
1.00	20.0	2.00	4.00

(a) The relationship between period and length is $T = 2\pi \sqrt{\frac{L}{g}}$. Explain how a graph of $T^2$ against $L$ can be used to determine $g$. [2]

(b) Using the data provided, plot a graph of $T^2$ against $L$ on the grid provided (not shown, assume plotted). Determine the gradient of the best-fit line. [2]

(c) Calculate the value of $g$ from your gradient. [2]

(d) The student repeats the experiment using a heavier bob of the same size. State and explain the effect, if any, on the value of $g$ determined. [2]

12. A car travels around a circular bend of radius 50 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 for the car. [1]

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

(c) The road is now banked at an angle $\theta$ to the horizontal. Explain how banking the road allows the car to travel at a higher speed without relying solely on friction. [2]

13. Two trolleys, A and B, are placed on a friction-compensated runway. Trolley A has a mass of 0.50 kg and is moving with a velocity of 0.80 m s⁻¹ towards stationary trolley B, which has a mass of 0.30 kg. The trolleys collide and stick together.

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

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

(c) State the type of collision that occurred. [1]

14. A rocket of mass $5.0 \times 10^4$ kg is launched vertically from the Earth's surface. The engines produce a constant upward thrust of $8.0 \times 10^5$ N. Assume the mass of the rocket remains constant for the first 10 seconds of flight and $g = 9.81$ m s⁻².

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

(b) Calculate the resultant force acting on the rocket at lift-off. [2]

(c) Calculate the acceleration of the rocket at lift-off. [2]

(d) Explain why the acceleration of the rocket increases as it rises, even if the thrust remains constant. [2]

15. A ball is attached to a string of length 0.80 m and whirled in a vertical circle. The mass of the ball is 0.20 kg.

(a) At the top of the circle, the speed of the ball is 4.0 m s⁻¹. Calculate the tension in the string at this point. [3]

(b) Determine the minimum speed required at the top of the circle for the ball to maintain circular motion. [2]

(c) Explain why the tension in the string is greater at the bottom of the circle than at the top, assuming constant speed. [2]

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Section C: Long Structured Questions

Answer all questions in this section.

16. A skier of mass 70 kg starts from rest at the top of a slope inclined at 20° to the horizontal. The slope is 100 m long. The coefficient of dynamic friction between the skis and the snow is 0.10.

(a) Draw a diagram showing the forces acting on the skier. [2]

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

(c) Calculate the frictional force acting on the skier. [2]

(d) Calculate the acceleration of the skier down the slope. [3]

(e) Calculate the speed of the skier at the bottom of the slope. [2]

(f) At the bottom of the slope, the skier moves onto a horizontal surface with the same coefficient of friction. Calculate the distance the skier travels before coming to rest. [3]

17. An elevator (lift) of mass 600 kg carries passengers of total mass 400 kg. The elevator is supported by a steel cable.

(a) Calculate the tension in the cable when the elevator is: (i) Stationary. [2] (ii) Moving upwards with a constant acceleration of 1.5 m s⁻². [3] (iii) Moving downwards with a constant deceleration of 1.5 m s⁻². [3]

(b) A passenger of mass 80 kg stands on a weighing scale inside the elevator. Calculate the reading on the scale when the elevator is accelerating upwards at 1.5 m s⁻². [3]

(c) Explain the sensation of "weightlessness" experienced by the passenger if the cable were to break and the elevator fell freely. [2]

18. A ball of mass 0.10 kg is thrown vertically upwards with an initial speed of 15 m s⁻¹. Air resistance is significant and can be modeled as a force $F_{air} = bv$, where $b$ is a constant.

(a) Describe the variation of the acceleration of the ball during its upward journey. [2]

(b) Compare the time taken for the ball to reach its maximum height with the time taken to fall back to the starting point. Explain your answer. [3]

(c) Sketch a velocity-time graph for the entire motion (up and down) until it returns to the starting height. Indicate the initial velocity, maximum height (where $v=0$), and final velocity. [3]

(d) Explain why the magnitude of the final velocity is less than the initial velocity. [2]

19. A uniform ladder of length 5.0 m and weight 200 N rests against a smooth vertical wall at an angle of 60° to the horizontal ground. The foot of the ladder is on rough horizontal ground.

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

(b) By taking moments about the foot of the ladder, calculate the normal contact force exerted by the wall on the ladder. [4]

(c) Calculate the frictional force exerted by the ground on the ladder. [2]

(d) Calculate the minimum coefficient of static friction required between the ladder and the ground to prevent slipping. [3]

20. A particle of mass $m$ moves in a horizontal circle of radius $r$ with constant speed $v$. The particle is attached to a fixed point by a light inextensible string.

(a) Derive the expression for the centripetal acceleration $a = \frac{v^2}{r}$. [3]

(b) Hence, show that the tension $T$ in the string is given by $T = \frac{mv^2}{r}$. [1]

(c) If the speed of the particle is doubled and the radius is halved, determine the factor by which the tension changes. [2]

(d) In a real scenario, the string has mass. Explain qualitatively how this would affect the tension along the length of the string. [2]

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End of Paper

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TuitionGoWhere Practice Paper - Physics H2 A-Level

Answer Key and Marking Scheme

Subject: Physics H2
Paper: Practice Paper – Version 5 (Mechanics Focus)

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Section A: Structured Questions

1. (a) $s = \frac{1}{2}at^2$ [1] (b) Comparing $s = \frac{1}{2}at^2$ with $y = mx$, the gradient $m = \frac{1}{2}a$. Therefore, $a = 2 \times \text{gradient}$. [2] (c) The distance $s$ was measured from a point other than the starting position (e.g., the front of the trolley instead of the back, or a zero error in the ruler). [1]

2. Linear momentum is the product of mass and velocity. ($p = mv$) [1]

3. In a closed system (or isolated system), [1] the total momentum before an interaction equals the total momentum after the interaction, provided no external forces act. [1]

4. (a) Using $v^2 = u^2 + 2as$: $v^2 = 0 + 2(9.81)(1.2)$ $v = \sqrt{23.544} = 4.85$ m s⁻¹ [2] (b) Using $v^2 = u^2 + 2as$ for upward motion (final $v=0$ at max height): $0 = u^2 - 2(9.81)(0.80)$ $u = \sqrt{15.696} = 3.96$ m s⁻¹ [2] (c) Taking downward as positive: $p_{initial} = 0.15 \times 4.85 = 0.7275$ N s $p_{final} = 0.15 \times (-3.96) = -0.594$ N s $\Delta p = p_{final} - p_{initial} = -0.594 - 0.7275 = -1.3215$ N s Magnitude = 1.32 N s [2]

5. (a) At constant speed, Driving Force = Resistive Force. $2500 = k(30)^2$ $k = \frac{2500}{900} = 2.78$ kg m⁻¹ [2] (b) At rest ($v=0$), $F_R = 0$. $F_{net} = 2500$ N $a = \frac{F}{m} = \frac{2500}{1200} = 2.08$ m s⁻² [2] (c) As speed increases, resistive force $F_R = kv^2$ increases. [1] Since Driving Force is constant, the resultant force ($F_D - F_R$) decreases. [1] Therefore, acceleration decreases ($a = F_{net}/m$).

6. (a) Gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ [1] $\frac{GM}{r} = v^2$ [1] $v = \sqrt{\frac{GM}{r}}$ [1] (b) Period $T = \frac{2\pi r}{v} = 2\pi r \sqrt{\frac{r}{GM}} = 2\pi \sqrt{\frac{r^3}{GM}}$. $T \propto r^{3/2}$ $\frac{T_B}{T_A} = (\frac{r_B}{r_A})^{3/2} = (\frac{4r}{r})^{3/2} = 4^{1.5} = 8$ [3]

7. (a) Normal reaction $R = mg = 2.0 \times 9.81 = 19.62$ N. Max static friction $F_{s,max} = \mu_s R = 0.40 \times 19.62 = 7.85$ N. Minimum force $F = 7.85$ N (or 7.8 N). [2] (b) Dynamic friction $F_d = \mu_d R = 0.30 \times 19.62 = 5.886$ N. Resultant force $F_{net} = F_{applied} - F_d = 7.848 - 5.886 = 1.962$ N. $a = \frac{F_{net}}{m} = \frac{1.962}{2.0} = 0.98$ m s⁻². [3]

8. (a) Vertical component $u_y = 20 \sin 30^\circ = 10$ m s⁻¹. At max height, $v_y = 0$. $0 = 10^2 - 2(9.81)h$ $h = \frac{100}{19.62} = 5.10$ m. [2] (b) Horizontal component $u_x = 20 \cos 30^\circ = 17.32$ m s⁻¹. Time of flight: $t = \frac{2u_y}{g} = \frac{20}{9.81} = 2.04$ s. Range $R = u_x t = 17.32 \times 2.04 = 35.3$ m. [2] (c) Velocity is horizontal only. Magnitude = $17.3$ m s⁻¹. Direction = Horizontal. [1]

9. (a) Diagram should show: Weight (downwards from center), Tension (along cable towards wall), Hinge reaction (horizontal and/or vertical components at hinge). [2] (b) Angle of cable with beam: $\tan \alpha = \frac{3}{4} \Rightarrow \alpha = 36.9^\circ$. Taking moments about hinge: Clockwise moment = Anticlockwise moment $W \times 2.0 = T \sin(36.9^\circ) \times 4.0$ $50 \times 2.0 = T \times 0.6 \times 4.0$ $100 = 2.4 T$ $T = 41.7$ N. [3]

10. (a) $k = \frac{F}{x} = \frac{10}{0.04} = 250$ N m⁻¹. [1] (b) $EPE = \frac{1}{2}kx^2 = 0.5 \times 250 \times (0.04)^2 = 0.20$ J. [2] (c) New extension for 20 N: $x_2 = \frac{20}{250} = 0.08$ m. Work done = Change in EPE = $\frac{1}{2}k(x_2^2 - x_1^2)$ $= 0.5 \times 250 \times (0.08^2 - 0.04^2)$ $= 125 \times (0.0064 - 0.0016) = 125 \times 0.0048 = 0.60$ J. [2]

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Section B: Data-Based and Contextual Questions

11. (a) Squaring the equation: $T^2 = \frac{4\pi^2}{g} L$. This is in the form $y = mx$. The gradient $m = \frac{4\pi^2}{g}$. Thus $g = \frac{4\pi^2}{\text{gradient}}$. [2] (b) Gradient calculation from points (e.g., (0,0) and (1.00, 4.00)): $m = \frac{4.00 - 0}{1.00 - 0} = 4.00$ s² m⁻¹. (Accept range 3.9–4.1 based on best fit). [2] (c) $g = \frac{4\pi^2}{4.00} = \pi^2 \approx 9.87$ m s⁻². [2] (d) No effect. [1] The period of a simple pendulum is independent of mass ($T = 2\pi \sqrt{L/g}$ does not contain $m$). [1]

12. (a) Frictional force between tires and road. [1] (b) Max friction provides centripetal force: $\mu mg = \frac{mv^2}{r}$ $v^2 = \mu gr = 0.80 \times 9.81 \times 50 = 392.4$ $v = \sqrt{392.4} = 19.8$ m s⁻¹. [3] (c) Banking allows the normal contact force to have a horizontal component. [1] This horizontal component contributes to the centripetal force, reducing the reliance on friction and allowing higher speeds before skidding occurs. [1]

13. (a) Conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B) v$ $0.50(0.80) + 0 = (0.50 + 0.30) v$ $0.40 = 0.80 v$ $v = 0.50$ m s⁻¹. [3] (b) Initial KE = $\frac{1}{2}(0.50)(0.80)^2 = 0.16$ J. Final KE = $\frac{1}{2}(0.80)(0.50)^2 = 0.10$ J. Loss = $0.16 - 0.10 = 0.06$ J. [3] (c) Inelastic collision. [1]

14. (a) Weight $W = mg = 5.0 \times 10^4 \times 9.81 = 4.905 \times 10^5$ N. [1] (b) Resultant Force $F_{net} = \text{Thrust} - \text{Weight}$ $F_{net} = 8.0 \times 10^5 - 4.905 \times 10^5 = 3.095 \times 10^5$ N. [2] (c) $a = \frac{F_{net}}{m} = \frac{3.095 \times 10^5}{5.0 \times 10^4} = 6.19$ m s⁻². [2] (d) As the rocket burns fuel, its mass $m$ decreases. [1] Since $F_{net}$ is roughly constant (or increases as drag decreases/thrust stays same), $a = F_{net}/m$ increases as $m$ decreases. [1]

15. (a) At top: $T + mg = \frac{mv^2}{r}$ $T = \frac{0.20(4.0)^2}{0.80} - 0.20(9.81)$ $T = \frac{3.2}{0.8} - 1.962 = 4.0 - 1.962 = 2.04$ N. [3] (b) Minimum speed when $T=0$: $mg = \frac{mv_{min}^2}{r} \Rightarrow v_{min} = \sqrt{gr}$ $v_{min} = \sqrt{9.81 \times 0.80} = \sqrt{7.848} = 2.80$ m s⁻¹. [2] (c) At bottom, Tension acts upwards, Weight downwards. $T - mg = \frac{mv^2}{r} \Rightarrow T = \frac{mv^2}{r} + mg$. At top, $T + mg = \frac{mv^2}{r} \Rightarrow T = \frac{mv^2}{r} - mg$. Thus, tension is greater at the bottom by $2mg$ (assuming constant $v$). [2]

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Section C: Long Structured Questions

16. (a) Diagram: Weight (vertical down), Normal Reaction (perpendicular to slope), Friction (up the slope). [2] (b) Component down slope = $mg \sin \theta = 70 \times 9.81 \times \sin 20^\circ = 234.6$ N. [2] (c) Normal Reaction $R = mg \cos \theta = 70 \times 9.81 \times \cos 20^\circ = 643.8$ N. Friction $F = \mu R = 0.10 \times 643.8 = 64.4$ N. [2] (d) Resultant force down slope = $234.6 - 64.4 = 170.2$ N. $a = \frac{170.2}{70} = 2.43$ m s⁻². [3] (e) $v^2 = u^2 + 2as = 0 + 2(2.43)(100) = 486$. $v = \sqrt{486} = 22.0$ m s⁻¹. [2] (f) On horizontal: Friction $F = \mu mg = 0.10 \times 70 \times 9.81 = 68.67$ N. Deceleration $a = -\frac{68.67}{70} = -0.981$ m s⁻². $0 = v^2 + 2as \Rightarrow 0 = 22.0^2 + 2(-0.981)s$. $s = \frac{484}{1.962} = 247$ m. [3]

17. (a) Total mass $M = 1000$ kg. (i) $T = Mg = 1000 \times 9.81 = 9810$ N. [2] (ii) $T - Mg = Ma \Rightarrow T = M(g+a) = 1000(9.81 + 1.5) = 11310$ N. [3] (iii) Moving down with deceleration means acceleration is upwards. Same as (ii). $T = 11310$ N. [3] (b) For passenger: $R - mg = ma$. $R = m(g+a) = 80(9.81 + 1.5) = 80(11.31) = 904.8$ N. [3] (c) If cable breaks, $a = g$ downwards. Normal reaction $R = m(g-a) = m(g-g) = 0$. [1] The passenger exerts no force on the scale/floor, creating the sensation of weightlessness. [1]

18. (a) Upward journey: Gravity acts down, Air resistance acts down (opposing motion). Resultant force $F = mg + bv$. [1] As $v$ decreases, air resistance decreases, so resultant force decreases. Acceleration decreases (but is always $> g$ initially and approaches $g$ as $v \to 0$). [1] (b) Time up < Time down. [1] Energy is lost to air resistance throughout the flight. [1] Therefore, at any given height, the speed on the way down is less than the speed on the way up. Since average speed is lower on the way down for the same distance, time taken is longer. [1] (c) Graph: Starts at $+15$. Curve concave up (decreasing gradient magnitude) to $v=0$. Then curve becomes negative, asymptotic to terminal velocity (straight line slope $g$ if $v$ small, but generally curved). Final $v$ magnitude $< 15$. [3] (d) Work is done against air resistance. [1] Mechanical energy is dissipated as heat. [1] Thus, final KE < initial KE, so final speed < initial speed.

19. (a) The wall is smooth, so there is no friction. [1] The only force is normal to the surface. (b) Let $N_w$ be normal force from wall. Ladder length $L=5$. Weight acts at $L/2 = 2.5$ m from foot. Horizontal distance of CG from foot = $2.5 \cos 60^\circ = 1.25$ m. Vertical height of top from foot = $5 \sin 60^\circ = 4.33$ m. Moments about foot: Clockwise (Weight): $200 \times 1.25 = 250$ Nm. Anticlockwise (Wall Normal): $N_w \times 4.33$. $N_w \times 4.33 = 250 \Rightarrow N_w = 57.7$ N. [4] (c) Horizontal equilibrium: Friction $F = N_w = 57.7$ N. [2] (d) Vertical equilibrium: Normal reaction from ground $N_g = \text{Weight} = 200$ N. Limiting friction $F_{max} = \mu N_g$. To prevent slipping, $F \le \mu N_g$. $57.7 \le \mu (200)$. $\mu \ge \frac{57.7}{200} = 0.29$. [3]

20. (a) Consider velocity vectors $\vec{v}_1$ and $\vec{v}_2$ at times $t$ and $t+\Delta t$. Angle between radii is $\Delta \theta$. Angle between velocity vectors is also $\Delta \theta$. Change in velocity $\Delta v \approx v \Delta \theta$ (for small $\theta$). Acceleration $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = v \frac{d\theta}{dt}$. Since $v = r \omega$ and $\omega = \frac{d\theta}{dt}$, then $\frac{d\theta}{dt} = \frac{v}{r}$. $a = v (\frac{v}{r}) = \frac{v^2}{r}$. [3] (b) Tension provides the centripetal force. $T = F_c = ma = \frac{mv^2}{r}$. [1] (c) $T_1 = \frac{mv^2}{r}$. $T_2 = \frac{m(2v)^2}{r/2} = \frac{m(4v^2)}{r/2} = \frac{8mv^2}{r} = 8 T_1$. Tension increases by a factor of 8. [2] (d) The string has mass, so each segment of the string requires centripetal force. [1] The tension must support the centripetal force for the particle PLUS the centripetal force for the portion of the string further out. Thus, tension is maximum at the fixed point and decreases towards the particle. [1]