A Level H1 Physics Practice Paper 4

TuitionGoWhere Practice Paper — Physics H1 A-Level

Answer Key — Mechanics Focus (Version 4 of 5)

Section A: Multiple Choice

1. C. $20.0 \text{ m s}^{-1}$ [1]
Explanation: Using $v = u + at$, where $u = 0$, $a = 2.5 \text{ m s}^{-2}$, $t = 8.0 \text{ s}$:
$v = 0 + (2.5)(8.0) = 20.0 \text{ m s}^{-1}$.
Common mistake: Forgetting the car starts from rest ($u = 0$).

2. D. Momentum [1]
Explanation: Momentum ($\mathbf{p} = m\mathbf{v}$) has both magnitude and direction, making it a vector. Energy, power, and speed are scalars — they have magnitude only.

3. C. Its horizontal velocity remains constant. [1]
Explanation: In projectile motion (ignoring air resistance), there is no horizontal force, so horizontal velocity is constant. The vertical motion is accelerated by gravity ($a = g$ downward), so vertical velocity increases.
Common mistake: Thinking the horizontal component changes because the path is curved.

4. B. $1.5 \text{ m s}^{-1}$ [1]
Explanation: Conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$
$(0.50)(6.0) + (1.5)(0) = (0.50 + 1.5)v$
$3.0 = 2.0v$
$v = 1.5 \text{ m s}^{-1}$.

5. C. 100 J [1]
Explanation: Work done $= F \times d = 20 \times 5.0 = 100 \text{ J}$.
Work is done when a force moves an object in the direction of the force.

6. B. $7.7 \text{ m s}^{-1}$ [1]
Explanation: Using conservation of energy: $mgh = \frac{1}{2}mv^2$, so $v = \sqrt{2gh}$
$v = \sqrt{2 \times 9.81 \times 3.0} = \sqrt{58.86} = 7.67 \approx 7.7 \text{ m s}^{-1}$ (2 s.f.).

7. C. 50 N [1]
Explanation: Total downward force $= 60 + 40 = 100 \text{ N}$. By symmetry (load at midpoint, beam uniform), each support carries half: $100 / 2 = 50 \text{ N}$.

8. C. When object A exerts a force on object B, object B exerts an equal and opposite force on object A. [1]
Explanation: Newton's Third Law states that forces come in equal and opposite pairs acting on different objects. Option A is Newton's First Law; Option B is Newton's Second Law.

9. C. $34.6 \text{ m s}^{-1}$ [1]
Explanation: Horizontal component: $u_x = u \cos\theta = 40 \cos 30° = 40 \times 0.866 = 34.6 \text{ m s}^{-1}$.
Common mistake: Using $\sin$ instead of $\cos$ for the horizontal component.

10. C. 792 N [1]
Explanation: Using Newton's Second Law: $R - mg = ma$, so $R = m(g + a)$
$R = 70(9.81 + 1.5) = 70 \times 11.31 = 791.7 \approx 792 \text{ N}$ (3 s.f.).
The normal force exceeds the student's weight because the lift accelerates upward.

Section B: Structured Questions

11. Kinematics [6 marks]

(a) Acceleration is defined as the rate of change of velocity with respect to time. [1]
Marking: Accept "change in velocity per unit time" or equivalent. Must convey rate of change.

(b)(i) Using $a = \frac{v - u}{t}$:
$a = \frac{0 - 25}{10} = -2.5 \text{ m s}^{-2}$
Deceleration $= 2.5 \text{ m s}^{-2}$ [2]
Marking: [1] for correct formula/substitution, [1] for correct answer with unit. Accept magnitude for deceleration.

(b)(ii) Using $s = \frac{(u + v)}{2} \times t$:
$s = \frac{(25 + 0)}{2} \times 10 = 125 \text{ m}$ [3]
Marking: [1] for correct formula, [1] for correct substitution, [1] for correct answer with unit.
Alternative: $s = ut + \frac{1}{2}at^2 = 25(10) + \frac{1}{2}(-2.5)(100) = 250 - 125 = 125 \text{ m}$.
Common mistake: Forgetting that final velocity is zero, or sign error on acceleration.

12. Newton's Laws and Dynamics [7 marks]

(a) Newton's Second Law: The net force acting on an object is equal to the rate of change of its momentum (or, for constant mass, $F = ma$). [1]
Marking: Must mention net/resultant force and acceleration (or rate of change of momentum).

(b)(i) Net force $= 3600 - 1200 = 2400 \text{ N}$
$a = \frac{F}{m} = \frac{2400}{1200} = 2.0 \text{ m s}^{-2}$ [2]
Marking: [1] for net force, [1] for acceleration with unit.

(b)(ii) Using $v = u + at$:
$v = 0 + (2.0)(6.0) = 12.0 \text{ m s}^{-1}$ [2]
Marking: [1] for formula/substitution, [1] for correct answer.

(b)(iii) Using $s = ut + \frac{1}{2}at^2$:
$s = 0 + \frac{1}{2}(2.0)(6.0)^2 = 36.0 \text{ m}$ [2]
Marking: [1] for formula/substitution, [1] for correct answer with unit.

13. Conservation of Momentum [8 marks]

(a) The principle of conservation of linear momentum states that the total momentum of a system remains constant (is conserved) provided no external forces act on the system (i.e., in a closed/isolated system). [2]
Marking: [1] for stating total momentum before = total momentum after (or momentum is conserved). [1] for mentioning the condition of no external forces / closed system.

(b)(i) Conservation of momentum:
$m_A u_A + m_B u_B = (m_A + m_B)v$
$(3.0)(4.0) + (2.0)(0) = (3.0 + 2.0)v$
$12.0 = 5.0v$
$v = 2.4 \text{ m s}^{-1}$ [3]
Marking: [1] for correct equation, [1] for correct substitution, [1] for correct answer with unit.

(b)(ii) Initial kinetic energy:
$KE_i = \frac{1}{2}(3.0)(4.0)^2 + 0 = 24.0 \text{ J}$

Final kinetic energy:
$KE_f = \frac{1}{2}(5.0)(2.4)^2 = \frac{1}{2}(5.0)(5.76) = 14.4 \text{ J}$

Kinetic energy lost:
$\Delta KE = 24.0 - 14.4 = 9.6 \text{ J}$ [3]
Marking: [1] for initial KE, [1] for final KE, [1] for energy lost with unit.
Teaching note: In a perfectly inelastic collision (objects stick together), kinetic energy is always lost. Momentum is conserved but kinetic energy is not — the lost KE is converted to heat, sound, and deformation energy.

14. Work, Energy and Power [9 marks]

(a) Work done by a force is defined as the product of the force and the displacement in the direction of the force ($W = Fd\cos\theta$). [1]
Marking: Must mention force and displacement/distance in the direction of force.

(b)(i) Work done by applied force:
$W = Fd = 40 \times 8.0 = 320 \text{ J}$ [1]

(b)(ii) Work done against friction:
$W_f = f \times d = 15 \times 8.0 = 120 \text{ J}$ [1]

(b)(iii) Net work done on block $= 320 - 120 = 200 \text{ J}$

By the work-energy principle: Net work $= \Delta KE$
$200 = \frac{1}{2}(5.0)v^2 - 0$
$v^2 = \frac{200 \times 2}{5.0} = 80$
$v = \sqrt{80} = 8.94 \approx 8.9 \text{ m s}^{-1}$ (2 s.f.) [3]
Marking: [1] for net work, [1] for applying work-energy principle, [1] for correct answer.

(c) Average power $= \frac{\text{Work done by applied force}}{\text{time}} = \frac{320}{4.0} = 80 \text{ W}$ [3]
Marking: [1] for using work = 320 J, [1] for correct formula, [1] for correct answer with unit.
Alternative: Power $= F \times v_{avg} = 40 \times \frac{8.94}{2} \approx 179 \text{ W}$ would be incorrect here since the question asks for average power over the 4.0 s interval, and $P = W/t$ is the appropriate method.
Note: $P = Fv$ gives instantaneous power; $P = W/t$ gives average power.

Section C: Free Response

15. Projectile Motion [10 marks]

Visual reference: Q15-fig1 — Tower of height 45 m, ball projected horizontally at 20 m s⁻¹, parabolic trajectory to ground. The diagram must show the tower, horizontal velocity arrow, parabolic path, and horizontal range R.

(a) The horizontal component of velocity remains constant because there is no horizontal force acting on the ball (air resistance is negligible). By Newton's First Law, an object continues at constant velocity unless acted on by a net force. Gravity acts vertically, so it does not affect the horizontal motion. [2]
Marking: [1] for stating no horizontal force / no air resistance, [1] for linking to Newton's First Law or explaining the consequence.

(b) Using vertical motion: $h = \frac{1}{2}gt^2$ (since initial vertical velocity $u_y = 0$)
$45 = \frac{1}{2}(9.81)t^2$
$t^2 = \frac{45 \times 2}{9.81} = \frac{90}{9.81} = 9.174$
$t = \sqrt{9.174} = 3.03 \approx 3.0 \text{ s}$ (2 s.f.) [3]
Marking: [1] for correct equation, [1] for correct substitution, [1] for correct answer with unit.

(c) Horizontal range: $R = u_x \times t = 20 \times 3.03 = 60.6 \approx 61 \text{ m}$ (2 s.f.) [2]
Marking: [1] for using $R = u_x t$, [1] for correct answer with unit.

(d) Vertical velocity just before impact:
$v_y = gt = 9.81 \times 3.03 = 29.7 \text{ m s}^{-1}$

Horizontal velocity: $v_x = 20 \text{ m s}^{-1}$ (constant)

Resultant speed:
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 29.7^2} = \sqrt{400 + 882.1} = \sqrt{1282.1} = 35.8 \approx 36 \text{ m s}^{-1}$ (2 s.f.) [3]
Marking: [1] for $v_y$ calculation, [1] for using Pythagoras to combine components, [1] for correct final speed with unit.
Common mistake: Adding $v_x$ and $v_y$ directly instead of using Pythagoras.

16. Forces and Equilibrium [10 marks]

Visual reference: Q16-fig1 — Horizontal beam (3.0 m, 2.0 kg) hinged at left end to wall, cable at right end at 37° above horizontal, 4.0 kg sign at midpoint. All forces labelled: tension T along cable, weight of beam (19.62 N) at midpoint, weight of sign (39.24 N) at midpoint, hinge reaction (R_x, R_y).

(a) Free-body diagram should show: [3]

• Weight of beam $W_b = 2.0 \times 9.81 = 19.62 \text{ N}$ acting downward at the midpoint (1.5 m from hinge)
• Weight of sign $W_s = 4.0 \times 9.81 = 39.24 \text{ N}$ acting downward at the midpoint (1.5 m from hinge)
• Tension $T$ in the cable, acting at the right end (3.0 m from hinge), at $37°$ above the horizontal
• Reaction force at the hinge, with horizontal component $R_x$ and vertical component $R_y$

Marking: [1] for each correctly labelled force (3 forces minimum: two weights + tension). [1] for correct positions/directions. Deduct for missing labels or incorrect directions.

(b) Taking moments about the hinge (anticlockwise positive):

Clockwise moments (weights):
$W_b \times 1.5 + W_s \times 1.5 = (19.62)(1.5) + (39.24)(1.5) = 29.43 + 58.86 = 88.29 \text{ Nm}$

Anticlockwise moment (tension):
The vertical component of tension provides the restoring moment. The perpendicular distance from hinge to line of action of $T$ is $L \sin(37°)$ vertically, or equivalently:
$T \sin(37°) \times 3.0$ (vertical component of T × perpendicular distance along beam)

For equilibrium: Sum of moments = 0
$T \sin(37°) \times 3.0 = 88.29$
$T \times 0.6018 \times 3.0 = 88.29$
$T \times 1.805 = 88.29$
$T = \frac{88.29}{1.805} = 48.9 \approx 49 \text{ N}$ (2 s.f.) [4]
Marking: [1] for correct moment equation, [1] for correct weight values, [1] for correct trigonometric resolution of T, [1] for correct answer with unit.

(c) Resolving forces horizontally:
$R_x = T \cos(37°) = 48.9 \times 0.7986 = 39.1 \text{ N}$

Resolving forces vertically:
$R_y + T \sin(37°) = W_b + W_s$
$R_y + (48.9)(0.6018) = 19.62 + 39.24$
$R_y + 29.43 = 58.86$
$R_y = 29.43 \text{ N}$

Magnitude of hinge reaction:
$R = \sqrt{R_x^2 + R_y^2} = \sqrt{39.1^2 + 29.43^2} = \sqrt{1528.8 + 866.1} = \sqrt{2394.9} = 48.9 \approx 49 \text{ N}$ (2 s.f.) [3]
Marking: [1] for horizontal resolution, [1] for vertical resolution, [1] for correct magnitude with unit.
Note: The reaction force at the hinge is approximately equal in magnitude to the tension but acts in a different direction. This is a good check for students.

End of Answer Key

Marks Summary:
Section A: Q1–Q10 = 10 × 1 = 10 marks
Section B: Q11 = 6, Q12 = 7, Q13 = 8, Q14 = 9 = 30 marks
Section C: Q15 = 10, Q16 = 10 = 20 marks
Total: 60 marks ✓