A Level H1 Physics Practice Paper 1

A-Level Physics H1 Quiz - Mechanics

Answer Key and Marking Scheme

Section A: Multiple Choice (10 marks)

1. D. Displacement
Displacement is a vector quantity because it has both magnitude and direction. Speed, distance, and energy are scalar quantities — they have magnitude only.
[1]

2. B. Velocity is zero and acceleration is $9.81 \text{ m s}^{-2}$ downward.
At the highest point, the ball momentarily stops before falling back down, so velocity is zero. However, gravity continues to act throughout the motion, so acceleration remains $g = 9.81 \text{ m s}^{-2}$ downward. A common mistake is to assume acceleration is zero at the top — this is incorrect because the gravitational force still acts.
[1]

3. B. $3.1 \text{ m s}^{-2}$
Using $a = \frac{v - u}{t} = \frac{25 - 0}{8.0} = 3.125 \approx 3.1 \text{ m s}^{-2}$.
[1]

4. B. $15 \text{ m s}^{-1}$
In projectile motion (neglecting air resistance), the horizontal velocity remains constant because there is no horizontal acceleration. The horizontal component does not change with time.
[1]

5. B. Momentum only
When two objects collide and stick together, this is a perfectly inelastic collision. Momentum is always conserved in collisions (provided no external forces act), but kinetic energy is not conserved in inelastic collisions — some is converted to thermal energy, sound, and deformation.
[1]

6. B. $3.3 \text{ m s}^{-2}$
Horizontal component of applied force: $F_x = 10 \cos 30^\circ = 10 \times 0.866 = 8.66 \text{ N}$.
Net horizontal force: $F_{\text{net}} = 8.66 - 2.0 = 6.66 \text{ N}$.
Acceleration: $a = \frac{F_{\text{net}}}{m} = \frac{6.66}{2.0} = 3.33 \approx 3.3 \text{ m s}^{-2}$.
[1]

7. C. $100 \text{ J}$
Work done: $W = F \times d = 20 \times 5.0 = 100 \text{ J}$. Since the force and displacement are in the same direction, $W = Fd\cos 0^\circ = Fd$.
[1]

8. C. $49 \text{ J}$
By conservation of energy, the gravitational potential energy lost equals the kinetic energy gained: $KE = mgh = 0.50 \times 9.81 \times 10 = 49.05 \approx 49 \text{ J}$.
[1]

9. D. $9600 \text{ N}$
Centripetal force: $F_c = \frac{mv^2}{r} = \frac{1200 \times 20^2}{50} = \frac{1200 \times 400}{50} = \frac{480000}{50} = 9600 \text{ N}$.
[1]

10. B. A straight line with positive gradient passing through the origin
For constant acceleration from rest, $v = u + at = 0 + at = at$, so $v$ is directly proportional to $t$. This gives a straight line through the origin with gradient equal to the acceleration.
[1]

Section B: Structured Questions (30 marks)

11. (a) The principle of conservation of linear momentum states that the total momentum of a closed/isolated system remains constant, provided no external forces act on the system. Equivalently: the total momentum before a collision equals the total momentum after the collision in the absence of external forces.
[2] — Award [B1] for "total momentum is constant/unchanged" and [B1] for "no external forces" or "closed/isolated system."

(b)(i) Using conservation of momentum:
$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$
$(0.80)(2.0) + (1.2)(0) = (0.80 + 1.2) v$
$1.6 = 2.0 v$
$v = 0.80 \text{ m s}^{-1}$
[3] — [M1] for correct equation, [M1] for correct substitution, [A1] for correct answer with unit.

(b)(ii) Initial kinetic energy: $KE_i = \frac{1}{2}(0.80)(2.0)^2 + 0 = 1.6 \text{ J}$.
Final kinetic energy: $KE_f = \frac{1}{2}(2.0)(0.80)^2 = 0.64 \text{ J}$.
Since $KE_f < KE_i$ (kinetic energy is not conserved), the collision is inelastic.
[3] — [M1] for calculating initial KE, [M1] for calculating final KE, [A1] for correct conclusion with justification.

(c) The system must be closed/isolated (no external forces acting), or the net external force on the system must be zero.
[1]

12. (a) Using the vertical motion equation (taking downward as positive):
$s = ut + \frac{1}{2}gt^2$
$45 = 0 + \frac{1}{2}(9.81)t^2$
$t^2 = \frac{90}{9.81} = 9.174$
$t = 3.03 \approx 3.0 \text{ s}$
[3] — [M1] for selecting correct equation, [M1] for correct substitution, [A1] for correct answer.

(b) Horizontal distance: $x = u_x \times t = 15 \times 3.03 = 45.4 \approx 45 \text{ m}$.
[2] — [M1] for using $x = u_x t$, [A1] for correct answer.

(c) Vertical velocity at impact: $v_y = gt = 9.81 \times 3.03 = 29.7 \text{ m s}^{-1}$.
Speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 29.7^2} = \sqrt{225 + 882.1} = \sqrt{1107.1} = 33.3 \approx 33 \text{ m s}^{-1}$.
[3] — [M1] for finding $v_y$, [M1] for using Pythagoras, [A1] for correct answer.

13. (a) Free-body diagram should show:

• Weight $mg$ acting vertically downward
• Normal reaction $R$ acting perpendicular to the plane (upward from the surface)
• Frictional force $f$ acting up the plane (opposing motion)
• Component of weight down the plane $mg\sin\theta$ (may be shown as resolved component)
  [2] — [B1] for correct forces, [B1] for correct directions.

(b) Resolving perpendicular to the plane: $R = mg\cos 30^\circ = 5.0 \times 9.81 \times 0.866 = 42.47 \text{ N}$.
Frictional force: $f = \mu_k R = 0.25 \times 42.47 = 10.62 \text{ N}$.
Resolving down the plane: $ma = mg\sin 30^\circ - f = 5.0 \times 9.81 \times 0.50 - 10.62 = 24.53 - 10.62 = 13.91 \text{ N}$.
$a = \frac{13.91}{5.0} = 2.78 \approx 2.8 \text{ m s}^{-2}$
[4] — [M1] for resolving forces, [M1] for friction calculation, [M1] for Newton's second law, [A1] for correct answer.

(c) Work done against friction: $W = f \times d = 10.62 \times 4.0 = 42.5 \approx 42 \text{ J}$.
[3] — [M1] for using $W = fd$, [M1] for correct friction value, [A1] for correct answer.

14. (a) Using $v^2 = u^2 + 2as$:
$20^2 = 0 + 2a(200)$
$400 = 400a$
$a = 1.0 \text{ m s}^{-2}$
[2] — [M1] for correct equation, [A1] for correct answer.

(b) $F = ma = 1000 \times 1.0 = 1000 \text{ N}$.
[2] — [M1] for using $F = ma$, [A1] for correct answer.

(c) $F_{\text{net}} = F_{\text{driving}} - F_{\text{resistive}}$, so $F_{\text{driving}} = 1000 + 500 = 1500 \text{ N}$.
[2] — [M1] for correct relationship, [A1] for correct answer.

(d) Time taken: $v = u + at \Rightarrow 20 = 0 + 1.0t \Rightarrow t = 20 \text{ s}$.
Average power: $P = \frac{W}{t} = \frac{F_{\text{driving}} \times d}{t} = \frac{1500 \times 200}{20} = 15000 \text{ W} = 15 \text{ kW}$.
Alternatively: $P = F_{\text{driving}} \times v_{\text{avg}} = 1500 \times 10 = 15000 \text{ W}$.
[3] — [M1] for finding time or using average velocity, [M1] for power formula, [A1] for correct answer.

15. (a) At the lowest point, the net force toward the center provides the centripetal force:
$T - mg = \frac{mv^2}{r}$
$T = mg + \frac{mv^2}{r} = 0.20 \times 9.81 + \frac{0.20 \times 5.0^2}{0.80}$
$T = 1.962 + \frac{5.0}{0.80} = 1.962 + 6.25 = 8.21 \approx 8.2 \text{ N}$
[3] — [M1] for correct equation, [M1] for substitution, [A1] for correct answer.

(b) Using conservation of energy between lowest and highest points:
$\frac{1}{2}mv_{\text{bottom}}^2 = \frac{1}{2}mv_{\text{top}}^2 + mg(2r)$
$\frac{1}{2}(5.0)^2 = \frac{1}{2}v_{\text{top}}^2 + 9.81 \times 1.60$
$12.5 = 0.5v_{\text{top}}^2 + 15.696$
$0.5v_{\text{top}}^2 = 12.5 - 15.696 = -3.196$

Since this gives a negative value, the ball does not have enough energy to reach the top of the circle. The maximum height reached is found by:
$mgh_{\text{max}} = \frac{1}{2}mv_{\text{bottom}}^2$
$h_{\text{max}} = \frac{v_{\text{bottom}}^2}{2g} = \frac{25}{19.62} = 1.27 \text{ m}$

Since $2r = 1.60 \text{ m} > 1.27 \text{ m}$, the ball does not complete the full circle. The speed at the highest point of the circle is not achievable — the ball would leave the circular path before reaching the top.

However, if we assume the question intends for the ball to complete the circle (perhaps with a rigid rod instead of a string), then:
$v_{\text{top}} = \sqrt{v_{\text{bottom}}^2 - 4gr} = \sqrt{25 - 39.24}$
This is not possible with a string. The question likely assumes a rigid support. In that case, the speed at the top would be found from energy conservation, and the tension would be:
$T' + mg = \frac{mv_{\text{top}}^2}{r}$

Given the context, the expected answer assumes the ball completes the circle:
$v_{\text{top}} = \sqrt{25 - 4(9.81)(0.80)} = \sqrt{25 - 31.39}$
This is imaginary, indicating the ball cannot complete the circle with a string.

Revised interpretation: The question likely contains values that should allow the ball to complete the circle. With the given values, the ball reaches a maximum height of $1.27 \text{ m}$ (below the top at $1.60 \text{ m}$).

For the purpose of this answer key, assuming the question intends a solvable scenario:
If we adjust to make it work (e.g., $v_{\text{bottom}} = 6.0 \text{ m s}^{-1}$):
$v_{\text{top}} = \sqrt{36 - 31.39} = \sqrt{4.61} = 2.15 \text{ m s}^{-1}$

With the original values, the ball does not reach the top. The answer should state this.
[3] — [M1] for energy conservation equation, [M1] for substitution, [A1] for correct conclusion.

(c) If the ball reaches the top: $T' = \frac{mv_{\text{top}}^2}{r} - mg$. With the given values, this is not achievable.
[2]

16. (a) Graph should show:

• Axes correctly labeled with units ($t$ / s and $s$ / m)
• Appropriate scale chosen
• All six points plotted accurately
• Smooth best-fit curve drawn (concave upward, indicating increasing velocity)
  [3] — [B1] for axes and scale, [B1] for plotting points, [B1] for best-fit curve.

(b) For constant acceleration from rest, $s = \frac{1}{2}at^2$, so $a = \frac{2s}{t^2}$.
Using the data point at $t = 3.00$ s, $s = 4.48$ m:
$a = \frac{2 \times 4.48}{(3.00)^2} = \frac{8.96}{9.00} = 0.996 \approx 1.0 \text{ m s}^{-2}$

Alternatively, from the graph, the gradient of $s$ vs $t^2$ gives $\frac{1}{2}a$.
[4] — [M1] for correct method, [M1] for using data/graph, [M1] for calculation, [A1] for correct answer.

(c)(i) The acceleration would increase because the component of gravitational force down the plane ($mg\sin\theta$) increases with angle, while the normal force (and hence friction) decreases.
[1]

(c)(ii) The graph would still be a curve (parabola) but would be steeper — the distance covered in the same time would be greater due to the larger acceleration. The curve would show a greater rate of increase.
[2] — [B1] for stating the curve is steeper/greater gradient, [B1] for explaining why.

Section D: Extended Response (10 marks)

17. (a) $a = \frac{v - u}{t} = \frac{0 - 30}{6.0} = -5.0 \text{ m s}^{-2}$.
Deceleration = $5.0 \text{ m s}^{-2}$.
[2] — [M1] for correct equation, [A1] for correct answer.

(b) $F = ma = 1500 \times 5.0 = 7500 \text{ N}$.
[2] — [M1] for using $F = ma$, [A1] for correct answer.

(c) $s = \frac{u + v}{2} \times t = \frac{30 + 0}{2} \times 6.0 = 90 \text{ m}$.
[2] — [M1] for correct equation, [A1] for correct answer.

(d) Thermal energy = initial kinetic energy = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1500 \times 30^2 = 675000 \text{ J} = 675 \text{ kJ}$.
[2] — [M1] for using KE formula, [A1] for correct answer.

(e) Kinetic energy is proportional to $v^2$. If speed doubles from $30$ to $60 \text{ m s}^{-1}$, KE increases by a factor of 4. By the work-energy principle, the work done by the braking force equals the change in KE: $W = Fd = \Delta KE$. Since the braking force is constant, $d \propto \Delta KE$. Therefore, if KE is 4 times greater, the braking distance is also 4 times greater.
[2] — [B1] for stating KE $\propto v^2$, [B1] for linking to work-energy principle.

18. (a) Taking right as positive:
$p_{\text{total}} = m_A u_A + m_B u_B = (2.0)(4.0) + (3.0)(-1.0) = 8.0 - 3.0 = 5.0 \text{ kg m s}^{-1}$
[2] — [M1] for correct signs, [A1] for correct answer.

(b) Conservation of momentum: $m_A u_A + m_B u_B = m_A v_A + m_B v_B$
$5.0 = (2.0)(-1.0) + (3.0)v_B$
$5.0 = -2.0 + 3.0v_B$
$v_B = \frac{7.0}{3.0} = 2.33 \approx 2.3 \text{ m s}^{-1}$ (to the right)
[3] — [M1] for conservation equation, [M1] for substitution, [A1] for correct answer.

(c) Initial KE: $KE_i = \frac{1}{2}(2.0)(4.0)^2 + \frac{1}{2}(3.0)(1.0)^2 = 16 + 1.5 = 17.5 \text{ J}$.
Final KE: $KE_f = \frac{1}{2}(2.0)(1.0)^2 + \frac{1}{2}(3.0)(2.33)^2 = 1.0 + 8.17 = 9.17 \text{ J}$.
Since $KE_f < KE_i$, the collision is inelastic.
[3] — [M1] for initial KE, [M1] for final KE, [A1] for conclusion.

(d) Impulse on A: $\Delta p_A = m_A(v_A - u_A) = 2.0(-1.0 - 4.0) = -10 \text{ kg m s}^{-1}$.
Magnitude of impulse = $10 \text{ N s}$.
$F = \frac{\Delta p}{\Delta t} \Rightarrow \Delta t = \frac{\Delta p}{F} = \frac{10}{500} = 0.020 \text{ s}$.
[2] — [M1] for impulse calculation, [A1] for correct answer.

19. (a) $u_x = 20 \cos 37^\circ = 20 \times 0.80 = 16 \text{ m s}^{-1}$.
$u_y = 20 \sin 37^\circ = 20 \times 0.60 = 12 \text{ m s}^{-1}$.
[2] — [B1] for each component.

(b) At maximum height, $v_y = 0$:
$v_y^2 = u_y^2 - 2gh$
$0 = 12^2 - 2(9.81)h$
$h = \frac{144}{19.62} = 7.34 \approx 7.3 \text{ m}$
[3] — [M1] for correct equation, [M1] for substitution, [A1] for correct answer.

(c) Time to reach maximum height: $t = \frac{u_y}{g} = \frac{12}{9.81} = 1.22 \text{ s}$.
Total time of flight: $T = 2t = 2.44 \approx 2.4 \text{ s}$.
[2] — [M1] for correct method, [A1] for correct answer.

(d) Range: $R = u_x \times T = 16 \times 2.44 = 39.0 \approx 39 \text{ m}$.
[2] — [M1] for using range formula, [A1] for correct answer.

(e) The acceleration is vertically downward (due to gravity) throughout the motion, including at the highest point.
[1]

20. (a) Simple harmonic motion (SHM).
[1]

(b) Maximum acceleration occurs at maximum displacement:
$a_{\text{max}} = \omega^2 A = \frac{k}{m}A = \frac{200}{2.0} \times 0.10 = 100 \times 0.10 = 10 \text{ m s}^{-2}$
[2] — [M1] for correct formula, [A1] for correct answer.

(c) Maximum speed: $v_{\text{max}} = \omega A = \sqrt{\frac{k}{m}}A = \sqrt{100} \times 0.10 = 10 \times 0.10 = 1.0 \text{ m s}^{-1}$.
[2] — [M1] for correct formula, [A1] for correct answer.

(d) Total mechanical energy: $E = \frac{1}{2}kA^2 = \frac{1}{2} \times 200 \times (0.10)^2 = 1.0 \text{ J}$.
[2] — [M1] for correct formula, [A1] for correct answer.

(e) On a rough surface, friction acts as a damping force. The motion would be damped harmonic motion:

• The amplitude would decrease over time (energy is dissipated as thermal energy)
• The block would eventually come to rest
• The frequency of oscillation would remain approximately the same (for light damping)
• The total mechanical energy would decrease with each oscillation
  [3] — [B1] for stating amplitude decreases, [B1] for energy dissipation, [B1] for eventual rest.

END OF ANSWER KEY