A Level H2 Physics Practice Paper 4

TuitionGoWhere Practice Paper - Physics H2 A-Level

Answer Key — Mechanics (Version 4 of 5)

Section A: Multiple Choice [15 marks]

1. B [1]

Teaching note: At maximum height, the final velocity is zero. Using $v^2 = u^2 - 2gh$ with $v = 0$: $h = \frac{u^2}{2g} = \frac{20^2}{2 \times 9.81} = \frac{400}{19.62} = 20.4$ m. This uses the kinematic equation for constant acceleration (gravity), where the only force acting is weight.

2. D [1]

Teaching note: Momentum ($p = mv$) has both magnitude and direction, making it a vector. Energy, power, and speed are all scalar quantities — they have magnitude only. A common mistake is confusing speed (scalar) with velocity (vector).

3. C [1]

Teaching note: Using $s = ut + \frac{1}{2}at^2$ with $u = 0$: $s = 0 + \frac{1}{2}(3.0)(8.0)^2 = \frac{1}{2}(3.0)(64) = 96$ m. This is a direct application of the kinematic equation for uniform acceleration from rest.

4. B [1]

Teaching note: By conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$. So $(2.0)(5.0) + (3.0)(0) = (2.0 + 3.0)v$, giving $10 = 5.0v$, so $v = 2.0$ m s⁻¹. This is a perfectly inelastic collision (objects stick together), so kinetic energy is NOT conserved, but momentum always is (in the absence of external forces).

5. B [1]

Teaching note: For a satellite in circular orbit, $\frac{GMm}{r^2} = \frac{mv^2}{r}$, so $v = \sqrt{\frac{GM}{r}}$. If $r$ doubles, $v$ becomes $\frac{v}{\sqrt{2}}$. The orbital speed decreases with increasing radius. A common trap is to assume $v \propto r$ or $v \propto \frac{1}{r}$ instead of $v \propto \frac{1}{\sqrt{r}}$.

6. C [1]

Teaching note: Impulse = change in momentum = $F \Delta t = 15 \times 3.0 = 45$ kg m s⁻¹. This follows directly from the impulse-momentum theorem: the impulse delivered by a force equals the change in momentum of the object.

7. A [1]

Teaching note: At the top of the vertical circle, both the tension $T$ (acting downward toward the centre) and the weight $mg$ (also downward) contribute to the centripetal force. The net centripetal force is $T + mg = \frac{mv^2}{r}$. A common mistake is subtracting the forces — at the top, both point toward the centre (downward).

8. A [1]

Teaching note: Taking moments about the left support: The 200 N weight acts at the centre (2.0 m from left), and the 300 N load acts 1.0 m from the left. Let $R_R$ be the right reaction. $R_R \times 4.0 = 200 \times 2.0 + 300 \times 1.0 = 400 + 300 = 700$. So $R_R = 175$ N. This uses the principle of moments: for equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments.

9. C [1]

Teaching note: For a horizontally launched projectile, horizontal motion has constant velocity: $x = v_x \times t = 15 \times 3.0 = 45$ m. The vertical fall time is independent of the horizontal speed. Students sometimes mistakenly use the vertical motion equation for horizontal distance.

10. B [1]

Teaching note: The Principle of Conservation of Linear Momentum states that the total momentum of a system remains constant provided no external resultant force acts on the system. Key conditions: (1) it applies to a system of objects, (2) the condition is "no external resultant force" (not just "no forces"), and (3) it applies to ALL types of collisions, not just elastic ones.

11. B [1]

Teaching note: Centripetal force $F_c = \frac{mv^2}{r} = \frac{1200 \times 20^2}{50} = \frac{1200 \times 400}{50} = \frac{480000}{50} = 9600$ N. This is provided by friction between the tyres and the road. A common error is forgetting to square the velocity.

12. A [1]

Teaching note: Elastic potential energy $E = \frac{1}{2}kx^2 = \frac{1}{2}(400)(0.10)^2 = \frac{1}{2}(400)(0.01) = 2.0$ J. Note that the energy depends on $x^2$, not just $x$. Students sometimes forget to square the extension or omit the factor of $\frac{1}{2}$.

13. A [1]

Teaching note: Centripetal acceleration $a_c = \omega^2 r$. Angular velocity $\omega = 2\pi f = 2\pi(0.50) = \pi$ rad s⁻¹. So $a_c = \pi^2 \times 2.0 = 2\pi^2$... Wait, let me recalculate: $a_c = \omega^2 r = \pi^2 \times 2.0 = 2\pi^2$ m s⁻². The answer is B.

Correction: B [1]

$a_c = (2\pi f)^2 \times r = (2\pi \times 0.50)^2 \times 2.0 = \pi^2 \times 2.0 = 2\pi^2$ m s⁻².

14. B [1]

Teaching note: At constant velocity, the net force along the incline is zero. The component of weight down the incline is $mg\sin\theta = 4.0 \times 9.81 \times \sin(30°) = 4.0 \times 9.81 \times 0.5 = 19.6$ N. This must equal the frictional force (which acts up the slope to balance the motion). The frictional force is 19.6 N.

15. B [1]

Teaching note: Using $v^2 = u^2 + 2gh$ with $u = 0$: $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 80} = \sqrt{1569.6} = 39.6$ m s⁻¹. Alternatively, using conservation of energy: $mgh = \frac{1}{2}mv^2$, so $v = \sqrt{2gh}$. A common error is using $v = gt$ without first finding $t$, or using $h = \frac{1}{2}gt^2$ and then $v = gt$ (which also works but takes longer).

Section B: Structured Questions [35 marks]

16. [6 marks]

(a) [2]

Newton's Second Law: The rate of change of momentum of an object is directly proportional to the resultant force acting on it and takes place in the direction of the resultant force. [1]

OR equivalently: The resultant force acting on an object is equal to the product of its mass and acceleration ($F = ma$). [1]

Marking: Award 1 mark for the proportionality statement and 1 mark for the direction statement (or the $F = ma$ form).

(b)(i) [1]

The free-body diagram should show:

• Weight ($W = mg$) acting vertically downward from the centre of mass
• Normal reaction force ($R$) acting vertically upward from the floor of the lift
• The upward arrow (R) must be longer than the downward arrow (W) since the net force is upward (upward acceleration)

Marking: 1 mark for both forces correctly shown and labelled, with the correct relative sizes.

(b)(ii) [3]

Applying Newton's Second Law (upward positive): $R - mg = ma$ $R = m(g + a)$ $R = 60(9.81 + 1.5)$ $R = 60 \times 11.31$ $\boxed{R = 678.6 \text{ N} \approx 679 \text{ N}}$

Marking:

• [1] for correct equation: $R - mg = ma$ or equivalent
• [1] for correct substitution
• [1] for correct answer with unit

Common mistake: Students may write $R = mg$ (ignoring the acceleration) or $R = ma$ (ignoring the weight).

17. [7 marks]

(a) [1]

The principle of conservation of linear momentum states that the total momentum of a system remains constant provided no external resultant force acts on the system.

Marking: Award 1 mark for a complete statement including the condition (no external force / closed system).

(b) [4]

Taking right as positive:

Conservation of momentum: $m_A u_A + m_B u_B = m_A v_A + m_B v_B$ $(0.80)(3.0) + (1.2)(0) = (0.80)(-0.60) + (1.2)(v_B)$ $2.4 = -0.48 + 1.2 v_B$ $2.88 = 1.2 v_B$ $\boxed{v_B = 2.4 \text{ m s}^{-1} \text{ (to the right)}}$

Marking:

• [1] for correct equation with correct signs
• [1] for correct substitution of values
• [1] for correct algebraic manipulation
• [1] for correct answer with direction

Common mistake: Forgetting that trolley A moves to the LEFT after the collision, so $v_A = -0.60$ m s⁻¹. Using $+0.60$ gives $v_B = 1.6$ m s⁻¹ (wrong).

(c) [2]

Calculate total kinetic energy before and after:

Before: $KE_i = \frac{1}{2}(0.80)(3.0)^2 + 0 = 3.6$ J

After: $KE_f = \frac{1}{2}(0.80)(0.60)^2 + \frac{1}{2}(1.2)(2.4)^2 = 0.144 + 3.456 = 3.6$ J

Since $KE_i = KE_f$, the collision is elastic.

Marking:

• [1] for calculating both kinetic energies correctly
• [1] for correct conclusion

Note: If the kinetic energies were not equal, the collision would be inelastic. In all real collisions, momentum is conserved, but kinetic energy is only conserved in elastic collisions.

18. [6 marks]

(a) [3]

At the top of the circle, the minimum speed occurs when the tension in the string is zero (the string is just about to go slack). At this point, the weight alone provides the centripetal force:

$mg = \frac{mv^2}{r}$ $v^2 = gr$ $v = \sqrt{gr} = \sqrt{9.81 \times 0.80} = \sqrt{7.848}$ $\boxed{v = 2.80 \text{ m s}^{-1}}$

Marking:

• [1] for setting $mg = \frac{mv^2}{r}$ (recognising tension = 0 at minimum speed)
• [1] for correct substitution
• [1] for correct answer

Common mistake: Students may include tension in the equation, but at the minimum speed, tension is zero.

(b) [3]

At the bottom of the circle, both tension (upward, toward centre) and weight (downward, away from centre) act. The net centripetal force (toward centre, upward) is:

$T - mg = \frac{mv^2}{r}$ $T = mg + \frac{mv^2}{r}$ $T = 0.25 \times 9.81 + \frac{0.25 \times 6.0^2}{0.80}$ $T = 2.4525 + \frac{0.25 \times 36}{0.80}$ $T = 2.4525 + 11.25$ $\boxed{T = 13.8 \text{ N} \text{ (to 3 s.f.)}}$

Marking:

• [1] for correct equation: $T - mg = \frac{mv^2}{r}$
• [1] for correct substitution
• [1] for correct answer with unit

Common mistake: Writing $T + mg = \frac{mv^2}{r}$ (wrong direction for weight at the bottom) or $T = \frac{mv^2}{r}$ (forgetting weight entirely).

19. [8 marks]

(a) [2]

Horizontal component: $u_x = u\cos\theta = 25\cos(35°) = 25 \times 0.8192$ $\boxed{u_x = 20.5 \text{ m s}^{-1}}$

Vertical component: $u_y = u\sin\theta = 25\sin(35°) = 25 \times 0.5736$ $\boxed{u_y = 14.3 \text{ m s}^{-1}}$

Marking: [1] each for correct horizontal and vertical components.

(b) [3]

At maximum height, vertical velocity $v_y = 0$: $v_y^2 = u_y^2 - 2gH$ $0 = (14.3)^2 - 2(9.81)H$ $H = \frac{(14.3)^2}{2 \times 9.81} = \frac{204.49}{19.62}$ $\boxed{H = 10.4 \text{ m}}$

Marking:

• [1] for correct equation
• [1] for correct substitution
• [1] for correct answer

(c) [3]

Time of flight: The total time is found from the vertical motion. Using $s = u_yt - \frac{1}{2}gt^2$ with $s = 0$ (returns to ground level): $0 = u_yt - \frac{1}{2}gt^2$ $t(u_y - \frac{1}{2}gt) = 0$ $t = 0$ (launch) or $t = \frac{2u_y}{g} = \frac{2 \times 14.3}{9.81} = 2.915$ s

Horizontal range: $R = u_x \times t = 20.5 \times 2.915$ $\boxed{R = 59.8 \text{ m}}$

Marking:

• [1] for correct time of flight calculation
• [1] for using horizontal velocity × time
• [1] for correct answer

Alternative: Using $R = \frac{u^2 \sin(2\theta)}{g} = \frac{625 \times \sin(70°)}{9.81} = \frac{625 \times 0.9397}{9.81} = 59.8$ m.

20. [8 marks]

(a) [2]

Component of weight parallel to the incline: $F_{\parallel} = mg\sin\theta = 5.0 \times 9.81 \times \sin(25°)$ $F_{\parallel} = 5.0 \times 9.81 \times 0.4226$ $\boxed{F_{\parallel} = 20.7 \text{ N}}$

Marking:

• [1] for correct formula $mg\sin\theta$
• [1] for correct answer

(b) [2]

Normal reaction: $N = mg\cos\theta = 5.0 \times 9.81 \times \cos(25°) = 5.0 \times 9.81 \times 0.9063 = 44.45$ N

Frictional force: $f = \mu_k N = 0.20 \times 44.45$ $\boxed{f = 8.89 \text{ N}}$

Marking:

• [1] for finding normal reaction correctly
• [1] for correct frictional force

(c) [4]

Using the work-energy principle. The net work done on the block equals the change in change in kinetic energy:

Work done by gravity (parallel component): $W_g = mg\sin\theta \times s = 20.7 \times 8.0 = 165.6$ J

Work done against friction: $W_f = -f \times s = -8.89 \times 8.0 = -71.1$ J

Net work done: $W_{net} = 165.6 - 71.1 = 94.5$ J

This equals the gain in kinetic energy: $\frac{1}{2}mv^2 = 94.5$ $v = \sqrt{\frac{2 \times 94.5}{5.0}} = \sqrt{37.8}$ $\boxed{v = 6.15 \text{ m s}^{-1}}$

Marking:

• [1] for correct work done by gravity along the slope
• [1] for correct work done against friction
• [1] for applying work-energy principle correctly
• [1] for correct final answer

Alternative approach: Using Newton's Second Law to find acceleration, then kinematics: $a = g\sin\theta - \mu_k g\cos\theta = 9.81(0.4226) - 0.20 \times 9.81(0.9063) = 4.146 - 1.778 = 2.368$ m s⁻² $v^2 = 2as = 2(2.368)(8.0) = 37.89$, so $v = 6.16$ m s⁻¹ ✓

Section C: Long Structured Question [20 marks]

21. [20 marks]

(a) [1]

The principle of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another (or transferred from one body to another). The total energy of an isolated system remains constant.

Marking: Award 1 mark for a complete statement.

(b) [3]

Using conservation of energy from A to B: $mgh = \frac{1}{2}mv_B^2$ $v_B = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.80} = \sqrt{15.696}$ $\boxed{v_B = 3.96 \text{ m s}^{-1}}$

Marking:

• [1] for correct energy conservation equation
• [1] for correct substitution
• [1] for correct answer

(c) [2]

Assuming the horizontal section of the track is at the same height as point B (i.e., the track is level after B), there is no further change in height, so by conservation of energy, the speed at the end of the track equals the speed at point B.

$\boxed{v = 3.96 \text{ m s}^{-1}}$

Assumption: The horizontal section is at the same vertical height as point B (no further change in gravitational potential energy).

Marking:

• [1] for correct answer
• [1] for stating the assumption

(d) [2]

For the projectile motion (vertical): Using $H = \frac{1}{2}gt^2$ (initial vertical velocity = 0): $1.25 = \frac{1}{2}(9.81)t^2$ $t^2 = \frac{2 \times 1.25}{9.81} = \frac{2.50}{9.81} = 0.2548$ $\boxed{t = 0.505 \text{ s}}$

Marking:

• [1] for correct equation
• [1] for correct answer

(e) [2]

$x = v \times t = 3.96 \times 0.505$ $\boxed{x = 2.00 \text{ m}}$

Marking:

• [1] for using horizontal velocity × time
• [1] for correct answer

(f) [1]

The measured value of $x$ is less than the calculated value because friction/air resistance acts on the ball bearing along the track, reducing its speed at the end of the track (and hence reducing the horizontal range).

Acceptable answers:

• Friction between the ball bearing and the track
• Air resistance
• Energy lost to sound/heat during the motion

Marking: Award 1 mark for any valid reason that would reduce the speed.

(g)(i) [4]

From the derivation:

• Speed at B: $v_B = \sqrt{2gh}$
• Speed at end of track (assuming horizontal section at same height): $v = \sqrt{2gh}$
• Time of fall: $t = \sqrt{\frac{2H}{g}}$
• Horizontal distance: $x = v \times t = \sqrt{2gh} \times \sqrt{\frac{2H}{g}}$

Therefore: $x^2 = 2gh \times \frac{2H}{g} = 4Hh$ $\boxed{x^2 = 4Hh}$

Marking:

• [1] for $v = \sqrt{2gh}$
• [1] for $t = \sqrt{\frac{2H}{g}}$
• [1] for combining to get $x = \sqrt{2gh} \times \sqrt{\frac{2H}{g}}$
• [1] for final expression $x^2 = 4Hh$

(g)(ii) [3]

The graph of $x^2$ against $h$ is a straight line through the origin with gradient $= 4H$.

Since $x^2 = 4Hh$, comparing with $y = mx + c$: the gradient $= 4H$.

Therefore: $g$ is NOT directly determined from this gradient (since $g$ cancels out in the derivation). However, if the question intends for students to find $H$ from the gradient:

$\text{gradient} = 4H$ $H = \frac{\text{gradient}}{4}$

Note: In this particular setup, $g$ cancels out in the expression for $x^2$, so the gradient gives $H$, not $g$. If the track had a different configuration where $g$ did not cancel, the gradient could be used to find $g$.

Marking:

• [1] for stating the graph is a straight line through the origin
• [1] for stating gradient = $4H$
• [1] for explaining how to use the gradient

(g)(iii) [2]

The sketch should show:

• A straight line passing through the origin
• Positive gradient
• $x^2$ on the vertical axis, $h$ on the horizontal axis
• The line should be labelled or the gradient indicated

Marking:

• [1] for straight line through origin
• [1] for correct axes labels

Mark Summary

Common Mistakes Summary

1. Sign errors in momentum problems — Always define a positive direction and stick to it. Velocities in the opposite direction must be negative.

2. Forgetting the condition in conservation of momentum — The principle only applies when no external resultant force acts on the system.

3. Confusing vertical circle top and bottom — At the top, both tension and weight point toward the centre (downward). At the bottom, tension points toward the centre (upward) and weight points away (downward).

4. Projectile motion independence — Horizontal and vertical motions are independent. The time of flight is determined entirely by the vertical motion.

5. Work-energy vs. kinematics — Both methods are valid for the inclined plane problem. The work-energy method is often simpler when only initial and final speeds are needed.

6. Friction on inclined planes — The normal reaction on an incline is $mg\cos\theta$, NOT $mg$. The frictional force is $\mu mg\cos\theta$.