Mechanics — Diagnostic Tests

Unit Tests

UT-1: Kinematics

Question:

(a) State the three equations of motion (SUVAT equations) and define each variable.

(b) A car accelerates uniformly from rest at $2.5\,\text{m s}^{-2}$ for $8$ seconds. Calculate the distance travelled.

(c) A ball is thrown vertically upwards with an initial velocity of $15\,\text{m s}^{-1}$. Calculate: (i) the maximum height reached, (ii) the time taken to reach the maximum height, (iii) the total time the ball is in the air before returning to its starting point.

(d) Explain why the horizontal and vertical components of projectile motion can be treated independently.

Solution:

(a) The SUVAT equations (for constant acceleration):

1. $v = u + at$
2. $s = ut + \frac{1}{2}at^2$
3. $v^2 = u^2 + 2as$
4. $s = \frac{1}{2}(u + v)t$

Where: $s$ = displacement, $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $t$ = time.

(b) Given: $u = 0$ (from rest), $a = 2.5\,\text{m s}^{-2}$, $t = 8\,\text{s}$.

Using $s = ut + \frac{1}{2}at^2$:

$s = 0 + \frac{1}{2}(2.5)(64) = 80\,\text{m}$

The car travels $80\,\text{m}$.

(c) Taking upwards as positive, $u = 15\,\text{m s}^{-1}$, $a = -g = -9.8\,\text{m s}^{-2}$.

(i) At maximum height, $v = 0$:

$v^2 = u^2 + 2as$ $0 = 15^2 + 2(-9.8)s$ $s = \frac{225}{19.6} = 11.48\,\text{m} \approx 11.5\,\text{m}$

(ii) Using $v = u + at$:

$0 = 15 + (-9.8)t$ $t = \frac{15}{9.8} = 1.53\,\text{s}$

(iii) By symmetry (or calculating the time for the downward journey), the total time is $2 \times 1.53 = 3.06\,\text{s}$.

Alternatively, setting $s = 0$: $0 = 15t - \frac{1}{2}(9.8)t^2$, so $t = \frac{2 \times 15}{9.8} = 3.06\,\text{s}$.

(d) The horizontal and vertical components of projectile motion can be treated independently because they are perpendicular to each other. Gravity acts only vertically (downwards), producing a constant vertical acceleration of $g = 9.8\,\text{m s}^{-2}$, but has no effect on the horizontal motion. Therefore, the horizontal component of velocity remains constant (assuming no air resistance), while the vertical component changes uniformly due to gravity. The two motions are independent and do not affect each other.

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UT-2: Forces and Newton’s Laws

Question:

(a) State Newton’s three laws of motion.

(b) A block of mass $5\,\text{kg}$ rests on a horizontal surface. A horizontal force of $30\,\text{N}$ is applied, and the block accelerates at $4\,\text{m s}^{-2}$. Calculate the magnitude of the frictional force acting on the block.

(c) Two forces act on an object: $F_1 = 12\,\text{N}$ at $30^\circ$ above the horizontal and $F_2 = 8\,\text{N}$ at $45^\circ$ below the horizontal. Calculate the resultant horizontal and vertical components of force.

(d) A lift (elevator) of mass $800\,\text{kg}$ carries a person of mass $70\,\text{kg}$. Calculate the tension in the lift cable when the lift is: (i) accelerating upwards at $1.5\,\text{m s}^{-2}$, (ii) moving downwards at constant velocity.

Solution:

(a)

• First law: An object will remain at rest or continue to move at constant velocity unless acted upon by a resultant (unbalanced) external force.
• Second law: The resultant force acting on an object is equal to the rate of change of momentum, or equivalently, $F = ma$ where $F$ is the resultant force, $m$ is mass, and $a$ is acceleration.
• Third law: If object A exerts a force on object B, then object B exerts an equal and opposite force on object A. These forces are of the same type, act on different objects, and are equal in magnitude and opposite in direction.

(b) Using Newton’s second law: $F_{\text{resultant}} = ma = 5 \times 4 = 20\,\text{N}$.

The resultant force is the applied force minus friction: $F_{\text{resultant}} = F_{\text{applied}} - F_{\text{friction}}$.

$20 = 30 - F_{\text{friction}}$ $F_{\text{friction}} = 10\,\text{N}$

(c) Resolving $F_1$:

• Horizontal: $12\cos(30^\circ) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \approx 10.39\,\text{N}$
• Vertical: $12\sin(30^\circ) = 12 \times 0.5 = 6\,\text{N}$ (upward)

Resolving $F_2$:

• Horizontal: $8\cos(45^\circ) = 8 \times \frac{\sqrt{2}}{2} = 4\sqrt{2} \approx 5.66\,\text{N}$
• Vertical: $8\sin(45^\circ) = 4\sqrt{2} \approx 5.66\,\text{N}$ (downward)

Resultant horizontal: $6\sqrt{3} + 4\sqrt{2} \approx 10.39 + 5.66 = 16.05\,\text{N}$ Resultant vertical: $6 - 4\sqrt{2} \approx 6 - 5.66 = 0.34\,\text{N}$ (upward)

(d) Total mass $= 800 + 70 = 870\,\text{kg}$.

(i) Accelerating upwards: resultant force $F = ma = 870 \times 1.5 = 1305\,\text{N}$ (upward). The forces on the lift are tension $T$ (up) and weight $W = 870 \times 9.8 = 8526\,\text{N}$ (down).

$T - W = F_{\text{resultant}}$ $T = 8526 + 1305 = 9831\,\text{N}$

(ii) Moving at constant velocity ($a = 0$): no resultant force, so $T = W = 8526\,\text{N}$.

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UT-3: Work, Energy, and Power

Question:

(a) Define work, kinetic energy, and gravitational potential energy, giving the formula for each.

(b) A crate of mass $50\,\text{kg}$ is pushed up a frictionless slope inclined at $20^\circ$ to the horizontal through a distance of $10\,\text{m}$. Calculate the work done against gravity.

(c) A car of mass $1200\,\text{kg}$ is travelling at $20\,\text{m s}^{-1}$. The brakes apply a constant force to bring the car to rest over a distance of $40\,\text{m}$. Calculate the braking force.

(d) A motor lifts a load of $200\,\text{kg}$ vertically at a constant speed of $0.5\,\text{m s}^{-1}$. Calculate the power output of the motor.

Solution:

(a)

• Work done: $W = F \times d \times \cos\theta$, where $F$ is force, $d$ is displacement, and $\theta$ is the angle between the force and the direction of motion. Work is the energy transferred when a force causes displacement. Unit: joules (J).
• Kinetic energy: $E_k = \frac{1}{2}mv^2$, where $m$ is mass and $v$ is speed. Unit: joules (J).
• Gravitational potential energy: $E_p = mgh$, where $m$ is mass, $g$ is gravitational field strength, and $h$ is height. Unit: joules (J).

(b) The vertical height gained: $h = 10\sin(20^\circ) = 10 \times 0.342 = 3.42\,\text{m}$.

Work done against gravity $= mgh = 50 \times 9.8 \times 3.42 = 1675.8\,\text{J} \approx 1676\,\text{J}$.

(c) Using the work-energy principle: work done by braking force = change in kinetic energy.

$F \times d = \frac{1}{2}mv^2 - 0$ $F \times 40 = \frac{1}{2}(1200)(20)^2$ $F \times 40 = 240000$ $F = 6000\,\text{N}$

The braking force is $6000\,\text{N}$.

(d) Power $= \frac{\text{work done}}{\text{time}} = \frac{mgh}{t} = mgv$ (since $h/t = v$).

$P = 200 \times 9.8 \times 0.5 = 980\,\text{W}$

The power output of the motor is $980\,\text{W}$.

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Integration Tests

IT-1: Combined Mechanics Problems

Question:

(a) A projectile is launched from ground level with an initial velocity of $25\,\text{m s}^{-1}$ at an angle of $35^\circ$ above the horizontal. Calculate: (i) the horizontal and vertical components of the initial velocity, (ii) the maximum height reached, (iii) the horizontal range, (iv) the time of flight.

(b) A car of mass $1000\,\text{kg}$ starts from rest at the top of a hill $30\,\text{m}$ high and rolls down to the bottom, reaching a speed of $22\,\text{m s}^{-1}$. Calculate the energy lost due to friction.

(c) A block of mass $8\,\text{kg}$ is placed on a rough surface inclined at $30^\circ$ to the horizontal. The coefficient of friction between the block and the surface is $0.4$. Determine whether the block will slide down the slope. If it does, calculate its acceleration.

(d) Evaluate the statement: “An object in free fall experiences zero air resistance.” Discuss the conditions under which this approximation is valid and the consequences of ignoring air resistance in calculations.

Solution:

(a) $u = 25\,\text{m s}^{-1}$, $\theta = 35^\circ$.

(i) Horizontal: $u_x = 25\cos(35^\circ) = 25 \times 0.8192 = 20.48\,\text{m s}^{-1}$

Vertical: $u_y = 25\sin(35^\circ) = 25 \times 0.5736 = 14.34\,\text{m s}^{-1}$

(ii) Maximum height: $v_y^2 = u_y^2 + 2(-g)s$

$0 = 14.34^2 - 2(9.8)s$ $s = \frac{205.6}{19.6} = 10.49\,\text{m}$

(iii) Time of flight: $t = \frac{2u_y}{g} = \frac{2 \times 14.34}{9.8} = 2.93\,\text{s}$.

Range: $R = u_x \times t = 20.48 \times 2.93 = 60.0\,\text{m}$.

(b) Initial energy (at top of hill): $E_p = mgh = 1000 \times 9.8 \times 30 = 294000\,\text{J}$.

Final kinetic energy (at bottom): $E_k = \frac{1}{2}mv^2 = \frac{1}{2}(1000)(22)^2 = 242000\,\text{J}$.

Energy lost to friction $= 294000 - 242000 = 52000\,\text{J}$.

(c) Forces acting on the block:

Weight component down the slope: $mg\sin(30^\circ) = 8 \times 9.8 \times 0.5 = 39.2\,\text{N}$

Normal reaction: $R = mg\cos(30^\circ) = 8 \times 9.8 \times 0.866 = 67.9\,\text{N}$

Maximum friction: $F_{\text{friction}} = \mu R = 0.4 \times 67.9 = 27.16\,\text{N}$

Since $39.2 > 27.16$, the gravitational component exceeds maximum friction, so the block will slide.

Acceleration down the slope:

$F_{\text{resultant}} = mg\sin(30^\circ) - F_{\text{friction}} = 39.2 - 27.16 = 12.04\,\text{N}$

$a = \frac{F_{\text{resultant}}}{m} = \frac{12.04}{8} = 1.51\,\text{m s}^{-2}$

(d) The approximation of zero air resistance is valid for objects that are: compact (low surface area to mass ratio), moving at low speeds, falling through short distances, or in contexts where air resistance is negligible compared to gravitational force (e.g., a dense metal ball dropped from a few metres). The consequences of ignoring air resistance include overestimating the terminal velocity of falling objects, overestimating the range of projectiles, and calculating unrealistically high kinetic energies for long falls. In real-world scenarios like parachuting or high-speed projectiles, air resistance is significant and cannot be ignored.

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IT-2: Energy and Momentum Applications

Question:

(a) A pendulum of mass $0.5\,\text{kg}$ is pulled aside so that it is $0.3\,\text{m}$ above its lowest point and then released. Calculate its speed at the lowest point, assuming no energy losses.

(b) A force of $40\,\text{N}$ acts on an object of mass $10\,\text{kg}$ for $3$ seconds. The object was initially at rest. Calculate: (i) the impulse, (ii) the final velocity of the object.

(c) A skier of mass $75\,\text{kg}$ skis down a slope of length $200\,\text{m}$ inclined at $25^\circ$ to the horizontal. The coefficient of kinetic friction is $0.1$. Calculate the skier’s speed at the bottom of the slope, starting from rest.

(d) Explain the principle of conservation of energy and describe how it applies to a bungee jumper undergoing simple harmonic motion at the lowest point of their jump.

Solution:

(a) Using conservation of energy: $E_p = E_k$ at the lowest point.

$mgh = \frac{1}{2}mv^2$ $9.8 \times 0.3 = \frac{1}{2}v^2$ $v^2 = 5.88$ $v = 2.42\,\text{m s}^{-1}$

(b)

(i) Impulse $J = F \times t = 40 \times 3 = 120\,\text{N s}$.

(ii) Impulse = change in momentum: $J = mv - mu = m(v - 0) = 10v$.

$10v = 120$ $v = 12\,\text{m s}^{-1}$

(c) Height of slope: $h = 200\sin(25^\circ) = 200 \times 0.4226 = 84.52\,\text{m}$.

Normal reaction: $R = mg\cos(25^\circ) = 75 \times 9.8 \times 0.9063 = 666.1\,\text{N}$.

Frictional force: $F_f = \mu R = 0.1 \times 666.1 = 66.61\,\text{N}$.

Net force down the slope: $F_{\text{net}} = mg\sin(25^\circ) - F_f = 75 \times 9.8 \times 0.4226 - 66.61 = 311.6 - 66.61 = 245.0\,\text{N}$.

Acceleration: $a = \frac{F_{\text{net}}}{m} = \frac{245.0}{75} = 3.27\,\text{m s}^{-2}$.

Using $v^2 = u^2 + 2as$: $v^2 = 0 + 2(3.27)(200) = 1308$.

$v = \sqrt{1308} = 36.2\,\text{m s}^{-1}$

(d) The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. The total energy of a closed system remains constant. For a bungee jumper, at the highest point all energy is gravitational potential energy. As they fall, this is converted to kinetic energy. At the lowest point, the kinetic energy is momentarily zero as the bungee cord reaches maximum extension, and the energy has been converted to elastic potential energy stored in the stretched cord. The total energy (gravitational potential + kinetic + elastic potential) remains constant throughout the jump. Energy losses to air resistance and internal friction in the cord reduce the total mechanical energy, producing heat.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Forgetting that weight acts vertically downwards when resolving forces on inclined planes; the component along the slope is $mg\sin\theta$ and the normal component is $mg\cos\theta$.
• Confusing displacement and distance in kinematics equations, which require displacement (a vector quantity with direction).
• Sign errors in SUVAT equations: always define a positive direction and be consistent with signs throughout the calculation.
• Applying Newton’s third law incorrectly: the action-reaction pair acts on different objects, not on the same object.
• Forgetting to include all forces in free body diagrams, particularly the normal reaction force.