Comprehensive Study Guide on Motion, Forces, Energy, and Electricity

Distance-Time Graph Interpretation (Fig. 1.1):

• To determine the distance travelled by the cyclist between specific points (e.g., points C and E), one must identify the distance values on the vertical axis (y-axis) corresponding to those points and subtract the initial value from the final value.

• Motion Analysis between Points B and C: If the line on a distance-time graph is perfectly horizontal (flat), it indicates that the distance is not changing as time passes. Therefore, the cyclist is stationary (at rest) and has zero speed.

• Identifying Maximum Speed: The speed of an object on a distance-time graph is represented by the gradient (slope) of the line. The section with the steepest gradient (e.g., AB, BC, CD, or DE) represents the period where the cyclist is moving at their fastest speed.

• Average Speed Calculation: Average speed is determined by the formula:
      $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

• For the journey between points A and E, the total distance is the final y-value at point E, and the total time is the final x-value at point E. The result must be expressed with units such as $\text{m/s}$.

Speed-Time Graph Interpretation (Fig. 1.1):

• Determining Instantaneous Speed: At any specific time (e.g., $t = 6.0\,s$), the speed can be read directly from the vertical axis.

• Maximum Speed: The highest point reached on the plot represents the maximum speed attained during the motion.

• Describing Motion Trends:

• A straight line with a constant positive gradient (e.g., from A to B) represents uniform (constant) acceleration.

• If the motion between points B and C differs from A to B, it may be because the gradient is changing (non-uniform acceleration) or the line has become horizontal (constant speed/zero acceleration).

• Calculating Distance from Speed-Time Graphs: The distance travelled is equal to the area under the graph line. For a section representing uniform acceleration from rest (a triangle shape), the formula is:
      $\text{Distance} = \frac{1}{2} \times \text{base (time)} \times \text{height (speed)}$

Dynamics: Forces and Resultant Motion:

• Resultant Horizontal Forces:

• The resultant force is the single force that produces the same effect as all individual forces acting on an object.

• If a pushchair wheel (Fig. 2.1) has a $30\,N$ force acting in one direction and a $10\,N$ force acting in the opposite direction, the resultant force is calculated by subtraction:
      $\text{Resultant Force} = 30\,N - 10\,N = 20\,N$

• The direction of the resultant force is always in the direction of the larger force.

Work Done and Energy Transfer:

• Work done is calculated when a force moves an object through a distance in the direction of the force:
      $\text{Work Done} = \text{Force} \times \text{Distance}$

• Example: A force of $40\,N$ moving a trolley $50\,m$ performs work:
      $\text{Work Done} = 40\,N \times 50\,m = 2000\,J$

• As the trolley moves, energy is transferred from the person into other stores, such as:

1. Kinetic Energy Store: Associated with the movement of the object.

2. Thermal Energy Store: Due to friction between the trolley and the ground or in the wheel bearings.

Force Balance and Motion Effects:

• In Fig. 2.2, if a cyclist experiences forces in opposite directions (e.g., $220\,N$ forward and $160\,N$ backward), the resultant force is:
      $\text{Resultant Force} = 220\,N - 160\,N = 60\,N$

• Effect on Motion: If the resultant force is not zero, the cyclist will accelerate (change speed) in the direction of the resultant force. If the resultant force were zero, the cyclist would move at a constant speed or remain stationary.

Acceleration and Energy Transformations:

• Aeroplane Acceleration (Fig. 3.1):

• Given a mass ($m$) of $3.4 \times 10^5\,kg$, a change in speed ($\Delta v$) from $0$ to $65\,m/s$, and a time ($t$) of $26\,s$:

• Acceleration ($a$):
      $a = \frac{\Delta v}{t} = \frac{65\,m/s - 0\,m/s}{26\,s} = 2.5\,m/s^2$

• Resultant Force ($F$): Using Newton's Second Law:
      $F = m \times a = (3.4 \times 10^5\,kg) \times 2.5\,m/s^2 = 8.5 \times 10^5\,N$

• Energy Changes during Take-off: As the aeroplane gains height and continues to accelerate:

1. Kinetic Energy ($K.E.$) Increases: Because the speed is increasing.

2. Gravitational Potential Energy ($g.p.e.$) Increases: Because the height is increasing.

3. Chemical Energy Decreases: As fuel is burned to provide the energy for the engine.

Kinetic and Potential Energy of a Salmon:

• Mass ($m$) = $2.0\,kg$, Initial $K.E.$ = $16.2\,J$.

• Speed Calculation:
      $K.E. = \frac{1}{2}mv^2$
      $16.2 = \frac{1}{2} \times 2.0 \times v^2$
      $v = \sqrt{16.2} \approx 4.02\,m/s$

• Maximum Height Calculation: Ignoring air resistance, the gain in $g.p.e.$ equals the initial $K.E.$.
      $\text{Gain in } g.p.e. = mgh = 16.2\,J$
      Assuming $g = 10\,m/s^2$:
      $2.0 \times 10 \times h = 16.2$
      $h = \frac{16.2}{20} = 0.81\,m$

Skier Physics (Fig. 3.1):

• Mass = $75\,kg$, vertical height change = $880\,m$.

• Decrease in $g.p.e.$:
      $\Delta g.p.e. = mgh = 75\,kg \times 10\,m/s^2 \times 880\,m = 660,000\,J$

• Work Done Against Resistive Forces:
      $\text{Work} = \text{Resistive Force} \times \text{Total Distance} = 220\,N \times 2800\,m = 616,000\,J$

• Final Kinetic Energy:
      $\text{Final } K.E. = \text{Initial } g.p.e. - \text{Work done against resistive forces}$
      $\text{Final } K.E. = 660,000\,J - 616,000\,J = 44,000\,J$

• Aerodynamics: Skiers bend their bodies to minimize their surface area facing the wind, which reduces air resistance (drag).

Electricity and Circuit Components:

• Circuit Measurement:

• Ammeter: Connected in series to measure the current in a circuit.

• Voltmeter: Connected in parallel across a component to measure the potential difference ($p.d.$).

• Component Y / Component X: Both refer to a variable resistor (or rheostat), which allows the user to change the resistance and thereby control the current in the circuit.

Ohm's Law and Resistance:

• Resistance ($R$) is calculated using the formula:
      $R = \frac{V}{I}$

• Example: If $V = 4.5\,V$ and $I = 0.25\,A$, then $R = \frac{4.5}{0.25} = 18.0\,\Omega$.

• Combined Resistance:

• Series: The total resistance is the sum of individual resistors ($R\_{total} = R\_1 + R\_2$). For resistors of $5.0\,\Omega$ and $7.0\,\Omega$, the series total is $12.0\,\Omega$.

• Parallel: The combined resistance of resistors in parallel is always less than the resistance of the smallest individual resistor.

• Calculating a specific resistor in parallel: If the total resistance of two resistors (one being $6.0\,\Omega$ and the other $R$) is $4.0\,\Omega$:
      $\frac{1}{R\_{total}} = \frac{1}{R\_1} + \frac{1}{R}$
      $\frac{1}{4.0} = \frac{1}{6.0} + \frac{1}{R} \implies \frac{1}{R} = \frac{1}{4.0} - \frac{1}{6.0} = \frac{3-2}{12} = \frac{1}{12}$
      $R = 12\,\Omega$

Electrical Energy Transfers:

• Energy transferred ($E$) by a heater over time ($t$) is given by:
      $E = V \times I \times t$

• Example: $V = 8.0\,V$, $I = 1.4\,A$, $t = 30\,s$:
      $E = 8.0 \times 1.4 \times 30 = 336\,J$

Charge and Current:

• Current ($I$) is the rate of flow of charge ($Q$):
      $I = \frac{Q}{t} = \frac{90\,C}{45\,s} = 2.0\,A$

• Electromotive Force (e.m.f.): It is the energy supplied by a source (like a battery) to each unit of charge to move it around a complete circuit.

Electromagnetic Induction and Transformers:

• Working Principle of a Transformer (Fig. 10.1):

• Primary Coil: An alternating current ($A.C.$) flows through the primary coil. This current produces a constantly changing magnetic field around the coil.

• Core: The soft iron core provides a low-reluctance path that channels and concentrates this changing magnetic flux, linking it to the secondary coil.

• Secondary Coil: The changing magnetic flux passsing through the secondary coil induces an alternating electromotive force ($e.m.f.$) or voltage across its ends through the process of electromagnetic induction.

Design Considerations:

• Step-up Transformer: To make the output voltage higher than the input voltage, the secondary coil must have more turns ($N\_s$) than the primary coil ($N\_p$).

• Core Continuity: If the halves of the core were separated (e.g., by $30\,cm$), the magnetic flux linkage would be lost. The magnetic field would not effectively reach the secondary coil, and no voltage would be induced.

Transformer Calculations (100% Efficient):

• In an ideal transformer, input power equals output power:
      $V\_p I\_p = V\_s I\_s$

• If a transformer steps up voltage from $100\,V$ to $200\,V$ and the primary current is $0.4\,A$:
      $100\,V \times 0.4\,A = 200\,V \times I\_s$
      $40 = 200 \times I\_s$
      $I\_s = \frac{40}{200} = 0.2\,A$