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Edexcel A Level (IAL) Physics-1.11 Core Practical 1: Investigating the Acceleration of Freefall- Study Notes- New Syllabus

Edexcel A Level (IAL) Physics -1.11 Core Practical 1: Investigating the Acceleration of Freefall- Study Notes- New syllabus

Edexcel A Level (IAL) Physics -1.11 Core Practical 1: Investigating the Acceleration of Freefall- Study Notes -Edexcel A level Physics – per latest Syllabus.

Key Concepts:

  • 1.11 Determine the acceleration of a freely-falling object

Edexcel A level Physics-Study Notes- All Topics

CORE PRACTICAL 1: Determine the Acceleration of a Freely-Falling Object

This core practical aims to measure the acceleration due to gravity \( g \) by analysing the motion of an object falling freely under gravity.

 Aim of the Experiment

To determine the value of gravitational acceleration \( g \) using the equation of motion:

\( s = ut + \dfrac{1}{2}gt^2 \)

For a freely falling object: \( u = 0 \) → initial velocity is zero.

Apparatus

  • Light gate(s)
  • Electronic timer
  • Data logger (optional)
  • Clamp stand
  • Ball bearing (or similar small dense object)
  • Metre ruler

 Method (Using One Light Gate)

  • Set up a light gate connected to an electronic timer.
  • Measure the height \( h \) from the bottom of the ball to the top of the light gate.
  • Hold the ball at the measured height and release it so it falls freely.
  • The light gate measures the time \( t \) it takes for the ball to fall through the gate.
  • Repeat several times for the same height, then for different heights.

Using the equation:

\( h = \dfrac{1}{2}gt^2 \)

Rearrange to find \( g \):

\( g = \dfrac{2h}{t^2} \)

 Alternative Method (Using Two Light Gates)

  • Place two light gates a known distance \( s \) apart.
  • As the object falls, each gate records its velocity at that point.
  • The average acceleration is calculated using:

\( g = \dfrac{v_2 – v_1}{t} \)

Where:

  • \( v_1 \) = velocity at first gate
  • \( v_2 \) = velocity at second gate
  • \( t \) = time between passing the two gates

 Data Processing

  • Record several values of height and corresponding time.
  • Plot a graph of \( h \) against \( t^2 \).
  • The gradient of the straight line is \( \dfrac{1}{2}g \).
  • Multiply the gradient by 2 to obtain \( g \).

Sources of Error

  • Reaction time error if using manual timing (avoid by using light gates).
  • Air resistance slowing the object slightly.
  • Height measurement inaccuracies.
  • The ball may not fall straight if released inconsistently.

 Improvements

  • Use light gates instead of a stopwatch.
  • Use a dense object (reduces air resistance effect).
  • Ensure the ball is released without any initial push.
  • Take multiple readings for each height and average them.

Example (Easy)

A ball is dropped from a height of \( 0.80\, \mathrm{m} \). The measured fall time is \( 0.40\, \mathrm{s} \). Calculate \( g \).

▶️ Answer / Explanation

\( g = \dfrac{2h}{t^2} = \dfrac{2(0.80)}{0.40^2} = \dfrac{1.6}{0.16} = 10\, \mathrm{m\,s^{-2}} \)

Close to accepted value \( 9.8\, \mathrm{m\,s^{-2}} \).

Example (Medium)

Two light gates measure velocities \( v_1 = 1.2\, \mathrm{m\,s^{-1}} \) and \( v_2 = 3.4\, \mathrm{m\,s^{-1}} \). The time between gates is \( 0.25\, \mathrm{s} \). Find \( g \).

▶️ Answer / Explanation

\( g = \dfrac{v_2 – v_1}{t} = \dfrac{3.4 – 1.2}{0.25} = \dfrac{2.2}{0.25} = 8.8\, \mathrm{m\,s^{-2}} \)

Example (Hard)

A student collects height–time data and plots a graph of \( h \) vs \( t^2 \). The gradient of the line is \( 4.92\, \mathrm{m\,s^{-2}} \). Find the value of \( g \).

▶️ Answer / Explanation

The gradient of the graph = \( \dfrac{1}{2}g \).

\( g = 2 \times 4.92 = 9.84\, \mathrm{m\,s^{-2}} \)

This is very close to the accepted value.

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