In a school experiment, a particle, of mass m kilograms, is released from rest at a point h metres above the ground. At the instant it reaches the ground, the particle has velocity v m s -1 Show that $$v = \\sqrt{2gh}$$ A student correctly used h = 18 and measured v as 20. The student's teacher claims that the machine measuring the velocity must have been faulty. Determine if the teacher's claim is correct. Fully justify your answer.

A-Level AQA Maths Pure Modelling involving Numerical Methods: In a school experiment, a partic

In a school experiment, a particle, of mass m kilograms, is released from rest at a point h metres above the ground - AQA - A-Level Maths Pure - Question 13 - 2019 - Paper 2

Question 13

In a school experiment, a particle, of mass m kilograms, is released from rest at a point h metres above the ground. At the instant it reaches the ground, the parti... show full transcript

Worked Solution & Example Answer:In a school experiment, a particle, of mass m kilograms, is released from rest at a point h metres above the ground - AQA - A-Level Maths Pure - Question 13 - 2019 - Paper 2

Step 1

Show that $v = \sqrt{2gh}$

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Answer

To derive the formula for the velocity of the particle when it reaches the ground, we can use the principles of energy conservation.

Identify Initial and Final States: The particle is released from rest, so its initial velocity $u = 0$ m/s. As it falls, it converts potential energy into kinetic energy.
Write the Energy Conservation Equation: The potential energy (PE) at height $h$ is given by:

$PE = mgh$

where $m$ is the mass of the particle and $g$ is the acceleration due to gravity.

The kinetic energy (KE) when it reaches the ground is:

$KE = \frac{1}{2} mv^2$
Set PE equal to KE: At the instant it reaches the ground, all potential energy is converted into kinetic energy:

$mgh = \frac{1}{2} mv^2$
Simplifying: Cancel the mass $m$ from both sides (assuming $m \neq 0$ ):

$gh = \frac{1}{2} v^2$
Rearranging for v: Multiply both sides by 2:

$2gh = v^2$

Taking the square root gives:

$v = \sqrt{2gh}$

Thus, the desired formula is shown.

Step 2

Determine if the teacher's claim is correct.

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Answer

Given:

Height, $h = 18$ m
Measured velocity, $v = 20$ m/s

Calculate the Expected Velocity: Using the derived formula:

$v = \sqrt{2gh}$

Substituting in the values:

$v = \sqrt{2 \times 9.8 \times 18}$

Calculate:

$v \approx \sqrt{352.8} \approx 18.8 \, \text{m/s}$
Comparison of Results:

The measured velocity of 20 m/s is greater than the expected velocity of approximately 18.8 m/s. Since the measured value is significantly higher, it suggests a discrepancy.
Justification of Teacher's Claim: The student’s measurement exceeds the physical limits set by the potential energy conversion, validating the teacher’s assertion that the measuring device might be faulty. Thus, the conclusion is that the machine measuring the velocity is indeed faulty.

In a school experiment, a particle, of mass m kilograms, is released from rest at a point h metres above the ground - AQA - A-Level Maths Pure - Question 13 - 2019 - Paper 2

Worked Solution & Example Answer:In a school experiment, a particle, of mass m kilograms, is released from rest at a point h metres above the ground - AQA - A-Level Maths Pure - Question 13 - 2019 - Paper 2

Show that $v = \sqrt{2gh}$

Determine if the teacher's claim is correct.

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