GCSE Edexcel Physics Conservation of energy: Figure 15 shows a ‘Mars rover’ d

Figure 15 shows a ‘Mars rover’ descending to the surface of the planet Mars - Edexcel - GCSE Physics - Question 7 - 2022 - Paper 1

Question 7

Figure 15 shows a ‘Mars rover’ descending to the surface of the planet Mars. Mass of rover = 1100 kg Gravitational field strength on Mars = 3.7 N/kg (i) Calculate... show full transcript

Worked Solution & Example Answer:Figure 15 shows a ‘Mars rover’ descending to the surface of the planet Mars - Edexcel - GCSE Physics - Question 7 - 2022 - Paper 1

Step 1

Calculate the change in gravitational potential energy of the rover as it descends from position P to position Q.

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Answer

To find the change in gravitational potential energy (GPE), we can use the formula:

$\text{GPE} = mgh$

Where:

$m$ is the mass of the rover (1100 kg)
$g$ is the gravitational field strength (3.7 N/kg)
$h$ is the height difference between positions P and Q.

The height difference ( $h$ ) can be calculated as:

$h = 1.80 \text{ km} - 1.60 \text{ km} = 0.20 \text{ km} = 200 \text{ m}$

Now substituting the values into the GPE formula:

$\text{GPE} = 1100 \text{ kg} \times 3.7 \text{ N/kg} \times 200 \text{ m}$

Calculating this gives:

$\text{GPE} = 1100 \times 3.7 \times 200 = 814000 \text{J}$

To 2 significant figures, the change in gravitational potential energy is:

$\text{GPE} = 810000 \text{ J}$

Step 2

Use data from Figure 15 to calculate the change in kinetic energy of the rover as it descends from position P to position Q.

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Answer

The change in kinetic energy (KE) can be calculated using the kinetic energy formula:

$\text{KE} = \frac{1}{2} mv^2$

Where:

$m$ is the mass of the rover (1100 kg)
$v$ is the speed of the rover.

At position P, the speed is 88 m/s, and at position Q, the speed is 0 m/s. Therefore, the change in kinetic energy from P to Q will be:

$\Delta\text{KE} = \text{KE}_{Q} - \text{KE}_{P}$

Calculating the kinetic energy at position P:

$\text{KE}_{P} = \frac{1}{2} \times 1100 \text{ kg} \times (88 \text{ m/s})^2 = 0.5 \times 1100 \times 7744 = 4269200 \text{ J}$

And at position Q:

$\text{KE}_{Q} = \frac{1}{2} \times 1100 \text{ kg} \times (0 \text{ m/s})^2 = 0 ext{ J}$

Thus, the change in kinetic energy is:

$\Delta\text{KE} = 0 - 4269200 \text{ J} = -4269200 \text{ J}$

Therefore, the change in kinetic energy as it descends from position P to position Q is:

$\Delta\text{KE} = -4269200 \text{ J}$

Figure 15 shows a ‘Mars rover’ descending to the surface of the planet Mars - Edexcel - GCSE Physics - Question 7 - 2022 - Paper 1

Worked Solution & Example Answer:Figure 15 shows a ‘Mars rover’ descending to the surface of the planet Mars - Edexcel - GCSE Physics - Question 7 - 2022 - Paper 1

Calculate the change in gravitational potential energy of the rover as it descends from position P to position Q.

Use data from Figure 15 to calculate the change in kinetic energy of the rover as it descends from position P to position Q.

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