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

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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

(i) 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 calculate the change in gravitational potential energy (GPE), use the formula:

GPE=mgh\text{GPE} = mgh

Where:

  • m = mass of the rover = 1100 kg
  • g = gravitational field strength = 3.7 N/kg
  • h = change in height = 1.80 km - 1.60 km = 0.20 km = 200 m

Substituting the values:

GPE=1100×3.7×200\text{GPE} = 1100 \times 3.7 \times 200

Calculating:

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

Expressing the answer to 2 significant figures:

change in gravitational potential energy=8.1×105 J\text{change in gravitational potential energy} = 8.1 \times 10^5 \text{ J}

Step 2

(ii) 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

To calculate the change in kinetic energy (KE), use the formula:

KE=12mv2\text{KE} = \frac{1}{2}mv^2

Where:

  • m = mass of the rover = 1100 kg
  • v = initial speed at position P = 88 m/s

Calculating the KE at position P:

KEP=12×1100×(88)2\text{KE}_P = \frac{1}{2} \times 1100 \times (88)^2

Calculating:

KEP=0.5×1100×7744=4268800 J\text{KE}_P = 0.5 \times 1100 \times 7744 = 4268800 \text{ J}

The final speed at position Q is 0 m/s, so:

KEQ=0\text{KE}_Q = 0

Now, the change in kinetic energy:

change in KE=KEQKEP\text{change in KE} = \text{KE}_Q - \text{KE}_P

Therefore:

change in KE=04268800=4268800 J\text{change in KE} = 0 - 4268800 = -4268800 \text{ J}

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