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Figure 9 is a diagram showing a rocket that is sent into space to try and change the path of a small asteroid - Edexcel - GCSE Physics Combined Science - Question 6 - 2020 - Paper 1

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Figure 9 is a diagram showing a rocket that is sent into space to try and change the path of a small asteroid. (i) The rocket has a mass of 5.5 × 10^6 kg and is tra... show full transcript

Worked Solution & Example Answer:Figure 9 is a diagram showing a rocket that is sent into space to try and change the path of a small asteroid - Edexcel - GCSE Physics Combined Science - Question 6 - 2020 - Paper 1

Step 1

Calculate the momentum of the rocket

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Answer

To find the momentum (pp) of the rocket, we can use the formula:

p=m×vp = m \times v

Here, the mass (mm) is 5.5 × 10^6 kg and the velocity (vv) is 14 km/s. First, we need to convert the velocity to m/s:

14 km/s=14,000 m/s14 \text{ km/s} = 14,000 \text{ m/s}

Now, substituting the values:

p=5.5×106 kg×14,000 m/s=7.7×1010 kg m/sp = 5.5 \times 10^6 \text{ kg} \times 14,000 \text{ m/s} = 7.7 \times 10^{10} \text{ kg m/s}

Therefore, the correct answer is: A. 7.7 × 10^10 kg m/s.

Step 2

Calculate the speed of the asteroid

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Answer

To find the speed of the asteroid, we use the formula for momentum again:

p=m×vp = m \times v

We can rearrange this to solve for speed (vv):

v=pmv = \frac{p}{m}

Given that the momentum (pp) is 7.5 × 10^6 kg m/s and the mass (mm) is 8.0 × 10^6 kg:

v=7.5×106 kg m/s8.0×106 kg=0.9375extm/sv = \frac{7.5 \times 10^6 \text{ kg m/s}}{8.0 \times 10^6 \text{ kg}} = 0.9375 ext{ m/s}

Thus, the speed of the asteroid is approximately 0.94 m/s.

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