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Figure 10 shows a football kicked against a wall - Edexcel - GCSE Physics Combined Science - Question 6 - 2023 - Paper 1

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Figure 10 shows a football kicked against a wall. The football has a mass of 0.42 kg. (i) The football gains 11 J of gravitational potential energy as it moves fro... show full transcript

Worked Solution & Example Answer:Figure 10 shows a football kicked against a wall - Edexcel - GCSE Physics Combined Science - Question 6 - 2023 - Paper 1

Step 1

Calculate the height at which the ball hits the wall.

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Answer

To find the height, we will use the formula for change in gravitational potential energy:

ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta h

Given that:

  • The change in gravitational potential energy (ΔGPE) is 11 J,
  • The mass (m) of the football is 0.42 kg,
  • The gravitational field strength (g) is 10 N/kg.

Rearranging the formula to find height (Δh):

Δh=ΔGPEm×g\Delta h = \frac{\Delta GPE}{m \times g}

Substituting the known values:

Δh=11 J0.42 kg×10 N/kg=114.22.62 m\Delta h = \frac{11 \text{ J}}{0.42 \text{ kg} \times 10 \text{ N/kg}} = \frac{11}{4.2} \approx 2.62 \text{ m}

Thus, the height at which the ball hits the wall is approximately 2.62 m.

Step 2

Calculate the kinetic energy of the football when it is moving at a velocity of 12 m/s.

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Answer

To calculate the kinetic energy (KE) of the football, we will use the kinetic energy formula:

KE=12mv2KE = \frac{1}{2} m v^2

Where:

  • m is the mass of the football (0.42 kg),
  • v is the velocity (12 m/s).

Substituting the values into the formula:

KE=12×0.42 kg×(12 m/s)2KE = \frac{1}{2} \times 0.42 \text{ kg} \times (12 \text{ m/s})^2

Calculating the results:

KE=12×0.42×144=30.24 JKE = \frac{1}{2} \times 0.42 \times 144 = 30.24 \text{ J}

Thus, the kinetic energy of the football when moving at 12 m/s is 30.24 J.

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