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Figure 17 shows a football kicked against a wall - Edexcel - GCSE Physics - Question 9 - 2023 - Paper 1

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Figure 17 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 f... show full transcript

Worked Solution & Example Answer:Figure 17 shows a football kicked against a wall - Edexcel - GCSE Physics - Question 9 - 2023 - Paper 1

Step 1

(i) Calculate the height at which the ball hits the wall.

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Answer

To calculate the height at which the ball hits the wall, we start with the equation for gravitational potential energy (GPE):

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

Given:

  • Change in gravitational potential energy, (\Delta GPE = 11 J)
  • Mass of the football, (m = 0.42 kg)
  • Gravitational field strength, (g = 10 N/kg)

We can rearrange the formula to solve for height, (\Delta h):

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

Substituting in the values:

Δh=11J0.42kg×10N/kg\Delta h = \frac{11 J}{0.42 kg \times 10 N/kg}

Calculating gives:

Δh=114.22.62m\Delta h = \frac{11}{4.2} \approx 2.62 m

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

Step 2

(ii) 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 use the kinetic energy formula:

KE=12×m×v2KE = \frac{1}{2} \times m \times v^2

Where:

  • Mass, (m = 0.42 kg)
  • Velocity, (v = 12 m/s)

Substituting in the values:

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

Calculating gives:

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

Therefore, the kinetic energy of the football at that velocity is 30.24 J.

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