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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 can rearrange the equation for gravitational potential energy:

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

Substituting the given values:

  • (m = 0.42 , \text{kg})
  • (g = 10 , \text{N/kg})
  • (\Delta GPE = 11 , \text{J})

We rearrange the formula to solve for h:

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

Substituting the values:

h=110.42×10=114.22.619 mh = \frac{11}{0.42 \times 10} = \frac{11}{4.2} \approx 2.619 \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), we use the formula:

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

Substituting the given values:

  • (m = 0.42 , \text{kg})
  • (v = 12 , \text{m/s})

Substituting into the equation:

KE=12×0.42×(12)2KE = \frac{1}{2} \times 0.42 \times (12)^2

Calculating the velocity squared:

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

Now, performing the multiplication:

KE=0.21×144=30.24JKE = 0.21 \times 144 = 30.24 \, \text{J}

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

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