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Hailstones are small balls of ice - AQA - GCSE Physics - Question 4 - 2022 - Paper 1
Question 4
Hailstones are small balls of ice. Hailstones form in clouds and fall to the ground. Figure 7 shows different-sized hailstones. A hailstone falls from a cloud and a... show full transcript
Worked Solution & Example Answer:Hailstones are small balls of ice - AQA - GCSE Physics - Question 4 - 2022 - Paper 1
Step 1
Step 2
2. Explain why the hailstone reaches terminal velocity.
Answer
A hailstone reaches terminal velocity when the force of air resistance becomes equal to its weight. Initially, as the hailstone falls, its speed increases, leading to increased air resistance. Eventually, the air resistance increases until it balances the weight of the hailstone, resulting in a net force of zero. When this balance is achieved, the hailstone falls at a constant speed, termed the terminal velocity.
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3. Why does terminal velocity increase with mass?
Answer
Terminal velocity increases with mass because as the mass of a hailstone increases, its weight increases. Since air resistance is dependent on the speed and cross-sectional area, a heavier hailstone requires a higher speed to reach a point where the weight is balanced by air resistance, thus increasing its terminal velocity.
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4. Explain the difference in the maximum kinetic energy of a hailstone with a mass of 10 g and a hailstone with a mass of 20 g.
Answer
The kinetic energy ( E_k ext{)} of an object depends on its mass and velocity, expressed by the formula: egin{equation} E_k = rac{1}{2} mv^2
ewline ext{As the mass increases, kinetic energy increases because it is directly proportional to the mass.}
A hailstone with a mass of 20 g will thus have a higher maximum kinetic energy compared to a 10 g hailstone, given that both are falling at the same terminal velocity. In fact, if the 20 g hailstone is moving at the same speed, its kinetic energy will be double that of the 10 g hailstone.
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6. Determine the average force on the hailstone as it hit the ground.
Answer
To calculate the average force on the hailstone as it hit the ground, we can use the impulse-momentum principle:
egin{equation} F_{avg} = rac{ ext{Change in momentum}}{ ext{time}}
ewline ext{First, we calculate the change in momentum.}
ewline ext{Initial momentum} = m imes v = 0.0185 ext{ kg} imes 25 ext{ m/s} = 0.4625 ext{ kg m/s}
ewline ext{Final momentum} = 0 ext{ (after hitting the ground)}
ewline ext{Change in momentum} = 0 - 0.4625 = -0.4625 ext{ kg m/s}
ewline ext{Now, substituting into the force equation:}
ewline F_{avg} = rac{-0.4625 ext{ kg m/s}}{0.060 ext{ s}} = 7.708 ext{ N}
ewline Thus, the average force exerted on the hailstone as it hit the ground is approximately 7.71 N.