Hailstones are small balls of ice - AQA - GCSE Physics - Question 4 - 2022 - Paper 1

As the hailstone falls, it accelerates until it reaches terminal velocity when the force of gravity equals the force of air resistance acting on it. At this point, there is no longer a resultant force, and the hailstone descends at a constant velocity.

Terminal velocity increases with mass because as mass increases, the weight of the hailstone also increases. This means that a heavier hailstone will have a greater gravitational force acting on it, which allows it to overcome air resistance more effectively, resulting in a higher terminal velocity.

The kinetic energy (KE) of an object is given by the formula:

$KE = \frac{1}{2} mv^2$

where m is the mass and v is the velocity. For a hailstone with a mass of 20 g, all else being equal, the maximum kinetic energy will be greater than that of a 10 g hailstone. This is due to the fact that as mass doubles, and if velocity remains constant, kinetic energy also doubles, indicating that the heavier hailstone can generate more kinetic energy during its fall.

The correct answer is: 1 N m. This is because 1 joule is defined as the work done when a force of 1 newton displaces an object by 1 meter in the direction of the force.

To find the average force acting on the hailstone, we can use the formula:

$F = m \cdot a$

where m is the mass and a is the acceleration. First, we need the mass in kg, which is 0.0185 kg for a hailstone returning at a terminal velocity of 25 m/s. The time taken to stop is 0.060 s.

We can find acceleration using the change of velocity:

$a = \frac{\Delta v}{t} = \frac{25 m/s}{0.060 s} = 416.67 m/s^2$

Now, substituting into the force equation:

$F = 0.0185 kg \cdot 416.67 m/s^2 ≈ 7.708 N$

Thus, the average force on the hailstone as it hit the ground is approximately 7.708 N.