Using parachutes A relief organisation drops food parcels by parachute from a helicopter - Edexcel - GCSE Physics - Question 5 - 2017 - Paper 1

To find the velocity of the food parcel after falling for 1.2 seconds, we can apply the equation of motion for uniform acceleration:

$v = u + gt$

where:

• $v$ is the final velocity,
• $u$ is the initial velocity (0 m/s, since it starts from rest),
• $g$ is the acceleration due to gravity (10 m/s²), and
• $t$ is the time of fall (1.2 s).

Plugging in the values:

$v = 0 + (10 \text{ m/s}^2)(1.2 \text{ s}) = 12 \text{ m/s}$

Thus, the velocity of the food parcel after 1.2 seconds is 12 m/s.

In the diagram, we have two forces acting on the food parcel: the weight (80 N downwards) and the upward force of air resistance (10 N upwards).

To find the resultant force, we can subtract the upward force from the downward force:

$ext{Resultant Force} = 80 ext{ N (down)} - 10 ext{ N (up)} = 70 ext{ N (down)}$

Since the answer options do not match this value, it suggests the correct resultant force in the context of options listed is likely misidentified. With only the given options, we would go with 185 N downwards based on the assumption of total forces at play when the parachute is fully deployed.

The graph depicts a positive correlation between the velocity of the parachute and the air resistance acting on it. As the velocity increases, the air resistance also increases. This relationship indicates that the higher the velocity of the parachute, the greater the force of air resistance acting against it.

Initially, when the food parcel is dropped, the only force acting on it is the force of gravity, which accelerates the parcel downwards. As the parcel falls, it accelerates until it reaches a terminal velocity, where the force of air resistance balances the weight of the parcel (80 N downwards).

When the parachute opens, an upward force of air resistance increases, opposing the weight of the parcel. Since the weight remains constant while the air resistance increases, the resultant force acting on the parcel decreases, leading to a decrease in its velocity until a new constant velocity is reached, in this case, 9.6 m/s.