• Support PF! Buy your school textbooks, materials and every day products Here!

Terminal Velocity & Acceleration Question

  • Thread starter Molly1235
  • Start date
77
4
1. Homework Statement

Hi there. I have previously asked questions on here and found it very helpful so would be grateful if someone could maybe help me again? I am currently studying for my Physics P2 GCSE unit and came across this question on the topic of Terminal Velocity.

"Two identical crates are dropped from an aircraft. One is empty and the other is full. Which will reach the ground first?"


2. Homework Equations

Force = Mass x Acceleration


3. The Attempt at a Solution

Here is the answer I have written up. I would just like some clarification as to whether I am right, along the right lines, or completely wrong please.

The heavier crate will reach the ground first. This is because there is more air resistance acting on the empty crate than the full one. After being dropped, both crates will accelerate, but because the heavier crate has more mass the downwards force on it (weight) will be greater than that on the lighter crate (f=m x a) even though both objects will accelerate by the same amount due to the pull of gravity. This means that it will take a shorter time for air resistance to balance with the weight of the lighter crate than the heavier crate, stopping it from accelerating before the heavier. This gives it a slower terminal velocity, therefore meaning that the heavier crate, with a greater terminal velocity, will reach the ground first.

I would be really grateful for some clarification on this, as I have been teaching myself practically the whole of the syllabus due to having a less than reliable teacher, so have no other way of knowing if what I have learnt is correct.

Many thanks,
Molly.
 

Answers and Replies

6,049
390
You could consider that air resistance (drag) is proportional to the square of velocity. Then write down the equation of forces.
 
77
4
Thanks for the reply - I'm pretty sure this is not mentioned in my syllabus - maybe this is more A Level stuff? Is everything I have written down so far correct though?
 
6,049
390
You state that that there will be more air resistance acting on the lighter crate. That is incorrect. The forces will be equal. Unlike the force of gravity, they do NOT depend on the mass of the crate. The force of gravity DOES depend on the mass. The what is the net result?
 
77
4
Ah, I thought the force of gravity would be the same on both crates? As the gravitational field strength on earth is approximately 10N (according to my textbook)? But then I see what you mean as because force = mass x gravity, gravity must = force/mass. So surely these contradict each other...oh man, I'm so confused.
 
77
4
Will everything I have said be correct if I take out the bit that says there is more air resistance on the empty crate? I think I understand what you're saying...the air resistance itself is equal on both, but has more of an effect on the lighter crate, right?
 
6,049
390
Ah, I thought the force of gravity would be the same on both crates? As the gravitational field strength on earth is approximately 10N (according to my textbook)? But then I see what you mean as because force = mass x gravity, gravity must = force/mass. So surely these contradict each other...oh man, I'm so confused.
The acceleration by gravity alone - when there are no other forces - is independent of the mass. Because the force of gravity is proportional to the mass: ma = mg, so a = g.

But if you have some other force, you have ma = mg + F. Now think about two objects with masses m and M, M bigger than m, which are falling because of gravity and there is resistance that does not depend on the mass. What are their accelerations?
 
CAF123
Gold Member
2,888
88
In the absence of air resistance, both crates will fall at the acceleration due to gravity which is about 10m/s2. But because we have air resistance in your question, the heavier will fall first, as you pointed out. The contribution of air resistance is the same on both crates, as voko said, it is just the effect that is different.
 
77
4
In the absence of air resistance, both crates will fall at the acceleration due to gravity which is about 10m/s2. But because we have air resistance in your question, the heavier will fall first, as you pointed out. The contribution of air resistance is the same on both crates, as voko said, it is just the effect that is different.
Ok, I think I understand now. So the force of air resistance is equal on both crates, but because the lighter one has less mass the forces are balance out more quickly than on the heavier crate, causing it to stop accelerating and reach it's terminal velocity quicker. This therefore means that the light crate's terminal velocity will be slower than that of the heavy crate. Correct?
 
77
4
The acceleration by gravity alone - when there are no other forces - is independent of the mass. Because the force of gravity is proportional to the mass: ma = mg, so a = g.

But if you have some other force, you have ma = mg + F. Now think about two objects with masses m and M, M bigger than m, which are falling because of gravity and there is resistance that does not depend on the mass. What are their accelerations?
Oh goodness, this goes way beyond me! Um, surely the acceleration of M is greater than that of m if ma = mg + F and Ma = Mg + F because g and F are the constants? Or I could be totally along the wrong lines here...
 
I like Serena
Homework Helper
6,553
176
Welcome to PF, Molly!

Ok, I think I understand now. So the force of air resistance is equal on both crates, but because the lighter one has less mass the forces are balance out more quickly than on the heavier crate, causing it to stop accelerating and reach it's terminal velocity quicker. This therefore means that the light crate's terminal velocity will be slower than that of the heavy crate. Correct?
Correct! :smile:
 
CAF123
Gold Member
2,888
88
@voko
But if you have some other force, you have ma = mg + F.
Is this other force you are talking about the effect of air resistance? If so, would this not give [itex] mg - kv^2 = ma, [/itex] taking down as positive?
 
6,049
390
@voko


Is this other force you are talking about the effect of air resistance? If so, would this not give [itex] mg - kv^2 = ma, [/itex] taking down as positive?
That was what I proposed in my first post on the subject. From this, terminal velocity is easily determined. Since it is constant, we must have [itex]a = 0[/itex], so [itex]mg = kv^2[/itex] (in other words, the two forces balance each other), from where the formula for terminal velocity follows immediately, and it is easily seen that it depends on the mass.
 

Related Threads for: Terminal Velocity & Acceleration Question

Replies
11
Views
3K
Replies
4
Views
2K
Replies
9
Views
3K
  • Last Post
Replies
2
Views
589
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
1
Views
841
Replies
6
Views
1K
Top