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Comparing elastic and inelastic collision unknown mass

See below:

IMG_1608.jpeg


I’ve spent all day trying to make sense of these formulas because they’re completely new to me at the start of this question other than, f=ma, p=mv and change in velocity.

It’s the unknown mass value, that’s got me quite confused.

Part ii and iii, involve change in momentum and Impulse. I have discovered that they mean the same.
It seems redundant that it’s asking me the same thing but can someone confirm for me please?

Part iiii. Which has the greatest force applied. Assumption time is the same in both.

Since a=change in velocity/change in time.

I have applied a variable to time as 1 and mass as the same in both cases. As in physics, sources say this can be done.

Is this fine to do? Or, should I leave the solution with unknown variables?

Thank you
(edited 1 month ago)
Reply 1
Original post by KingRich
See below:
IMG_1608.jpeg
I’ve spent all day trying to make sense of these formulas because they’re completely new to me at the start of this question other than, f=ma, p=mv and change in velocity.
It’s the unknown mass value, that’s got me quite confused.
Part ii and iii, involve change in momentum and Impulse. I have discovered that they mean the same.
It seems redundant that it’s asking me the same thing but can someone confirm for me please?
Part iiii. Which has the greatest force applied. Assumption time is the same in both.
Since a=change in velocity/change in time.
I have applied a variable to time as 1 and mass as the same in both cases. As in physics, sources say this can be done.
Is this fine to do? Or, should I leave the solution with unknown variables?
Thank you

Id guess its a reasoning problem so what you have is about right. One typo, for iii) impulse = force*time and for iv) Id probably leave it as (m/Delta t) rather than assuming Delta t = 1, but thats minor.

Momentum (change in velocity) and impulse (force*time) are indeed related as you say because
v-u = at
or
mv-mu = (ma)t
Reply 2
Original post by mqb2766
Id guess its a reasoning problem so what you have is about right. One typo, for iii) impulse = force*time and for iv) Id probably leave it as (m/Delta t) rather than assuming Delta t = 1, but thats minor.
Momentum (change in velocity) and impulse (force*time) are indeed related as you say because
v-u = at
or
mv-mu = (ma)t

Ah, cool. So, I’ve seen impulse as f=mt and f=mΔt. Is this a common error or are they used interchangeably or in different cases?

See khan academy: https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/a/what-are-momentum-and-impulse
Reply 3
Original post by KingRich
Ah, cool. So, I’ve seen impulse as f=mt and f=mΔt. Is this a common error or are they used interchangeably or in different cases?
See khan academy: https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/a/what-are-momentum-and-impulse

Impulse is force*time or mass*acceleration*time or ... but neither of the forms you say. t is often really Delta t in the context of these problems.
Reply 4
Original post by mqb2766
Impulse is force*time or mass*acceleration*time or ... but neither of the forms you say. t is often really Delta t in the context of these problems.

Apologise. I meant J=fΔt… so it should be J=ft but it still relates the same way with change of momentum? J=Δp?
Reply 5
Original post by KingRich
Apologise. I meant J=fΔt… so it should be J=ft but it still relates the same way with change of momentum? J=Δp?

yes, its basically just saying newton 2
force = mass * acceleration
so if you integrate with respect to time on both sides you get
impulse = change in momentum
Reply 6
Original post by mqb2766
yes, its basically just saying newton 2
force = mass * acceleration
so if you integrate with respect to time on both sides you get
impulse = change in momentum

Okay, great. 🙂. So, see if I can understand this, if f=ma , and (a=Δ/t), f=m(Δv/t)
ft=mΔv, therefore, because (j=ft) and(Δp=mΔv)….
J =Δp. 🙂
If it helps, my physics teachers strictly say to use Δt rather than just "t", I think it otherwise tends to be misleading.
I dont think ive seen solutions use just "t", its quite uncommon from my experience.
Reply 8
Original post by mosaurlodon
If it helps, my physics teachers strictly say to use Δt rather than just "t", I think it otherwise tends to be misleading.
I dont think ive seen solutions use just "t", its quite uncommon from my experience.

It’s why I was confused. I saw physics websites use Δt and mathematics sites use t. Perhaps, I’ll keep it in the physics approach then, or just write ‘also seen as’ note.

I do know physicians and mathematicians always have different approaches in such cases. So, I’m unsure which way to approach it.
Original post by KingRich
It’s why I was confused. I saw physics websites use Δt and mathematics sites use t. Perhaps, I’ll keep it in the physics approach then, or just write ‘also seen as’ note.
I do know physicians and mathematicians always have different approaches in such cases. So, I’m unsure which way to approach it.

Firstly, to talk about the original question. I really think these were supposed to be "almost no thought" answers that are really just checking if you know the definitions.

So (i) is fine, but then:
(ii) momentum = mass x velocity. The mass doesn't change, so change in momentum = mass x change in velocity. So bigger velocity change means bigger change in momentum. So answer is same as (i).
(iii) impulse = change in momentum. So answer is same as (ii)
(iv) impulse = force x time (that the force is applied). So if the time is the same, a bigger impulse means a bigger force. So answer is same as (iii).

Regarding using tt v.s. δt\delta t or Δt\Delta t. In all 3 cases, you're talking about a measure of time. Fundamentally, you're talking about the time during which the force is applied; you can either talk directly about that time and say "the force is applied for a period of duration t", or you can think of it in terms of "the force starts being applied at time t and stops being applied at time t+δtt+\delta t, so the duration is δt\delta t".

But worrying about which of these is "the correct approach" or "mathematicians do it differently from physicists" is, I think, getting bogged down in minutae. At degree level (and frankly you're not really talking about "mathematicians" or "physicists" below degree level), neither of them really care. Do whatever suits your current task.

[To put it in terms of something you may be more familiar with: in projectile questions you can often decide whether "up" or "down" is the +ve y-coordinate. (i.e. acceleration due to gravity is "g" if down is +ve or "-g" if up is +ve). Physicists probably tend to take "+ve y is up" more than mathematicians. But which ever you choose is hardly a "different approach".]

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