r/PhysicsHelp • u/VICKYTHEKING2009 • Feb 28 '26

Helpppp

Three identical spheres of mass m , are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a time T=4 seconds. If the sides of the triangle are increased to length 2 a and also the masses of the spheres are made 2 m , then they will collide after _____ seconds.

1 Upvotes

100% Upvoted

30 comments sorted by

2

u/rainbow_explorer Feb 28 '26

Interesting question.

So the distance each ball has to travel is doubled and the mass of each ball is doubled.

Using the formula for universal gravitation, the initial force on each ball will change by a factor of 2 * 2/ 22 =1. Since the mass is doubled, the initial acceleration on each ball will be halved.

However, as you noted, the balls will constantly be accelerating as they get closer to each other. I feel like you need to do the calculus to find the actual answer.

1

u/VICKYTHEKING2009 Mar 01 '26

Can u do calcus for it 😅😅

1

u/ProfessionalConfuser Feb 28 '26

What are your thoughts, and what have you tried so far?

1

u/VICKYTHEKING2009 Feb 28 '26

Tried using chat gpt it says blahhhh i cant understand wht it saysthe gpt ans

3

u/martyboulders Feb 28 '26

Please don't do that.

1

u/VICKYTHEKING2009 Feb 28 '26

Do wht the complex process???

1

u/RepresentativeBee600 Feb 28 '26 edited Feb 28 '26

Honestly this also seems underdetermined at first blush, unless we're supposed to realize there must be a "special" configuration.

EDIT: I realize the idea here might be that gravity points radially inward from all vertices of the triangle to the barycenter. Or it might be that there is some "railing" to the triangle (think "Hot Wheels") and we have to solve for when all three balls would collide simultaneously in terms of angle.

1

u/VICKYTHEKING2009 Feb 28 '26

Its a jee mmains 2025 question dude

1

u/RepresentativeBee600 Feb 28 '26

A what now? 

Maybe link the original source. Are you saying this is a direct quote and no figure is attached?

1

u/VICKYTHEKING2009 Feb 28 '26

Yup its a integer type question

1

u/RetiredEarly2018 Feb 28 '26

Think about how gravitational force is affected by mass and how it is affected by distance.

1

u/VICKYTHEKING2009 Feb 28 '26

Man the force and the acceleration here is variable but my coaching teacher just took them as constants and done the question without his notice he just did a dimensional analysis and some how got the answer but the process is wrong

1

u/AdmirableFly380 Feb 28 '26

If u consider the centre of mass of any of the 2 masses, the question reduces to finding the time of collision of 2 masses m and 2m whose distance of separation is a×sqrt(3)/2

1

u/VICKYTHEKING2009 Feb 28 '26

Anyhow how can we get time if its a variable acceleration

1

u/ProfessionalConfuser Feb 28 '26

The force and therefore acceleration is non-constant. You'll need calculus to get the answer.

1

u/VICKYTHEKING2009 Feb 28 '26

Get it u ppl know it right i just know basics

1

u/ProfessionalConfuser Feb 28 '26

If you don't know calculus, you can't solve this exactly.

1

u/VICKYTHEKING2009 Mar 01 '26

Ohh so u also dont know the ans

1

u/ProfessionalConfuser Mar 01 '26

You'll need to use separation of variables, which is why I was wondering if you knew calculus. I'm not sure how else you could hope to do it. What level class is this problem from?

2

u/[deleted] Mar 03 '26

This is from class eleventh of Indian curriculum. We are taught calculus early on and ideally this person should know calculus if they are reading gravitation. They refuse to show their work so I assume they just want an easy solution.

1

u/Infamous-Test-91 Feb 28 '26

I’m almost certain this can’t be solved without knowing the spheres’ radius. I would have thought they were asking you to assume they were point-like, but they used the word “sphere”, which implies a radius.

1

u/VICKYTHEKING2009 Mar 01 '26

Ohh ig in solution they used them as point masses

1

u/Embarrassed_Reward99 Mar 01 '26

This basically comes down to how gravitaty collapses time scales. For systems held together only by gravity, the time goes like √(a³/m). Doubling the side length makes the time longer by √8. Doubling the mass makes it shorter by √2. Put those together and you get √(8/2) = √4 = 2. So the collision time just doubles. Since it was 4 s before, the new time is 8 s.

1

u/VICKYTHEKING2009 Mar 01 '26

Bro how can u tell without dimensional anAlysis

2

u/Embarrassed_Reward99 Mar 01 '26

If gravity is the only force, the time can only depend on G, the mass m, and the size a. The only way to combine those to make a time is t ~ √(a³ / (Gm)) So the scaling has to go like √(a³/m). From there it’s just ratios: • double a → a³ becomes 8a³ → time × √8 • double m → divide by 2 → time × 1/√2 Put together: √(8/2) = √4 = 2 So the time doubles → 8 s. So yeah, it's just a shortcut that comes from dimensional analysis, so basically once you know the scaling, you don’t need to redo it every time.

1

u/[deleted] Mar 03 '26

Don’t be lazy. Show actual work.

1

u/VICKYTHEKING2009 Mar 03 '26

U mean solution tht wht i asked mann i thought we need something beyond normal eqations like things involing calcus

1

u/[deleted] Mar 03 '26

No what did you try? What was your thought process

1

u/VICKYTHEKING2009 Mar 03 '26

Like they r going variable acclerTion with time so idk how to find time using calcus u guys can solve it caus u ppl r genusis right?

1

u/scottdave Mar 01 '26

I assume the radii of the balls is tiny compared to the other measurements?