Calculating Force of Colliding Objects: A Newbie's Guide

  • Thread starter Thread starter jorge.s
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
To determine the force upon collision between two objects, it's essential to understand that colliding objects do not possess force in the traditional sense; instead, they have momentum and kinetic energy. The momentum of each object is calculated as the product of mass and velocity, while kinetic energy is derived from the formula mv²/2. The total momentum before and after the collision is conserved, meaning the combined momentum remains constant. To find the impact force, one must consider the time duration of the collision, as force is defined as the change in momentum over time. Understanding these principles is crucial for accurately calculating the effects of the collision between Object A and Object B.
jorge.s
Messages
8
Reaction score
0
I'm new to this forum, and during the 10+ hours of googling I've done I've found links to this site over and over. Since I can't figure it out on my own, perhaps this awesome community can help.

My problem is as follows:

I have Object A and Object B, and I want to know their force upon collision.


Object A weights 2.7 grams and is traveling at 54 Meters per second. (194.4 kph)

It's force equals to:

0.1458 Newton

14.867462385218195 gram force
0.014867462385218196 Kg Force
-----

Object B weights 200 grams and is traveling at 44 m/s

It's force equals to:

8.8 Newton

897.3502674205769 Gram force
0.8973502674205769 Kg Force


How do I calculate this ? Do I just add their gram forces and that's it? Doesn't seem right to me...

This is where I get confused.

I want to find out how much force is generated when this two objects collide in exact opposite directions. ----><----


If I calculate their kinetic energy.. I don't really know how to convert it back to force ? I'm soo confused. :)

Object A Kinetic energy = 3.9366 Joules
Object B Kinetic energy = 193.6 Joules

Something needs to be multiplied...


Any explanation as to how to calculate would be amazing! thanks!
 
Physics news on Phys.org
Welcome to PF!

Hi jorge.s! Welcome to PF! :smile:
jorge.s said:
I have Object A and Object B, and I want to know their force upon collision.

It would help if you gave us the exact question.

(colliding objects do not actually have forces).

Object A weights 2.7 grams and is traveling at 54 Meters per second. (194.4 kph)

It's force equals to:

0.1458 Newton

This is the momentum of Object A, not the "force". So it's in ks m s-1, not Newtons. :wink:

Momentum = mv, KE = mv2/2.

I can't help you with the rest of the question without knowing exactly what the question is. :confused:
 
Hi, thanks!

You said " (colliding objects do not actually have forces). "

What do they have?

Mainly my question is:

I want to know what how much weight is applied to object B from Object A, as clearly. object B would win over object A and continue it's path.

That's all i want to know. How much weight, force, energy ( don't know what it would be called ) does object B receive upon impact from object A.
 
Hi, thanks!

You said " (colliding objects do not actually have forces). "

What do they have?

Mainly my question is:

I want to know what how much weight is applied to object B from Object A, as clearly. object B would win over object A and continue it's path.

That's all i want to know. How much weight, force, energy ( don't know what it would be called ) does object B receive upon impact from object A.
 
Hi jorge.s! :smile:

Colliding objects have momentum (mv) and kinetic energy (mv2/2).

Total momentum is always conserved in a collision (ie total momentum before = total momentum after).

This is another way of saying that the centre of mass of the two particles has the same velocity before as after.

Since Object B has roughly 10 times the mass of Object A, their centre of mass is moving at very nearly the same speed as B, and the two objects may continue together at that speed, with B pushing A, or A may bounce off at high speed … it depends on whether energy is lost in the collision.
 
right, i can deduce that. but the exact numbers. the raw data, how much is it ?
 
As I said before, what is the exact question?

(i need to know how elastic the collision is, ie how much energy is lost)
 
We get this kind of question repeatedly. You cannot tell how much force is involved in a collision just from the mass and speeds of the objects- even if you know "how elastic the collision is". When two objects collide each has a change in momementum. Force is "change in momentum per unit time". How much force is involved also depends on the time required for the collision.
 
so there is no standard calculation to find out how much is exerted upon object b when object b tries to move object a that is moving at 54m/s and weights 2.7 grams...

if i understand this right. object b would have to apply 14 grams to stop object a.

All I'm missing is how much extra little bit of force does it need to then move it back the way it came at 44m/s.
 
  • #10
jorge.s said:
if i understand this right. object b would have to apply 14 grams to stop object a.

I'm sorry, but this makes no sense.

Force is not measured in grams. Nor is energy. Nor is momentum.

Where did you get it from? :confused:
 
  • #11
  • #13
ah, that's an F = ma (force = mass times acceleration) calculator …

it only works when a force is applied over a known time or distance …

it does not apply to an "instantaneous" event such as a collision. :wink:
 
  • #14
tiny tim, pardon my imprudence, but do you just go to each thread and say everything that is not? and don't provide any actual helpful information ? Or correct those who are wrong .. Is that what you're good at ?
 
  • #15
jorge.s said:
tiny tim, pardon my imprudence, but do you just go to each thread and say everything that is not? and don't provide any actual helpful information ? Or correct those who are wrong .. Is that what you're good at ?

jorge, I'm trying to help you …

you have somehow got the wrong idea about collisions, and it's best to tell you so

also, i provided plenty of "actual helpful information" in my post #5 :smile:

if you want more information, just ask​
 
Back
Top