Average force in totally inelastic collision

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Homework Help Overview

The discussion revolves around a problem involving a totally inelastic collision between two bodies, A and B. The participants are tasked with determining the average force vectors acting on each body during the collision, given specific masses and velocities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore various attempts to calculate the average force using momentum equations, but encounter issues with their calculations. Some participants suggest breaking down the problem into components and focusing on individual bodies rather than the combined system.

Discussion Status

There is ongoing exploration of the problem, with some participants providing guidance on how to approach the calculations by emphasizing the importance of considering individual momentum changes. Multiple interpretations of the problem are being discussed, and no consensus has been reached yet.

Contextual Notes

Participants note that the question provides more information than necessary and question the approach of using combined momentum for both bodies. There is a focus on impulse and its vector components in the context of the collision.

uk9999
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Homework Statement



Two bodies A and B collide in a totally inelastic collision. Using the relevant equations and given that the mass of body A is 1200kg and the collision lasts for 0.2s, determine the average force vectors acting on each body during the collision.

Homework Equations



vA=5i+3j m/s
vB=-i+4j m/s
mA=(3/2)mB
Common velocity after collision v=2.6i+3.4j m/s


The Attempt at a Solution


attempt 1
Ft = (mA+mB)v - (mAvA+mBvB)
calculation checked and rechecked... leads to wrong answer
attempt 2
Ft = (mA+mB)v - (mAvA-mBvB)
calculation checked and rechecked... leads to wrong answer
attempt 3
FAt=0.5(mA+mB)v - mAvA
FBt=0.5(mA+mB)v - mBvB
calculation checked and rechecked... leads to wrong answer

Answer is stated as "FA= -FB= (-14400i+2400j) N in the mark scheme
 
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Hi there uk9999. Well i think you need to rethink the way you looking at the question. First let's have a look back at how we define impulse and momentum:

[tex]\textbf{I} = \textbf{F}t = m\textbf{v} - m\textbf{u}[/tex]

now I, F, v and u can all be vectors as we have in this question, but in order to solve a problem involving vectors we should brake it down into its components, in this scenario in 2D so well equate the x comonets and the y comonents seperatly.

Now in actual fact the question gives you far more information than you actually need, and it is not nessesary to consider both bodies, as the question tells us the initial and final velocity of both, and Impulse is described for a single body, so you can take you pick as to which one you want to use. so looking back at the first equation we need to modify this so that we can consider componets:

[tex]I_x = F_x t = mv_x - mu_x[/tex]
[tex]I_y = F_y t = mv_y - mu_y[/tex]

now i decided to use body A, no reason just was the first one I came to :D, so if wee input those values into you componentised Impluse equations we get:

[tex]I_x = F_x t = 1200(2.6) - 1200(5)[/tex]
[tex]I_y = F_y t = 1200(3.4) - 1200(3)[/tex]

now have a go from there, I think there is enough info there for you to finish off the question :D have fun
 
uk9999 said:

The Attempt at a Solution


attempt 1
Ft = (mA+mB)v - (mAvA+mBvB)
calculation checked and rechecked... leads to wrong answer
attempt 2
Ft = (mA+mB)v - (mAvA-mBvB)
calculation checked and rechecked... leads to wrong answer
attempt 3
FAt=0.5(mA+mB)v - mAvA
FBt=0.5(mA+mB)v - mBvB
calculation checked and rechecked... leads to wrong answer
In all of your attempts you have tried to use the combined momentum of both masses. Don't!

To find the average force on mass A, you need the change in momentum of mass A. Similarly, to find the average force on mass B, you need the change in momentum of mass B. (Of course, you don't have to calculate the force twice. Use Newton's 3rd law.)
 
Ah thank you knew I was doing something wrong :blushing:
 

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