Impulse when 2 bodies stick together?

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SUMMARY

The discussion centers on the calculation of impulse when two bodies collide and remain in contact post-collision. It is established that impulse can still be calculated using the change in momentum, even if the bodies stick together. The key formula I=F(Δt) applies, where Δt represents the time during which the forces act, regardless of whether the bodies rebound or remain stuck. The conversation also emphasizes that while kinetic energy may convert to internal energy, some kinetic energy can still be retained in the new combined object's velocity.

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agentmm
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Hello,

I was just wondering if an Impulse was exerted by one body on another if they collided but remained in total contact post-collision. The reason I ask is because my little I=F(delta)t indicates that there should be a change in time to calculate an impulse but in this case the 2 bodies are stuck together forever (so time change is 0 or infinite or something?)

The example I had in my head was GSP vs Matt Hughes: They rush each other and upon collision, morph into one body...can you calc. the Impulse of GSP on Hughes?

Thanks for your consideration.
 
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agentmm said:
Hello,

I was just wondering if an Impulse was exerted by one body on another if they collided but remained in total contact post-collision. The reason I ask is because my little I=F(delta)t indicates that there should be a change in time to calculate an impulse but in this case the 2 bodies are stuck together forever (so time change is 0 or infinite or something?)

The example I had in my head was GSP vs Matt Hughes: They rush each other and upon collision, morph into one body...can you calc. the Impulse of GSP on Hughes?

Thanks for your consideration.

There is still a delta t.
The delta t is the time during which the mutual forces act on the two bodies and cause their individual momenta to change. It doesn't matter if they stick or rebound; there was still a force; and it still acted over a (short) period of time.
 
The only difference here is that all the kinetic energy is internal after collision. The new object would heat up.
 
Thanks a lot for the responses...
Since I can't really use this "small" delta t, would it be acceptable to do a "mass * delta v" for each body (assuming we know the velocities of the bodies pre and post collision...)
Even though their masses have combined post impact, I guess you can still use them separately with the same final vel. but obviously different initial vel.
 
You can always find the impulse because it equals the change in momentum.
So if you have the momentum (from mass and velocity) of one body before the collision, and the momentum after; then the impulse was the change in the momentum.
Because the force on the other body was equal and opposite in direction (Newton 3) its change in momentum and impulse will be numerically the same as the first body.
As you don't know the Δt, you can't really calculate the magnitude of the force.
There may be ways of estimating Δt which would enable you to estimate the force.
 
LostConjugate said:
The only difference here is that all the kinetic energy is internal after collision. The new object would heat up.
You're right that much of the original KE will end up as internal energy, but not necessarily all. (That would be the case if the total momentum were zero.)

agentmm said:
Since I can't really use this "small" delta t, would it be acceptable to do a "mass * delta v" for each body (assuming we know the velocities of the bodies pre and post collision...)
Even though their masses have combined post impact, I guess you can still use them separately with the same final vel. but obviously different initial vel.
Absolutely. You should have no problem calculating the impulse on each body, since you have the change in momentum.
 
Doc Al said:
You're right that much of the original KE will end up as internal energy, but not necessarily all. (That would be the case if the total momentum were zero.)
.

Perhaps all the KE of the object with the lowest momentum, any KE difference would be maintained in the new velocity of the combined object.
 

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