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Why is it that when cars lock together after colliding, the collision is inelastic? (1 Viewer)

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Re: Why is it that when cars lock together after colliding, the collision is inelasti

ALSO another question.
A father and his young daughter are skiing on ice. the father is heavier than the daughter. the father pushes his daughter forward and she slides away at a speed of 3m/s.
which statement is true about the father's velocity
It's a multiple choice question and the answer is : it is smaller than 3ms^-1 in the opposite direction as his daughter

Can someone explain why this is so?
Thanks again
 

esaitchkay

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Re: Why is it that when cars lock together after colliding, the collision is inelasti

1. When cars lock together after colliding, some of the original kinetic energy of the cars is now transformed to other types of energy, such as deformation, sound, heat etc.
Now, an elastic collision is when kinetic energy is conserved within a closed system and an inelastic collision is when kinetic energy is NOT conserved within the closed system (e.g., it is transformed into different types).
Therefore, cars locking together is an inelastic collision.
On a sidenote, note that the Law of Conservation of Energy (LOCOE) ALWAYS applies.

2. Similar to the LOCOE, the Law of Conservation of Momentum (LOCOM) ALWAYS applies.
This refers to the idea that initial momentum = final momentum in a closed system.
Let's consider a scenario where both daughter and father are stationary.
The initial momentum can be assumed to be 0 (p=mv; p=0m=0).
Now, since the initial momentum is 0, the final momentum must also be 0.
Now, if the father pushes the daughter, and causes her to move at 3m/s, there is momentum.
Let's say the father and daughter have a mass of m1 and m2, respectively.
If we consider p = mv (momentum = mass x velocity) then there is p=3m2 in the direction of the daughter. In order to have a net momentum of 0 (in order to follow LOCOM and have a final momentum of 0), the father must also experience a momentum of SAME magnitude/size in the OPPOSITE direction.
Showing this quantitatively,
-vm1 = 3m2 --> v=-3m2/m1 (where v = velocity of father)
but m1>m2 (father weighs more than daughter) therefore m2/m1 <1 and therefore v < -3
The negative simply represents direction (opposite to the daughter), whilst the velocity must be smaller than 3m/s.

Hope this helped!
 
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kev-kun

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Re: Why is it that when cars lock together after colliding, the collision is inelasti

Just adding to above (but not that great) There are quite a few reasons. But from the top of my head:
Cars are constructed with crumple zones in both the front and rear well, 'crumple'. By crumpling, the car is brought to rest over a longer time interval and so the forces are reduced (impulse), as time is increased. Between the crumple zones lies a strong central cell that protects the occupants in the event of a crash.

Say if it were to be elastic, as it crashes, it would rebound after a certain of time and it would force the passenger backwards. Safety wise, the human body is fragile and after the elastic effect has given multiple oscillations, the spinal chord would be damaged in some way as the brain is moved back and forth due to the force of the collision.

This is generally why. There are also other safety measures like airbags, but a collision shouldn't be elastic due to safety concerns on the most part.

For your second question think about the addition of vectors. Also, the father is heavier than the daughter, so by F=ma, he has more force due to his mass being greater.

Don't know if I explained well.... hoped it helped though.
 
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Re: Why is it that when cars lock together after colliding, the collision is inelasti

THANK YOU both

1. When cars lock together after colliding, some of the original kinetic energy of the cars is now transformed to other types of energy, such as deformation, sound, heat etc.
Now, an elastic collision is when kinetic energy is conserved within a closed system and an inelastic collision is when kinetic energy is NOT conserved within the closed system (e.g., it is transformed into different types).
Therefore, cars locking together is an inelastic collision.
On a sidenote, note that the Law of Conservation of Energy (LOCOE) ALWAYS applies.

2. Similar to the LOCOE, the Law of Conservation of Momentum (LOCOM) ALWAYS applies.
This refers to the idea that initial momentum = final momentum in a closed system.
Let's consider a scenario where both daughter and father are stationary.
The initial momentum can be assumed to be 0 (p=mv; p=0m=0).
Now, since the initial momentum is 0, the final momentum must also be 0.
Now, if the father pushes the daughter, and causes her to move at 3m/s, there is momentum.
Let's say the father and daughter have a mass of m1 and m2, respectively.
If we consider p = mv (momentum = mass x velocity) then there is p=3m2 in the direction of the daughter. In order to have a net momentum of 0 (in order to follow LOCOM and have a final momentum of 0), the father must also experience a momentum of SAME magnitude/size in the OPPOSITE direction.
Showing this quantitatively,
-vm1 = 3m2 --> v=-3m2/m1 (where v = velocity of father)
but m1>m2 (father weighs more than daughter) therefore m2/m1 <1 and therefore v < -3
The negative simply represents direction (opposite to the daughter), whilst the velocity must be smaller than 3m/s.

Hope this helped!

danke! But just one small thing.
Why is it that they must be stationary in this case? Is it okay to infer that from the nature of the question?
 

esaitchkay

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Re: Why is it that when cars lock together after colliding, the collision is inelasti

Edit: read poster below
 
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ProtoStar

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Re: Why is it that when cars lock together after colliding, the collision is inelasti

Actually, pretty sure they do need to be stationary. Not sure where you got the question from, but it should specify that. If we were talking about relative velocity (ie. the relative velocity of the daughter from the father's perspective is less than 6m/s), then it wouldn't. But we aren't. So they do have to be stationary. Also, skating on ice would make more sense than skiing.

It would also still work if we were only talking about the change in velocity that occurred. But again, we aren't.
 

esaitchkay

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Re: Why is it that when cars lock together after colliding, the collision is inelasti

Thanks for clearing that up :)
 

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