Stunt people can fall from great heights

[RachelSSS]

Explain how momentum can be conserved in a collision when kinetic energy is not? Describe a situation in which it is desirable to have a highly inelastic collision. Explain why in this case. In motion pictures, stunt people can fall from great heights and land safely on giant airbags. Using the principles of momentum and impulse, explain how this is possible.

I have no idea. ANy ideas? please and thank you :)

 

[Shooting Star]

You have to give some ideas of your own, before we can give ours, because this is a homework help forum, not HW do forum.

 

[RachelSSS]

Well, i think that kinetic energie can't be destroy, but in an inelastic collision, the kinetic energy turns into different kinds of energy like internal energy.

i thoguht a car crash would be a good explain for a inelastic collision. Becasue they crumple up and stick together. Which happens in an inelastic colisn.

 

[Shooting Star]

That is correct. But what happened to the momentum conservation?

 

[RachelSSS]

well, I am not totally sure, but i think momentum is alwasy conserved, becasue it can't leave the objects the are in collision.

 

[Shooting Star]

Have you been taught the mathematics of momentun conservation and impulse etc.? From which level are you expected to give the answers?

 

[RachelSSS]

like... p = mv. impulse is the change in momentum so p=mv2-mv1. Grade 12 physics

 

[Shooting Star]

> Explain how momentum can be conserved in a collision when kinetic energy is not?

Try to write the answer in your own way.

 

[RachelSSS]

In a collision, momentum is conserved becasue the lose of one object is the gain of another object and vice versa. But kinetic energie is not conserved, but change into a different type of energie (thermal, internal, etc.)

 

[Shooting Star]

The way you should write it is: when there is no net external force on a system, the total linear momentum of the system is conserved.

What you have said is correct about the momentum. However, the KE may or may not be conserved. In perfectly elastic collisions, the KE is conserved.

> Describe a situation in which it is desirable to have a highly inelastic collision. Explain why in this case.

You have chosen the example of a car crash for this. While the incident itself is highly undesirable, why do you feel it is better for a crash to be inelastic rather than elastic?

 

[RachelSSS]

Ok, well... what about a hammer hitting a nail. An inelastic collision would transfer the energy into hitting the nail straight in

 

[Shooting Star]

Actually, no. The nail will suffer from more change of momentum if the hammer bounces back. Does this give you a clue for the car crash scenario?

 

[RachelSSS]

how is a car crash desirable tho. people will just get hurt and/or die. Well if the car crash is highly inelastic, would the cars would just bounce off each other kinda?, and if it was elastic they would get smashed even more

 

[Shooting Star]

The car crash was your idea. But if at all there is a crash between two cars or a car and a fixed object, it's better if it's inelastic. Suppose that the speed of a car is v, and the collision is perfectly elastic. If the car bounces back with speed v, that means that the total change in momentum is (final – initial) = mv-(-mv) = 2mv. But if the car does not bounce back, the change in momentum is simply mv - 0 = mv.

Also, if the time of impact is more, the impulsive force is reduced in magnitude, reducing damage to passengers. That's why cars have crumple zones.

> In motion pictures, stunt people can fall from great heights and land safely on giant airbags. Using the principles of momentum and impulse, explain how this is possible.

Now it’s your turn. You have enough clues now.

 

[RachelSSS]

Well for this one i was thinking that when the actor falls, he has a lot of momentum. so i wasnt sure how he was goign to stop and actually be safe. but went hits the air bag, there's an impulse with changes his momentum to 0?

 

[belliott4488]

Well for this one i was thinking that when the actor falls, he has a lot of momentum. so i wasnt sure how he was goign to stop and actually be safe. but went hits the air bag, there's an impulse with changes his momentum to 0?

You've almost got it. You're right: he has a lot of momentum, and then he has none, so there was a big change in momentum. On the other hand, what does your intuition tell you about landing on a hard sidewalk as opposed to landing on a nice big fluffy air bag? Which hurts more?

Remember, it takes a force to change momentum, and a big force to your face will hurt. How is force related to change in momentum? If change in momentum is fixed (the stunt guy is going to lose the same amount of momentum, no matter what he lands on), what else can vary in order to get a smaller force?

 

[Shooting Star]

(To RachelSSS)

If he hits concrete, that would also change his momnetum to zero, in an undesirable way. So, that can't be the answer. Read post #14 again, and think what's the difference between hitting a hard surface and an airbag.

 

[RachelSSS]

when your falling with a lot of momentum, you need a lot of force to stop you, the only forces acting on him would be the normal force and gravtity. The force of the air pushing him up after he hits the air bag.

 

[Shooting Star]

But both the times, that is, when he is hitting a hard surface or an airbag, the force acting upward stops him. What is the difference in the two cases?

 

[RachelSSS]

air is softer then concrete? I don't know. the airbag would the difference casue he'll bounce and not land on rock hard concrete

 

[Shooting Star]

Yes, it's softer. That means that the time taken to reduce the momentum to zero is more in the case of airbag. Change in momentum = F*t, and since the change in momentum is the same in both cases, F is less in the case of the air bag because t is more.

 

[RachelSSS]

so I am done that's it! the end! omg thanksssssssssssssssssssssss :):):):) oxoxoxox. 1 of 11 down. hahah thanks