Anyone please lend me a hand?

Question

Rank,the following 3 collisions in terms of extent of damage that the car would experience.Explain your reasons for ranking the collisions as you did.
(a)A car going 10 m/s striking in identical car that was stationary.
(b)A car going 10 m/s running into an immovable concrete wall.
(c)A head-on collision between identical car, both going 10 m/s

Please explain your reasons in detail.Really appreciate your answer.Thank you so much!!=)

Answer 1

We have the same amount of energy in each case, however


(A)Kinetic Energy (KE) car1 >0 , KE car2 =0
50% damage. Energy is being shared equally (since the cars are identical)

(B) KE car1 >0 , KE Wall =0
100% damage. All energy is observed by the car since the wall is inelastic and immovable.
(C) KE car1 = KE car2 >0
100% damage since both cars had the same amount of kinetic energy.

In reality the structure of the vehicles and the geometry of collision would determine the extent of damage.

Answer 2

The 3 collisions ranked from most damage to least damage is B, C, A. This is because the concrete wall has no "crunch zone." Similar to a egg being thrown at a hanging sheet vs. an egg being thrown at a wall, there is more damage when it hits the wall. If there is something to absorb the impact, there is less damage. Cars are designed to crunch in a head on collision, so the people inside the car stay alive. When the car hits the wall, it will completely crunch and have a lot of damage. When the car hits another moving car, they will both crunch, but less than the car hitting the wall because the other car will absorb some of the impact. Finally, when a car hits another stationary car, it will crunch but there is less momentum, therefore, less damage.

Answer 3

B is the worst. All of the energy that your car holds will NOT disperse anywhere but from itself. (the brick wall cannot move)The more energy a vehicle is forced to get rid of, the more damage is done.

C is second worse if it is head-on. Remember, cars are more elastic than brick walls. Even though these two cars, in theory, have the same amount of kinetic energy, they will also "give" a little on impact. The reason that "10m/s + 10m/s = 20m/s" doesn't hold true is because both cars carry the same velocity and the same (I assume) mass. This means that they will (in a perfect situation) stop upon impact. If the car hits the concrete wall, it will also stop, EXCEPT there will be no elasticity with the concrete wall.

A is the least. There is simply less energy involved AND the stationary vehicle will move in same the direction as the moving vehicle once struck.

This analysis is based strictly on the damage of one vehicle.

Hope that helps,
Jon
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Vishal has got it only slightly wrong. There is no fault in his logic or calculation. However, he is forgetting the cars' elasticity relative to the concrete wall.

it has been shown that the secondary objects in collisions B and C exert THE SAME FORCE on the primary object. Given the cars' elasticity, a car-to-car collision would slow the rate of deceleration (as compared to a stationary wall), thus causing less damage.

Sorry Vishal. I'm not trying to be offensive : )
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Actually, I withdraw my last comments. Vishal is CORRECT. I am wrong. The car will stop at the same spot at the same rate both times (B and C) if the opposite energies are equal regardless of elasticity. Why? Because the primary object (car 1) will always be the same elasticity. Also, the secondary car will absorb the shock at the same rate as the first, so will the wall. Therefore, the elasticity of vehicle 2 is without consequence.

Give it to Vishal

Answer 4

I'd rank them, in order from most damage to least damage:
c,b,a. Here's why...

In case c, the relative velocity of the two cars is twice as big as in the other two cases-- it's equivalent to one car going 20 mph hitting a stationary car. So, it'll cause the most damage.

In case b, all the energy in the collision goes into damaging the car. The wall won't be moved or damaged (it's immovable). So, it'll be the second most damaging collision.

In case a, there are two factors that make it the least damaging. First, the second car isn't immovable. So, it'll be moved a bit by the collision. And, in this case, both cars will be damaged, not just one. So, the damage to the moving car will be less than if it hits a wall or a moving car.

Answer 5

Well.. the collision with the stationairy car would cause the least damage. You have a relative velocity of 10 m/s plus whatever give was in the stationary car.

I believe the immovable wall would cause the second least amount of damage. It should be the same thing as running into another car, but without the give of the other car crumpling.

With the other car going the opposite direction, your relative velocity is 20 m/s, quite an impact. The other car would give of course. I suppose its potentially possible that the give of the other car would offset the damage from its speed (making the stationary wall the most damaging)... but I kind of doubt it.

Answer 6

A would be the best situation...

B and C turn out to be the same.

think of it this way...
If you hit a wall you will hit it and bounce off all of the momentum would be put in to the wall and then transfered directly back to you.

If you hit the other car all of your energy will be transferd to the other car, and all of his energy would be transfered to you, thus you would have the same energy transfered to you as if you hit the wall.

some might counter my argument by saying that in the two car case the cars will absorb some of the impacte (that it will be an inelastic collition)

Now this is true I was assuming an eleastic collition. If it wrecks your car (i do not think your car will get too much damage at a 10 m/h collition so I think that my previous answer is still valid) meaning it bends the frame or crushes the front of your car.

But if you hit the wall then the ammount of energy that your car will absorb will be subtracted from the ammount that will be put in to making your car bounce back.

In the two car case the same is true on both sides... The ammount of energy that is put in to the deformation of the two cars will be subtracted from the total.
Each car will deform the same amount, thus the ammount of deformation energy will be the same as if you hit the wall.

I maintain that choises B and C are identicle (in theory). ( in reality some funky stuff can happen if the two cars get tangled together and do not bounce back at all, but I am not spending the time to think about that too much)


modification:

I just thought of this. take one of those big steal plates that they put in the road (assume it is indistructable), stand it up strait ... Now have the two cars hit it at the same time. Clearly this should be the same as hitting an imovabel wall, and it is the same as the two cars hitting each other.

Again ignoring the entaglement thing.

Answer 7

(c)A head-on collision between identical car, both going 10 m/s
-- both having the same speed going at each other
(b)A car going 10 m/s running into an immovable concrete wall.
-- going into the concrete wall which cant be moved
(a)A car going 10 m/s striking in identical car that was stationary.
-- even stationary but can be moved damage is less

Answer 8

I would say:

from largest to smallest:

c. Two bodies moving towards each other. 10 m/s of energy both ways needs to be expelled somehow.

b. The wall stops the car completly, expelling energy from the motion of 10 m/s to 0 m/s. Wall suffers little damage, car takes most.

a. The car hits the other car, but since it is a car, it will give. It's also designed to take an impact most likely, resulting in less damage overall.

Answer 9

Well, striking a stationary car helps reduce impact because the car will give way. So this would be first.

Striking another car going at the same rate, increasing the force of impact. This would be the worst.

Striking a concrete wall would not be that bad. This would be second.

Answer 10

C. both cars going 10 m/s increases the damage because instead of the force of just the car hitting an stationary object, now you have two forces hitting each other
b. the wall offers not absorbtion on has no give, and is flat
a. a car, depends on where it is hit, offers a little bit of give