Safer Collision: Front-end Damage vs. No Damage - Impulse-Momentum Theorem

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A collision where the front end of a car is crumpled is considered safer than one with no damage, according to the Impulse-Momentum Theorem. This theorem indicates that a force applied over time reduces the impact force, thereby decreasing potential damage. In the first scenario, the energy is absorbed by the crumpled front end, which mitigates the force experienced by the occupants. Conversely, in a no-damage scenario, the cars bouncing off each other may result in a more abrupt change in momentum. Overall, the discussion emphasizes that energy absorption through deformation is crucial for safety in collisions.
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[SOLVED] What would you say...

Which is safer, a collision in which the front end of a car is completely crumpled, or a collision in which no damage is done to the car, but the two cars bounce backwards off one another. Use the Impulse-Momentum Theorem.

Hm. Well, the Impulse-Momentum Thm. says that a faorce exerted over some time, impulse, causes a change in momentum. So, by increasing the time over which a force must act to stop motion, the force can be reduced and the potential damage reduced. So, the scenario that is safer...logically I'm thinking the second scenario, but I'm not really thinking as a physicist. I'm just thinking that if there's no damage, it has to be safer.
 
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Would you rather be riding in a car with a spring supported bumper, or one with a rigid 'no - give' bumper? Or if you jump from a building, is it better to land with legs stiff and erect, or bend your knees as you hit the ground?
 
pebbles said:
Which is safer, a collision in which the front end of a car is completely crumpled, or a collision in which no damage is done to the car, but the two cars bounce backwards off one another. Use the Impulse-Momentum Theorem.

Use the Impulse-Momentum Theorem

Think on this line: In which case change in momentum is more? Then, assuming time of impact to be same, you can calculate in which case impulse is more.
 
now that you put it that way, it seems to make more sense. so, the first scenario would be safer because the energy is absorbed by the front end of the car...
 
pebbles said:
now that you put it that way, it seems to make more sense. so, the first scenario would be safer because the energy is absorbed by the front end of the car...

It's why those barrels at the entry to a highway exit ramp are filled with sand (to go 'splash' on the occasion someone's car rams them), rather than a nice big synthetic rubber bumper...
 
ok. thanks everyone. it makes sense now.
i greatly appreciate it!
:]
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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