Conservation of momentum of two cars

AI Thread Summary
In a collision between two cars, such as a Honda Civic and a Toyota Camry, momentum is conserved, but total energy is not necessarily conserved due to energy transformations. Real-life collisions are typically inelastic, resulting in energy being converted into other forms, such as thermal energy and deformation, rather than being conserved as kinetic energy. The discussion highlights the distinction between conservation of momentum and conservation of energy, emphasizing that momentum can be conserved while energy is dissipated. The effectiveness of safety features like airbags also illustrates the conversion of kinetic energy during a collision. Understanding these principles is crucial for accurately analyzing real-world car collisions.
Stipe
Messages
3
Reaction score
0
For my physics assignment i have a question on conservation of momentum that i am completely unsure of and have heard mixed answers to from teachers and fellow students. Here is the question and can ai have nice detailed answers please. Thanks in advance.

Section A - conservation of momentum


In a real life collision between your two chosen cars ( Honda civic and Toyota Camry (sp?) ) on a bitumen road on a dry day, is it reasonable to assume that the total energy and momentum of the cards is conserved? Explain your answer.
 
Physics news on Phys.org
In the title you address this as an issue of
"Conservation of momentum"
whereas in the bold face at the bottom you say:
"is it reasonable to assume that the total energy and momentum of the cards is conserved?"

...conserving momentum only is very different than
conserving (energy and momentum in combination).

Obviously momentum can convert to other kinds of
energy than kinetic energy e.g. thermal, stress, ...

Clearly you don't see elastic collisions between
cars other than bumper cars... and as for inelastic collisions,
well, usually on the highway I see more CRUNCH than
BOING/BOUNCE...

And how does an air-bag work again?...

So think of all the kinds of resultant MOMENTUM, but
then think of all the energy dissipated in other ways,
the conversion of kinetic energy to other kinds, etc...

What do you think?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top