Question

I'm confused... when should I use the momentum conservation law and when should I use the energy conservation law?

Answer 1

in elastic collision both momentum and kinetic energy ...but in inelastic collision only momentum is conserved

Answer 2

but how to solve certain problems effectively

Answer 3

Well.. of course.. isn't the definition of inelastic collision a sort of collision where energy isn't conserved? But what I want to know is.. elastic ones.

Answer 4

yes some of the energy is converts into heat ,light or sound

Answer 5

and... what about the elastic collisions?

Answer 6

@souvik

Answer 7

in elastic collision energy and linear momentum are conserved...you should follow some examples..

Answer 8

i'm curious about what I should choose when Im solving problems...generally.

Answer 9

and...there are times when you can't be so sure that it really is an inelastic one. Sometimes it isn't so obvious.

Answer 10

Your replies are quite vague! Please tell me clearly what are your problems. Also if possible try to show me a numerical problem in which u are facing problems. It is always easier to clear concepts using numericals.

Answer 11

This is for P VS E

Answer 12

So.. let's compare the momentum of the two balls blasted in total and the p of the remaining gun. and then the energy..

Answer 13

@Diwakar

Answer 14

I am thinking about the mini gun problem. So, we are to compare the momentums and energies of the two balls and the gun. The masses of the two balls is m and that of gun is 2m. Gun always shoots the bullet with a FIXED velocity with respect to ITSELF. I am assuming that that fixed velocity is "v". We conserve momentum in the frame of ground and not that of the gun. So, initially, the gun was at rest and it fired mass "m" with velocity"v" wrt to itself. The velocity wrt to ground will also be "v" (as the gun is at rest)[read about relative velocity if u have problem understanding it].

Conservation of momentum.

Initially momentum = pi=0
Final velocity of bullet = vb=v
Let final velocity of [gun+one more bullet of mass"m" inside it] = vg

pf=m*vb+3m*vg = pi=0

This gives vg=-v/3

So the gun + one more ball will move backwards with speed v/3
Now we fire the second ball again with speed v wrt gun. This time speed of bullet wrt ground will not be "v" but it will be v-v/3 = 2v/3

If u don't get this concept read about RELATIVE VELOCITY. It must be in ur text.

Now for conservation of momentum.

pi=-(3m)*(v/3)=-mv

ball of mass "m" is thrown with speed = vb=2v/3
Let gun (maass =2m) recoils with speed =vg

pf=2m*vg+2mv/3 =pi=-mv

This gives vg=-5v/6

Now we summarize:

First ball (mass m) was released with speed = v
Second ball (mass m) was released with speed=2v/3
Gun (mass=2m) finally had speed =5v/6

Now u can find K.E and momentum of each part using their respective formulas.
K.E.=1/2mv\(^{2}\)
p=mv

Answer 15

wow thanks

Answer 16

and... i don't understand how you can know whether it's an elastic collsion or not in general problems on collisions of two objects. It's too...vague.

Answer 17

Use conservation of energy whenever possible as in Classical Mechanics, it will generally be conserved. But in questions involving bullet (or any other object) getting embed into another object use conservation of momentum as there may be energy dissipated in the form of the heat and energy conservation may not be valid.

Answer 18

thanks