Question

HELP!GIVING OUT MEDALS!!
Which o the following statements is true about a collision between two cars?
A.The total momentum after the collision is the same as the total momentum before the collision
B.The total momentum after the collision is greater than the total momentum after the collision is greater than the total momentum before the collision
C.The total momentum after the collision is less than the total momentum before the collision
D.The total momentum after the collision is zero

Answer 1

It is A, conservation of momentum

It is only A if it is the total linear momentum

Answer 2

If this was cue balls bouncing off each other, the answer would be A, because they would rebound with the same momentum they had a collision with

But since this is cars...and they don't bounce after collisions and have zero velocity then I think the answer for this is D) The total momentum after the collision is zero

Answer 3

Yes yes, if it is a elastic car collision, then it is D

Answer 4

This time I'm going with C...

C.The total momentum after the collision is less than the total momentum before the collision

An elastic collision is one in which no kinetic energy is lost in the collision...like Newton's cradle balls smacking each other.

And an inelastic collision is one where KE is lost in the collision. Two cars colliding would lose kinetic energy because of the metal crumpling and absorbing shock and converting that energy to heat. But the cars probably wouldn't come to a complete stop in the collision, so some of the energy is saved as momentum

Look at these equations , m1 is car one and m2 is car two :
KEf = kinetic energy final, KEi , is kinetic energy initial

Ratio of kinetic energies before and after collision:
KEf / KEi = m1 / (m1 + m2)

Fraction of KE lost in the collision:
KEi - KEf / KEi = m / (m1 + m2)

So the momentum going into the collision is the same minus the KE lost in the collision, which absorbs some of that momentum.

Answer 5

.-. didn't mean elastic in that way, I just made a new 'term' where the KE would be neutralized. Well, I guess C answers it in a more general state, since if momentum is 0 then it is of course less than the beginning momentum

Answer 6

it doesnt matter whether the collision is elastic or inelastic,, momentum is ALWAYS conserved in any collision,,,,provided EXTERNAL forces dont act on the system ,,

Answer 7

answer must be A

Answer 8

@LastDayWork

Answer 9

loss of KE doesnt mean loss of momentum,,,,,

Answer 10

Yea A is the correct answer..unless someone is considering the friction on tires :P

Answer 11

Linear momentum remains conserved even in perfectly inelastic collisions..

Answer 12

Ah yeah, I forgot one little key thing, yes, my first answer was correct, it is A

Answer 13

I was rereading the difference between KE and conservation of momentum....and you guys are right..it is answer A)

This goes against everything common sense, meaning KE = 1/2mv^2 and momentum = m*v , so it appears with mass and velocity being the only two inputs into what KE and momentum are, that it would make sense if you have less KE after a collision, you would also have less momentum.

But you don't, momentum is conserved regardless of the collision type since momentum is considered a vector and by not squaring velocity like you do when calculating KE, that keeps momentum a linear component and not exponential like KE would be. In other words, you can double velocity on KE and KE increases by a factor of four, whereas doubling velocity for momentum only doubles the momentum.

Answer 14

yeah ,,u are right man

Answer 15

its quite intuistic to believe that if KE reduces, Momentum also reduces ,, because velocity reduced,,,,,,

Answer 16

but ,, though velocity reduced......the reduced velocity is with a more MASSIVE BODY ,, so MV remains constant compensating it

Answer 17

in simple words FORCE= d(P)/dt = constant (means P doesnt change) ONLY when F is a constant ,,,,,,i.e when no force acts on the system , momentum remains intact ,,conserved

Answer 18

"This goes against everything common sense, meaning KE = 1/2mv^2 and momentum = m*v , so it appears with mass and velocity being the only two inputs into what KE and momentum are, that it would make sense if you have less KE after a collision, you would also have less momentum."

There's a subtle difference (as you already pointed out), while calculating total KE of the system we add the magnitude of KE of individual parts ; whereas total momentum may be (and generally is) less than the summation of magnitudes of individual parts.

In other words, the fact that magnitude of individual velocities decreases after collision DOESN'T imply that the magnitude of their (weighted) difference reduces too.