COLLISION

A BALL STRIKES A WALL(ASSUME PERFECTLY ELASTIC COLLISION).THE BALL BOUNCES BACK WITH SAME VELOCITY(SAY V).
SO,WE SEE THAT KINETIC ENERGY REMAINS CONSERVED BUT THE MOMENTUM OF BALL CHANGES BY 2mV.

HOW CAN WE HAVE THAT MOMENTUM ALSO REMAINS CONSERVED AS WELL AS THE KINETIC ENERGY ALSO REMAINS CONSERVED??????

10 Answers

1
ith_power ·

didn't get your question

11
virang1 Jhaveri ·

See 2mV is the change in momentum but monemtum remain constant and conserved as it is MV before and after only the change is 2mv
Therefore both kinectic and momentum is conserved

11
Anirudh Narayanan ·

kinetic energy b4 collision = 1/2 m lvl2 = 1/2 m v.v

kinetic energy after collision = 1/2 m -v.-v = 1/2 m lvl2

1
JOHNCENA IS BACK ·

u din get my ques.

kinetic energy will remain conserved as wall has initial as well as final velocity=0.

so Ki=Kf=1mV2/2

right....

now the momentum shud also be conserved.

so,Pball i =Pball f+Pwall f

right...

this gives
mv=-mv+0(since final velocity of wall is 0)

now how will u account for this.......clearly momentum is not conserved..but in priniple it shud be conserved

1
JOHNCENA IS BACK ·

@ith_power ,anirudh,virang

what have u got to say bout #5

1
ith_power ·

@anirudh momentum is a VECTOR.
@john cena :

See that the ball applies a force on wall. but earth keeps the wall in place by applying a opposite force. so external impulsive force is present. So we cannot conserve momentum.

Ideally if m2 is too large than m1 then the force on m2 is too small to give it a perceptible acceleration. Now in this case the collision though assumued perfectly elastic is never realised in practice.
In fact the wall should have moved if earth were not there.

62
Lokesh Verma ·

to add to what ith power said

see john the thing is that we assume the wall to have infinite mass

so the change in velocity of the wall is there but we ignore that

24
eureka123 ·

Since wall applies an impulse on ball,so we cant conserve momentum along common normal for the ball.....but we can surely do that for wall+ball system becoz then impulses become internal...

@cena, ur post #5 is not correct either way...
taking ball +wall as system, eqn of momentum along common normal will be Pball i +Pwall i =Pball f+Pwall f

1
JOHNCENA IS BACK ·

@eureka

dude i know that but i din write Pwall i knowingly coz it is zero.

got it????

ith power thanx

24
eureka123 ·

pwall final is also zero ......
but when u want to conserve momentum of a system write all the components

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