If a ball of mass M moving with velocity u hits a massive vertical wall then after collision the momentum of b?

Question

If a ball of mass M moving with velocity u hits a massive vertical wall then after collision the momentum of ball will be -Mu but according to law of conservation of momentum the momentum should be Mu. Considering an isolated system,
Let the mass of wall be N and velocity v = 0, then
Mu + Nv (initial)= Mu' + Nv(final)
Mu = Mu'
u = u' but the aswer should be u = -u'
is this situation violates the law please explain

Answer 1

The fact that the wall is 'massive' (and fixed to the ground .. so effectively has the mass of the Earth) doesn't prevent equal and opposite forces (F) acting on both the ball and the wall for the same time (t) during their impact.



As (F x t) = change in momentum, the ball and wall suffer equal but opposite changes of momentum (-Ft for the ball and Ft for the wall/Earth) which causes a change in velocity for both.

Having a relatively small mass the ball has a noticeable change in velocity .. but due to enormous mass of the wall/Earth it's change in velocity is imperceptible .. but still exists.



You assumed the massive wall would remain stationary (or have no change of velocity) after impact .. but it will change by a minute amount .. ..



Mu + Nv (initial)= Mu' + Nv(final)

With Nvi = 0

Change in mom of ball M(u' - u) = change in mom of wall (Nvf - 0)



If u' = -u .. .. -2Mu (ball) = Nvf (wall)

Answer 2

Here it also depends on the inertia excerted by both to resist change and impulse at point of contact.one body is stationary but excerts an equal force to counter the force excerted by the ball.for the ball to have a momentum -mu,it must return withthe same velocity which is not possible.

Answer 3

I can detect the compression wave in the wall and in the foundations and the earth underneath as the original momentum of the ball is imparted to bits of it. In theory and with enough instrumentation I can determine to an arbitrary accuracy that momentum is conserved.

Answer 4

During collision the ball exerts a force to the wall Fbw and simultaneously the wall to the ball an equal to magnitude opposite direction force Fwb (3rd Newton's law) . This force is responsible to slow down - stop and reverse the ball changing the momentum of ball according 2nd Newton's law

Fwb=Δpb/Δt , Δt is collision's duration. So impulse to the ball by wall is Δpb = Fwb*Δt ,the Fwb varies during collision. Fbw = - Fwb, mb*ab = - mw*aw,for mw>>mb ab>>aw.

pb'+pw' =pb+pw , mbub'+mwuw' = mbub+mwuw (1)

Also cosidering completely elastic collision the kinetic energy is conserved, that is K'=K

1/2mbub'^2+ 1/2mwuw'^2 = 1/2mbub^2+ 1/2mwuw^2 (2)

Solving the system (1)(2) we get ub'=(mb-mw)/(mb+mw)*ub. Because mb<
mb-mw~-mw and mb+mw~mw so ub'~(-mw/mw)*ub =-ub