The impulse on a wall

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Solution 1

Let the mass of the ball be $m$ and that of the wall (and the Earth) be $M$.
The initial velocity of the wall is $\vec v_{\rm i}$ and that of the wall is zero.
The final velocity of the wall is $\vec v_{\rm f}$ and that of the wall is $\vec V_{\rm f}$.

Assume that the ball and wall system has no eternal forces acting on it then applying the conservation of linear momentum gives $m\vec v_{\rm i} = m\vec v_{\rm f} +\vec MV_{\rm f}$.

So the impulse on the ball is $m\vec v_{\rm f}- m\vec v_{\rm i} = m\Delta \vec v$ and the impulse on the wall is $M\vec V_{\rm f} = M\Delta \vec V$ and of the same magnitude as the impulse on the ball.

As $M\gg m$ then in terms of the magnitudes of the changes in velocities $\Delta V \ll \Delta v$ and in many cases the approximation $\Delta V \approx 0$ is made so to a very good degree of accuracy the ball does rebound with the same speed as it had before hitting the wall if the collision is elastic.

Solution 2

As the other answers have pointed out, its because $I=m\Delta v$ applies to the net impulse. The wall transfers nearly all of the impulse down into the ground, and into the earth, so the net impulse on the wall is nearly 0, even though the impulse on the wall from the ball is $16 \text N\cdot \text s$.

Technically speaking, this will actually change the rotation of the earth slightly, accounting for this transfer of momentum. Practically speaking, however, this change in velocity is so minuscule compared to our uncertainty in measurement that we handwave it away. It's not exactly 0, but its close enough for us to at least ignore it.

And, of course, we cant change the rate of rotation of the earth meaningfully with this trick. Nearly all ways we generate the momentum of the ball in the first place impart momentum in the planet one direction, and then the impact with the wall (transferred to the ground) negates it.

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Mohammad Alshareef
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Updated on July 08, 2020

Comments

  • Mohammad Alshareef
    Mohammad Alshareef over 3 years

    If a ball was thrown to a wall, How could we calculate the impulse on the wall ?

    For example, if I throw a $2kg$ ball with $ 4m/s$, and it returned with the same speed but in the opposite direction, so the impulse on the ball would be $16N.s$.

    According to the conservation to the momentum the impulse on the wall is also $16N.s$ ,But the wall is stable hence ,$Δv = 0$. In other word the impulse on the wall should be zero Depending on $I = mΔv$.

    This confuses me and I don't know what I am missing.

    Please Help

    • Paul Childs
      Paul Childs about 4 years
      You are dealing with zeros and infinities. A rigid wall is taken lim $m \to \infty$. So $I$ can be whatever it needs to be.
    • Mohammad Alshareef
      Mohammad Alshareef about 4 years
      @PaulChilds ,Sorry, But what you are trying to say in not clear to me, I can't see any limit here, Is it zero or 16 and why ?
  • Mohammad Alshareef
    Mohammad Alshareef about 4 years
    ,When you say that $\delta V$ is not zero ,This means that a very very very small change in momentum occur to the earth. $\delta V = \frac {16}{M}$.Is this right ?
  • Farcher
    Farcher about 4 years
    @MohammadAlshareef The magnitude of the momentum change of the wall/Earth is the same as that of the ball but because the wall/Earth has a mass so much greater than the ball the magnitude of the change in velocity of the wall/Earth is very much smaller than that of the ball.
  • Urb
    Urb over 3 years
    Welcome to PhysicsSE! Please use MathJax to type formulae.