Suppose we have object A , which has momentum mv. Then it elastically collides with a wall, so it's momentum now -mv, So how is momentum conserved here ?
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Let's assume that the wall is rigidly connected to the Earth, which has mass $\sim 10^{25}\rm\,kg$. Then you can use the usual two-body formalism for conservation of momentum. You discover that the wall plus Earth can absorb twice the ball's initial momentum by acquiring an unmeasureably small final velocity.
The less massive your wall is, the easier it becomes to detect its recoil.
rob
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object A has it's momentum change by $\Delta p_{A} = -2m_A v_A$. This means that the wall must gain momentum $\Delta p_{wall} = +2m_A v_A$. To conserve momentum. We can calculate the change in velocity of the wall now. $\Delta v_{wall} = \Delta p_{wall}/m_{wall} = +2 \frac{m_A}{m_{wall}}v_A$. The point here is that the mass of the wall us much greater than the mass of the object $m_{wall} \gg m_A$. so we basically treat $\Delta v_{wall}$ as being zero. This is why there is an apparent breaking of conservation of momentum, but as you can see, if you take into account the momentum of the wall you'll see momentum is conserved.
Jagerber48
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When the object A strike the wall and go back in different direction,you have interpreted right that wall must gain a momentum equal to 2mv( m is mass of object A),but when you think deeply you will find that mass of wall is much more than that of object A and it is rigidly connected to massive earth. So if the mass of wall(+earth) is M,the velocity of wall will be; (m/M)*Velocity of object A≈0. It means the wall gains the momentum,but the change is so small and it remain undetected.
Vidyanshu Mishra
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