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For ex if a ball of mass m undergoes an oscillatory motion in a bowl of mass M placed on planet where all surfaces are smooth( no friction ), while considering the part of the motion of the ball from bottom most point to one of the extremes of the bowl, say the velocity at bottom most point is V and if at that instant velocity of the system (bowl+ball) is say, Z then,

$m*V$ = $(M+m)*Z$

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  • $\begingroup$ is the instant velocity $Z$ given from the movement of planet because(thats where velocity could possibly generate) , bowl is at rest for the observer observing ball oscillating on the bowl ? $\endgroup$
    – Naveen V
    Commented Mar 24, 2023 at 15:25
  • $\begingroup$ The bowl moves due to the motion of the ball inside ( by conservation of momentum) $\endgroup$
    – Elizabeth Huffman
    Commented Mar 24, 2023 at 15:32
  • $\begingroup$ ah I get it , due to the frictionless surface it moves (my thinking was blocked friction) then surely the whole momentum with the which the system is moving would be equal to the instantaneous momentum transferred to the bowl-ball system by ball $\endgroup$
    – Naveen V
    Commented Mar 24, 2023 at 15:37
  • $\begingroup$ @NaveenV So like if there were 2 balls of m1 and m2 inside with velocities v1 and v2 resp, then the eq would be m1v1+m2v2=(M+m+m)Z right? $\endgroup$
    – Elizabeth Huffman
    Commented Mar 24, 2023 at 15:43
  • $\begingroup$ I do believe that is the case for n number of balls in the bowl (because the velocity which is resultant is represented in Z) $\endgroup$
    – Naveen V
    Commented Mar 24, 2023 at 15:47

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Velocity of the system is better translated as velocity of the centre of mass. We can use the definition of centre of mass: $$\mathbf R{cm} = \frac{M\mathbf R_M + m\mathbf R_m}{M+m}$$ Taking the derivative with respect to time: $$\mathbf v_{cm} = \frac{M\mathbf v_M + m\mathbf v_m}{M+m} \implies \mathbf p_{sys} = \mathbf p_M + \mathbf p_m$$

So yes, the momentum of the system is the sum of the momentum of its components.

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  • $\begingroup$ Thats a very intelligent way to do it ! $\endgroup$
    – Naveen V
    Commented Mar 24, 2023 at 16:28

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