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This is Example 14.5 from Young and Freedman's University Physics on energy and momentum in SHM.

A block of mass $M$ attached to a horizontal spring with force constant $k$ is moving in SHM with amplitude $A_1$. As the block passes through its equilibrium position, a lump of putty of mass $m$ is dropped from a small height and sticks to it. Find the new amplitude and period of the motion.

The solution to the example states that during the collision the $x-$component of momentum of the block-putty system is conserved. (Why?) Just before the collision this component is the sum of $Mv_1$ (for the block) and zero (for the putty). Just after the collision the block and putty move together with speed $v_2$, so their combined $x-$component of momentum is $(M+m)v_2$. From conservation of momentum, $$Mv_1 + 0 = (M+m)v_2\; \text{so} \: v_2=\frac{M}{M+m}v_1.$$

I can't understand why the $x$-component of momentum is conserved? To have conservation of momentum during collision, there should be no external force. But here isn't gravity an external force? And clearly we don't get conservation of momentum in the $y$-component as the $y$-velocity is $0$ after collision, so how do we have conservation of momentum? I am not aware of any theorem giving conservation of momentum for only a single component. I would greatly appreciate some explanation on why this works.

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  • $\begingroup$ The main thing you need to absorb is that momentum is a vector and it is conserved as a vector, which means all three components are conserved. Furthermore, forces acting in one direction have no effect on momentum components in directions orthogonal to the force. $\endgroup$
    – Andrew Steane
    Commented Jul 24 at 19:06
  • $\begingroup$ @AndrewSteane I am confused with your comment. So you are saying that momentum conservation means conservation of momentum in all three components. But if there is no force acting in one component then the momentum in that coordinate is conserved even if there is force in another direction? $\endgroup$
    – nomadicmathematician
    Commented Jul 24 at 21:29
  • $\begingroup$ Yes. And in a collision with $\mathrm dt\to0$ even if you have a large finite force acting on the system, momentum will still be conserved. The error is $\vec F\mathrm dt\to0$ because the time part went to zero. $\endgroup$
    – naturallyInconsistent
    Commented Jul 25 at 4:18

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Textbook problems are sometimes contrived in a way such that the authors are trying to get at a certain concept, but the student gets confused by irrelevant details.

The problem is basically the same but simpler to grasp if you assume the small blob suddenly "appears" at the equilibrium position and the block collides with and sticks to it as it passes in the x direction.

Conservation of momentum applies independently in three directions - x, y, and z. It's technically true that vertical momentum is NOT conserved, but I think the authors didn't want you to consider vertical momentum or gravity; they want you to just treat the problem as one-dimensional. In my re-posing, the only impulsive force is between the two elements of the system (blob and block). The spring force does not change instantaneously. So it's like a typical collision question but with the added complication of calculating the SHM with the new total mass.

If you want to ponder the original problem, then there is an impulsive force in the y direction and y-momentum is NOT conserved, as previously stated. But the impulse has no x-component so does not affect the x momentum.

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