This is a question about the conservation of momentum:

If there were a cluster of billiard balls floating in space and the cluster was struck by one moving ball, the cluster balls would scatter in random directions and they would move with a velocity in proportion to their mass and the P transferred from the striking ball. The first cluster ball(s) struck by the moving ball would collide with other balls in the cluster and those would collide with other balls, etc. There would be a "chain reaction" of collisions. I understand this so far.

I also understand that the sum of the cluster balls' mass and velocity are equal to the mass and velocity of the striking ball. My question is: how is the linear momentum is conserved if the cluster balls move in scattered vectors (i.e., not in the direction of the striking ball's movement)?


2 Answers 2


Because vector quantities add like vectors.

When adding vectors, you must take into account direction. For example what is the result of adding the two vectors below? One has a length of 4 units and the other 6 units.

<----  + ------>

The answer is 2 units to the right. Use the tip-to-tail method or an equivalent method to add vectors. Don't just add their magnitudes together.

In this way, two vectors below, each of which point diagonally can sum up to something point to the right.

    ^      \
   /        \
  /    +     \    =   -->
 /            v

Thanks for humoring the ascii art.

  • $\begingroup$ Up voting the ASCII art! $\endgroup$
    – Floris
    Jan 10, 2015 at 16:10

When a single ball hits another ball not quite straight-on, momentum is conserved because the sideways momentum of one will be equal and opposite to the sideways momentum of the other - in other words, their vector sum will equal the initial momentum.

This is nicely illustrated with this image (from http://blog.carriesegal.com/wp-content/uploads/2010/03/3-22-10-Graphic-6.jpg)

enter image description here

The same result can now be extended: each ball can hit another ball and conserve momentum, etc.

In other words - the lateral momentum of all the balls in the cluster will cancel exactly.

  • $\begingroup$ To clarify things for me, here is a further question: There are two objects, one enclosed box loosely filled with ball bearings and the other a solid block of wood. Both have the same mass and are on a low-friction table. A bullet strikes each object. (Bullet have identical mass & velocity). Bullet is retained inside both objects. The bullet striking the box with the ball bearings penetrates into the box and strikes the ball bearings, causing them to scatter inside the box. From what you have told me, both objects will move backward (direction of bullet travel) the same amount? $\endgroup$
    – A J
    Jan 10, 2015 at 7:43
  • $\begingroup$ Yes exactly. The balls moving inside the box get some of the kinetic energy but net momentum is the same. In the wooden block, the "balls" are smaller (atoms) but they too will jiggle more (get hot) because of the absorbed energy of an inelastic collision. Momentum before and after collision must be the same for both. $\endgroup$
    – Floris
    Jan 10, 2015 at 16:09

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