I'm reading the following question

Two identical frictionless billiard balls are symmetrically hit by a third identical ball with velocity $\vec{i}v_0$ as in Figure (1). Find all subsequent velocities following this elastic collision.

You can download the problem set here to see the figure: http://oyc.yale.edu/sites/default/files/problem_set_5_2.pdf

In the solution, the professor writes

Plugging back into solve for $v_1$ and realizing that $m_1$ will move off in the minus $x$ direction ...

My question is, how do we know that the first mass will move in the opposite direction? This is something I've never fully understood about collisions. If a small mass impacts a much larger mass at high speed it makes intuitive sense to me that it would move backward after colliding. If a large mass impacts a small mass, it makes sense to me that they would move in the same direction. But intermediate or mixed cases seem unclear, like this one.


2 Answers 2


The rules of elastic collisions are of course derived from the appropriate equations but conveptually you remember your three cases of a what happens when a moving object collides head on with a stationary object.

  1. If the impacting one has less mass it bounces back. (Light balls bouncing back from heavy walls)

  2. If the impacting object has more mass it will slow down but continue to move forward while setting the stationary object in motion with a higher velocity (Moving baseball bat hitting a ball and the swing continuing after the impact)

  3. If the objects have the same mass the impacting object completely stops and the stationary object begins to move at exactly the same speed the impacting one had (Billiard balls colliding dead on or two Newton's cradle balls)

If collisions are not head on but at an angle then momentum is affected according to these rules along the normal direction of impact but momentum in the direction tangent to impact is unchanged.

To actually figure out the precice velocities in your problems you have to set up the appropriate equations but there are some reasons to at least reasonably expact the impacting billiard ball to bounce backwards. Ideas which are not mathematically exact but conceptually reasonable:

Idea 1. The two stationary balls can be imagined as a compound object with twice the mass of the first object where now the moving ball hits an object with twice it's own mass and thus according to rule 1 should bounce back.

Idea 2. Track with a sketch what would happen if the two stationary balls would be ever so slightly offset so that the impacting ball hit the top one first instead of both at the same time. In this case you would have collisions of type 3 (equal mass) with the first collision deflecting the impacting ball downwards, hitting the lower ball and then ending up being deflected to the lower left with a component backwards. This is completely different from the perfect symmetry situation in that a minimal offset yields significant asymmetry and it's an interesting discussion on how to try and breach the gap but nevertheless contains backwards motion.

  • $\begingroup$ Why does the ball bounce back if it's lighter? Why isn't all the energy transferred (and only the heavier 'target' will mover forward, albeit at a lower velocity than the original ball due to it's greater mass) $\endgroup$ Nov 20, 2018 at 19:05

In the case where the interaction is frictionless, elastic (with no deformation)... I would expect the first mast to retreat because it experiences negative acceleration during the impact of the 2nd hard surface. A similar result would be seen using newtons cradle. The impacting object retreats as its momentum is largely transferred to the remaining objects.


  • 2
    $\begingroup$ Negative acceleration sure, but from a positive velocity that doesn't imply the final velocity will be negative. $\endgroup$
    – Addem
    Jan 5, 2017 at 4:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.