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I was practicing some Mechanics, and I came across a problem that involved an impact between two objects: enter image description here

I understand how to approach the question:

Apply the conservation of momentum, and use the fact that the ratio of the difference in speeds before and after is e. And then it's simple rearranging and substitution.

However, it's important to get the directions of A and B after the collision correct. According to the mark scheme of the question, B also changes direction, thus meaning the speed of separation after the collision would be $y + x$, where $y$ is the speed of A, and $x$ is the speed of B.

But, how is one to assume this from the question given? Why can't B continue moving in the same direction (therefore meaning the speed of separation would be $y-x$ or $x-y$, since they'll both be in the same direction.)?

So my question is, how do you tell which way two objects will travel after they collide, when given their masses and speeds?

EDIT: I previously used the wrong question. In this question, both A and B are moving.

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    $\begingroup$ Conservation of momentum, provided you're always consistent with signs, will always tell you which direction it's going in. A has velocity $+u$ in the $\rightarrow$ direction. Using cons. of momentum, $mu + 0 = mv_A + 4mv_B$. From this equation alone, it is possible that both $v_A$ and $v_B$ are both positive. When you see that the direction of A is reversed, we know that $v_A < 0$, so it's going in the $\leftarrow$ direction. In general, you solve for $v_A$ and $v_B$, and the equations will tell you if $v_A > 0$, or $< 0$. $\endgroup$
    – Tweej
    May 8, 2016 at 14:04
  • $\begingroup$ Pick a coordinate system, say $+x$ is in the forward direction. Then positive velocities move in the positive direction, and negative velocities mean the object is going the other way. The sign of the velocity that results from your calculation tells you which way it's moving. $\endgroup$
    – garyp
    May 8, 2016 at 14:13
  • $\begingroup$ @Tweej My bad, I used the wrong question, in this one a and b are BOTH moving $\endgroup$
    – 83457
    May 8, 2016 at 15:56
  • $\begingroup$ The same thing holds. Your starting equation would be $m(3u) + 2m(-u) = m(v_A) + 2m(v_B)$. Solving for $v_A$ and $v_B$ using conservation of energy will get you some values in terms of u, and if there's a positive sign, it's travelling in the same direction as A was initially. If there's a negative sign, it's travelling in the opposite direction of A, or in the same direction as B. $\endgroup$
    – Tweej
    May 8, 2016 at 16:11
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    $\begingroup$ You have two equations and two unknowns. The unknowns are, of course, the final velocities of the bodies. The equations are one for conservation of momentum and one for the bulk kinetic energy (which is not conserved by the loss of which in the CoM frame is encode in the coefficient of restitution). Presumably by telling you that the direction of A's motion is reversed they make the system easier to solve. $\endgroup$
    – dmckee --- ex-moderator kitten
    May 8, 2016 at 17:01

3 Answers 3

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You get the direction of travel if you are careful with sign conventions. Let us have the convention that body A moves in a positive sense with velocity $3u$ and body B in a negative sense with velocity $-u$.

When they impact, they exchange momentum, the magnitude of it $J$ is applied in equal and opposite sense to each body such that the total momentum is conserved.

After the impact the speed of A is $v_A^\star = \left(3u-\frac{J}{m}\right)$ and of B is $v_B^\star = \left(-u+\frac{J}{2m}\right)$. There is nothing at this point telling us if the final velocity is positive or negative for both A and B.

Use the law of impact to find $J$. Here there is another convention needed. The relative impact speed is $v_{rel} = v_A-v_B = (3u)-(-u) = 4u$. The relative speed after the impact is $v_{rel}^\star = v_A^\star-v_B^\star = \left(3u-\frac{J}{m}\right) - \left(-u+\frac{J}{2m}\right) = \frac{8 m u - 3 J}{2 m} $. The law of collisions says $v_{rel}^\star = -e v_{rel}$ which yields $J=\frac{8 m u (1+e)}{3}$.

Now you have find the final velocities and if they are positive it means movement to the right and negative is movement to the left.

$$\begin{align} v_A^\star & = \frac{1-8 e}{3} u \\ v_B^\star & = \frac{4 e+1}{3} u \end{align} $$

Given that $e=0\ldots1$ there is a change in sign for body A when $e = \frac{1}{8}$.

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"when it collides directly" is to be interpreted as a "head-on" collision? So the balls go off along the line of the original velocity direction.
The question tells you that the direction of $A$ is reversed so the direction of $B$ must be in the opposite direction ie in the original direction of $A$'s velocity otherwise momentum cannot be conserved.

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  • $\begingroup$ I've modified the question, I previously used the wrong question. In this case, surely the direction of motion of A can be reversed, but B's stays the same and only reduces in speed? $\endgroup$
    – 83457
    May 8, 2016 at 15:58
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Taking the initial direction of A as "to the left", the total momentum of A and B is $(-3mu+2mu) =-mu$ before the collision.

When A reverses direction, it has (some amount of) positive momentum. This means B MUST have negative momentum equal to the momentum of $-(p_a +mu)$

So it reverses direction too.

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