If you have an elastic collision between objects 1 and 2 and where 'kinetic energy is conserved', does this mean object 1 will always have the same velocity it had before the collision?

Or will object 2 have all the energy from object 1 transferred to it and used for velocity or will both the objects always 'join' and have the same common velocity?

  • 3
    $\begingroup$ This question might be better suited in the physics section $\endgroup$
    – user58953
    Commented Sep 22, 2014 at 21:12
  • $\begingroup$ Almost certainly. $\endgroup$
    – HDE 226868
    Commented Sep 22, 2014 at 23:02
  • $\begingroup$ Also, change velocity to speed everywhere. Velocity never remains the same after a 'collision'. $\endgroup$
    – Takku
    Commented Sep 23, 2014 at 17:07
  • $\begingroup$ Oh, woops good point (I'm referring to their magnitude, the speed) $\endgroup$
    – user58953
    Commented Sep 23, 2014 at 21:18
  • $\begingroup$ Item 1 = a glass marble floating in space. Item 2 = a steel bowling ball travelling towards the marble at 20 kps. I strongly suspect the marble's speed will change when the bowling ball bounces off of it. $\endgroup$
    – Wayfaring Stranger
    Commented Sep 24, 2014 at 15:06

2 Answers 2


In an elastic collision the masses of both objects, the total kinetic energy, and the total linear momentum are conserved. The kinetic energy has contributions from the motions of the objects as well as their rotations. If we assume that no exchange between these two forms of kinetic energy occurs, i.e. that both forms are separately conserved, we have $$ m_1\boldsymbol{v}_1 + m_2\boldsymbol{v}_2 = m_1\boldsymbol{w}_1 + m_2\boldsymbol{w}_2 $$ and $$ \tfrac{1}{2}m_1\boldsymbol{v}_1^2 + \tfrac{1}{2}m_2\boldsymbol{v}_2^2 = \tfrac{1}{2}m_1\boldsymbol{w}_1^2 + \tfrac{1}{2}m_2\boldsymbol{w}_2^2 $$ where $\boldsymbol{v}$ and $\boldsymbol{w}$ denote the velocities before and after the collision, respectively. Thus, we have 4 equations (3 components of momentum and the energy) for 6 unknowns (3+3 components of the post-collision velocities). Consequently, the above equations (on conjunction with $\boldsymbol{v}_{1,2}$) do not uniquely constrain the $\boldsymbol{w}_{1,2}$. Some other information is required to determine the relative direction. This depends on the properties of the objects and the point of impact.

The above system of equations is invariant under a Galilean transformation, i.e. a change of the origin of velocity. They become particularly simple in the frame in which the total momentum vanishes. Let a prime denote velocities in that frame, then $$ \boldsymbol{v}' = \boldsymbol{v} - \boldsymbol{V},\qquad \boldsymbol{V} = \frac{m_1\boldsymbol{v}_1+m_2\boldsymbol{v}_2}{m_1+m_2} $$ and we have $$ m_1 \boldsymbol{w}'_1+m_2 \boldsymbol{w}'_2=0,\quad |\boldsymbol{w}'_1| = |\boldsymbol{v}'_1|,\quad |\boldsymbol{w}'_2| = |\boldsymbol{v}'_2| $$ (these equations are not independent, there still only 4 independent scalar constraints). In particular, the speeds remain the same in this frame.

In 1D we have 2 equations for 2 unknowns and hence the solution is completely determined (also, there is no rotation in 1D). Since a collision requires $w_{1,2}\neq v_{1,2}$, we have $w_{1,2}=-v_{1,2}$. Transforming back to the original frame, this gives $$ w_{1,2} = 2 V - v_{1,2}. $$ Requiring the speeds to remain the same ($w_{1,2}=-v_{1,2}$) gives us $V=0$. Thus in 1D, the speeds reamin the same if and only if the total momentum vanishes.

  • $\begingroup$ Additionally I'd like to point to Walter Lewin's classical mechanics course at MIT. He has one lecture about elastic and inelastic collisions. Everyone who has never seen mr. Lewin teach: totally recommended! $\endgroup$
    – agtoever
    Commented Sep 27, 2014 at 6:30

If you have an elastic collision between objects 1 and 2 and where 'kinetic energy is conserved' ... will both the objects always 'join' and have the same common velocity?

Linear momentum will be conserved, so if the objects 'join' $$\vec{p}_1+\vec{p}_2=(m_1+m_2)\vec{V}\ $$ where $\vec{V}$ is the final common velocity.

The initial kinetic energy is $$K_i=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2} $$The final kinetic energy will be $$K_f=\frac{(m_1+m_2)V^2}{2} = \frac{p_1^2+p_2^2+2\vec{p}_1\cdot\vec{p}_2}{2(m_1+m_2)}$$ We can easily show that if the particles have oppositely directed momenta or erpendicular momenta, $K_f<K_i$. The maximum of K_f will occur for parallel momenta, and hence parallel velocity. $$K_{fmax}=\frac{p_1^2+p_2^2+2|p_1||p_2|}{2(m_1+m_2)}=\frac{(|p_1|+|p_2|)^2}{2(m_1+m_2)}.$$

A brief exercise in algebra will show that $K_i\geq K_f$, with the equality happening only if $v_1=v_2$, in which case, the particle never collide.

Conclusion: If the particles collide and stick together the collision cannot be elastic.


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