Perfectly Elastic Collision in 2D Elastic collision in two dimensions
Suppose two objects with masses m 1 m_1 m 1 m 2 m_2 m 2 v ⃗ 1 \vec{v}_1 v 1 v ⃗ 2 \vec{v}_2 v 2 x ⃗ 1 \vec{x}_1 x 1 x ⃗ 2 \vec{x}_2 x 2 u ⃗ 1 \vec{u}_1 u 1 u ⃗ 2 \vec{u}_2 u 2
Elastic collision of two objects conserves momentum:
m 1 v ⃗ 1 + m 2 v ⃗ 2 = m 1 u ⃗ 1 + m 2 u ⃗ 2 m_1\vec{v}_1+m_2\vec{v}_2=m_1\vec{u}_1+m_2\vec{u}_2 m 1 v 1 + m 2 v 2 = m 1 u 1 + m 2 u 2 and kinetic energy:
1 2 m 1 v ⃗ 1 2 + 1 2 m 2 v ⃗ 2 2 = 1 2 m 1 u ⃗ 1 2 + 1 2 m 2 u ⃗ 2 2 \frac{1}{2}m_1\vec{v}_1^2+\frac{1}{2}m_2\vec{v}_2^2=\frac{1}{2}m_1\vec{u}_1^2+\frac{1}{2}m_2\vec{u}_2^2 2 1 m 1 v 1 2 + 2 1 m 2 v 2 2 = 2 1 m 1 u 1 2 + 2 1 m 2 u 2 2 We can calculate and obtain
u ⃗ 1 = v ⃗ 1 − m 2 ⋅ 2 k ( x 1 ⃗ − x 2 ⃗ ) m 1 + m 2 \vec{u}_1=\vec{v}_1-m_2\cdot\frac{2k(\vec{x_1}-\vec{x_2})}{m_1+m_2} u 1 = v 1 − m 2 ⋅ m 1 + m 2 2 k ( x 1 − x 2 ) u ⃗ 2 = v ⃗ 2 + m 1 ⋅ 2 k ( x 1 ⃗ − x 2 ⃗ ) m 1 + m 2 \vec{u}_2=\vec{v}_2+m_1\cdot\frac{2k(\vec{x_1}-\vec{x_2})}{m_1+m_2} u 2 = v 2 + m 1 ⋅ m 1 + m 2 2 k ( x 1 − x 2 ) k = ( v ⃗ 1 − v ⃗ 2 ) ⋅ ( x ⃗ 1 − x ⃗ 2 ) ∥ x ⃗ 1 − x ⃗ 2 ∥ 2 k=\frac{(\vec{v}_1-\vec{v}_2)\cdot(\vec{x}_1-\vec{x}_2)}{\|\vec{x}_1-\vec{x}_2\|^2} k = ∥ x 1 − x 2 ∥ 2 ( v 1 − v 2 ) ⋅ ( x 1 − x 2 ) This is the predictive equation for elastic collision in high dimensions.
Special cases of elastic collision in two dimensions
Let us consider some special cases of elastic collision in two dimensions.
Case 1
Back to collision in one dimension: just "remove" the vector symbols
k = v 1 − v 2 x 1 − x 2 ⇒ 2 k ( x 1 ⃗ − x ⃗ 2 ) m 1 + m 2 = 2 v 1 m 1 + m 2 − 2 v 2 m 1 + m 2 k=\frac{v_1-v_2}{x_1-x_2}\Rightarrow\frac{2k(\vec{x_1}-\vec{x}_2)}{m_1+m_2}=\frac{2v_1}{m_1+m_2}-\frac{2v_2}{m_1+m_2} k = x 1 − x 2 v 1 − v 2 ⇒ m 1 + m 2 2 k ( x 1 − x 2 ) = m 1 + m 2 2 v 1 − m 1 + m 2 2 v 2 This makes the velocities after collision fall back to the equations below:
u 1 = m 1 − m 2 m 1 + m 2 v 1 + 2 m 2 m 1 + m 2 v 2 u_1=\frac{m_1-m_2}{m_1+m_2}v_1+\frac{2m_2}{m_1+m_2}v_2 u 1 = m 1 + m 2 m 1 − m 2 v 1 + m 1 + m 2 2 m 2 v 2 u 2 = 2 m 1 m 1 + m 2 v 1 + m 2 − m 1 m 1 + m 2 v 2 u_2=\frac{2m_1}{m_1+m_2}v_1+\frac{m_2-m_1}{m_1+m_2}v_2 u 2 = m 1 + m 2 2 m 1 v 1 + m 1 + m 2 m 2 − m 1 v 2 Case 2
Two objects have the same mass: m 1 = m 2 m_1=m_2 m 1 = m 2
One of the objects does not move initially: v 2 = 0 v_2=0 v 2 = 0
u ⃗ 1 = v ⃗ 1 − k ( x 1 ⃗ − x 2 ⃗ ) \vec{u}_1=\vec{v}_1-k(\vec{x_1}-\vec{x_2}) u 1 = v 1 − k ( x 1 − x 2 ) u ⃗ 2 = k ( x 1 ⃗ − x 2 ⃗ ) \vec{u}_2=k(\vec{x_1}-\vec{x_2}) u 2 = k ( x 1 − x 2 ) k = v ⃗ 1 ⋅ ( x ⃗ 1 − x ⃗ 2 ) ∥ x ⃗ 1 − x ⃗ 2 ∥ 2 ⇒ v ⃗ 1 ⋅ ( x ⃗ 1 − x ⃗ 2 ) = k ∥ x ⃗ 1 − x ⃗ 2 ∥ 2 k=\frac{\vec{v}_1\cdot(\vec{x}_1-\vec{x}_2)}{\|\vec{x}_1-\vec{x}_2\|^2}\Rightarrow \vec{v}_1\cdot(\vec{x}_1-\vec{x}_2)=k\|\vec{x}_1-\vec{x}_2\|^2 k = ∥ x 1 − x 2 ∥ 2 v 1 ⋅ ( x 1 − x 2 ) ⇒ v 1 ⋅ ( x 1 − x 2 ) = k ∥ x 1 − x 2 ∥ 2 Then: u ⃗ 1 ⋅ u ⃗ 2 = k v ⃗ 1 ⋅ ( x ⃗ 1 − x ⃗ 2 ) − k 2 ∥ x ⃗ 1 − x ⃗ 2 ∥ 2 = 0 \vec{u}_1\cdot \vec{u}_2=k\vec{v}_1\cdot(\vec{x}_1-\vec{x}_2)-k^2\|\vec{x}_1-\vec{x}_2\|^2=0 u 1 ⋅ u 2 = k v 1 ⋅ ( x 1 − x 2 ) − k 2 ∥ x 1 − x 2 ∥ 2 = 0
In other words, if two objects both move after collision, their moving directions must be a right angle. This is consistent with the 90-deg rule in billiards:
For a stun shot, where the cue ball has no top or bottom spin at impact with the object ball, the cue ball and the object ball separate at 90 degrees, regardless of the cut angle, except for a straight-in shot, in which case the cue ball stops in place.
Case 3
Two objects have the same mass: m 1 = m 2 m_1=m_2 m 1 = m 2
Two objects share the same speed but moving to opposite directions: v ⃗ 1 = − v ⃗ 2 \vec{v}_1=-\vec{v}_2 v 1 = − v 2
u ⃗ 1 = v ⃗ 1 − k ( x 1 ⃗ − x 2 ⃗ ) \vec{u}_1=\vec{v}_1-k(\vec{x_1}-\vec{x_2}) u 1 = v 1 − k ( x 1 − x 2 ) u ⃗ 2 = v ⃗ 2 + k ( x 1 ⃗ − x 2 ⃗ ) = − v ⃗ 1 + k ( x ⃗ 1 − x ⃗ 2 ) = − u ⃗ 1 \vec{u}_2=\vec{v}_2+k(\vec{x_1}-\vec{x_2})=-\vec{v}_1+k(\vec{x}_1-\vec{x}_2)=-\vec{u}_1 u 2 = v 2 + k ( x 1 − x 2 ) = − v 1 + k ( x 1 − x 2 ) = − u 1 In other words, after collision, they still share another same speed but moving to opposite directions.
