How is the elastic collision equation derived?

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Solution

Derivation of elastic collision equation:
Step 1: Considering
Let us assume that two bodies of masses $m_{1}$ and $m_{2}$ with initial velocities $u_{1}$ and $v_{1}$ are going into an elastic collision. After the collision, let the velocities of the bodies are $u_{2}$ and $v_{2}$ .
Step 2: Formula Used
The formula for conservation of linear momentum is given as, $m_{1} u_{1} + m_{2} v_{1} = m_{1} u_{2} + m_{2} v_{2}$ .
The formula for conservation of kinetic energy is given as, $\frac{1}{2} m_{1} {u_{1}}^{2} + \frac{1}{2} m_{2} {v_{1}}^{2} = \frac{1}{2} m_{1} {u_{2}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2}$ .
Step 3: Elastic collision:
A collision between two masses is said to be elastic when the total kinetic energy remains the same before and after the collision.
Angular momentum is conserved in an elastic collision.
The kinetic energy of a body of mass $m$ moving with a velocity $v$ defined by the form,
In an elastic collision where both the Kinetic Energy, KE, and momentum, p are conserved which means that KE₀ = KE_f and p_o = p_f. When we recall that $E_{k} = \frac{1}{2} m v^{2}$ , we will write the equation in the following manner: $\frac{1}{2} m_{1} {u_{1}}^{2} + \frac{1}{2} m_{2} {v_{1}}^{2} = \frac{1}{2} m_{1} {u_{2}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2}$
Diagram:
Step 4: Calculations
Simplify the formula for the conservation of linear momentum.
$m_{1} u_{1} + m_{2} v_{1} = m_{1} u_{2} + m_{2} v_{2} \Rightarrow m_{1} u_{1} - m_{1} u_{2} = m_{2} v_{2} - m_{2} v_{1} \Rightarrow m_{1} (u_{1} - u_{2}) = m_{2} (v_{2} - v_{1}) \Rightarrow \frac{m_{1} (u_{1} - u_{2})}{m_{2} (v_{2} - v_{1})} = 1 \dots (1)$
Simplify the formula for the conservation of kinetic energy.
$\frac{1}{2} m_{1} {u_{1}}^{2} + \frac{1}{2} m_{2} {v_{1}}^{2} = \frac{1}{2} m_{1} {u_{2}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2} \Rightarrow \frac{1}{2} m_{1} {u_{1}}^{2} - \frac{1}{2} m_{1} {u_{2}}^{2} = \frac{1}{2} m_{2} {v_{2}}^{2} - \frac{1}{2} m_{2} {v_{1}}^{2} \Rightarrow \frac{1}{2} m_{1} ({u_{1}}^{2} - {u_{2}}^{2}) = \frac{1}{2} m_{2} ({v_{2}}^{2} - {v_{1}}^{2}) \Rightarrow \frac{m_{1} ({u_{1}}^{2} - {u_{2}}^{2})}{m_{2} ({v_{2}}^{2} - {v_{1}}^{2})} = 1 \Rightarrow \frac{m_{1} (u_{1} - u_{2}) (u_{1} + u_{2})}{m_{2} (v_{2} - v_{1}) (v_{2} + v_{1})} = 1 \dots (2)$
Substitute equation (1) in equation (2)
$\frac{m_{1} (u_{1} - u_{2}) (u_{1} + u_{2})}{m_{2} (v_{2} - v_{1}) (v_{2} + v_{1})} = 1 \Rightarrow \frac{m_{1} (u_{1} - u_{2})}{m_{2} (v_{2} - v_{1})} \times \frac{(u_{1} + u_{2})}{(v_{2} + v_{1})} = 1 \Rightarrow \frac{(u_{1} + u_{2})}{(v_{2} + v_{1})} = 1 (From equation (1)) \Rightarrow u_{1} + u_{2} = v_{2} + v_{1} \Rightarrow u_{1} = v_{2} + v_{1} - u_{2}$
Since mass gets canceled from both the bodies we get elastic collision as, $u_{1} = v_{2} + v_{1} - u_{2}$
Thus, the equation for elastic collision is, $u_{1} = v_{2} + v_{1} - u_{2}$ .

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