# Develop The Theory Of One-dimensional Elastic Collision.

An elastic collision is defined as when two bodies collide but there is no loss in the overall kinetic energy.

Both momentum and kinetic energy are conserved in an elastic collision.

The Elastic Collision formula of momentum is given by:

$$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$$

An elastic collision can be either one-dimensional or two-dimensional

The Theory Of One-dimensional Elastic Collision

Consider two bodies with masses m1 and m2 moving along a straight line with velocity u1 and u2, which coincides with the line joining their centres of mass.

Let the collision be elastic in nature. After the collision, the two bodies move with final velocities v1 and v2 respectively in the same direction. Hence both momentum and kinetic energy are conserved.

According to the law of conservation of linear momentum.

$$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$$ $$m_{1}(u_{1} – v_{1} )$$ $$m_{2}(v_{2} – u_{2})—-[1]$$

According to the law of conservation of kinetic energy

$$\frac{1}{2}m_{1}u_{1}^{2} + \frac{1}{2}m u_{2}^{2} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}, m_{1}(u_{1}^{2} – v_{1}^{2} )$$ $$m_{2}(v_{2}^{2} – u_{2}^{2})—-[2]$$

Now,$$\frac{equ[1]}{equ[2]}$$ $$\frac{m_{1}(u_{1} – v_{1})(u_{1} + v_{1})}{m_{1}(u_{1}-v_{1})} = \frac{m_{2}(v_{2} – u_{2})(v_{2} + u_{2})}{m_{2}(v_{2}- u_{2})}$$ $$= u_{1} + v_{1} = v_{2} + u_{2}, u_{1} – u_{2} = v_{2} – u_{1}—-[3]$$

According to a one-dimensional elastic collision, the relative velocity before the collision is equal to the relative velocity of separation after the collision.

Therefore, From equation (3), we get

$$u_{1} – u_{2} = -v_{1} + v_{2}, v_{2} = u_{1} – u_{2} + v_{1}—-[4]$$

Now by substituting equation(4) in equation (1) we get

$$m_{1}(u_{1} – v_{1}) = m_{2}(u_{1} – u_{2} + v_{1} – u_{2}), m_{1}u_{1} – m_{1}v_{1}$$ $$= m_{2}u_{1} – 2m_{2}u_{2} + m_{2}v_{1}$$ $$m_{1}u_{1} – m_{2}u_{1} + 2m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{1}, (m_{1}-m_{2})u_{1} + 2m_{2}u_{2}$$ $$=(m_{1}+m_{2})v_{1}$$ $$v_{1} = \left ( \frac{m_{1}-m_{2}}{m_{1} + m_{1}} \right) u_{1} \left ( \frac{2m_{2}}{m_{1} + m_{2}} \right) u_{2}—-[5]$$

Now by substituting $$v_{1} = v_{2} – u_{1} + u_{2}$$from equation(4) in equation (1) we get

$$m_{1}(u_{1} – v_{2} + u_{1} – u_{2}) = m_{2}(v_{2} – u_{2})- 2 m_{1}u_{1} – m_{1}v_{2}-m_{1}u_{21} = m_{2}(v_{2} – m_{2}(u_{2}$$ $$2m_{1}(u_{1} + (m_{2} – m_{1}) = (m_{2} + m_{2})v_{2}$$

Therefore, $$v_{2} = [\frac{2m_{1}}{(m_{1} + m_{1})}]u_{1} + [\frac{m_{2} – m_{1}}{m_{1} + m_{2}}]u_{2}$$

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