Starting with equations\({m_1}{v_1} = {m_1}{v_1}^\prime \cos {\theta _1} + {m_2}{v_2}^\prime \cos {\theta _2}\)and\(0 = {m_1}{v_1}^\prime \sin {\theta _1} + {m_2}{v_2}^\prime \sin {\theta _2}\)for conservation of momentum in the x- and y-directions and assuming that one object is originally stationary, prove that for an elastic collision of two objects of equal masses, 1/2mv₁² = 1/2mv^’₁² + 1/2mv^’₂² + mv^’₁v^’₂cos (ɵ₁ - ɵ₂)as discussed in the text.

The law of conservation of momentum asserts that unless an external force is introduced, the total momentum of two or more bodies acting on each other in an isolated system remains constant. As a result, neither the creation nor the destruction of momentum is possible.

The Equations for the conservation of momentum gives,

\({m_1}{v_1} = {m_1}{v_1}^\prime \cos {\theta _1} + {m_2}{v_2}^\prime \cos {\theta _2}\)………(1)

\(0 = {m_1}{v_1}^\prime \sin {\theta _1} + {m_2}{v_2}^\prime \sin {\theta _2}\)…..(2)

Using the conservation of momentum along vertical direction we get,

\({m_1}{v'_1}\sin {\theta _1} = - {m_2}{v'_2}\sin {\theta _2}\)

Squaring and adding equation (1) and (2) for the equal masses, we get,

v₁² + 0 = (v₁^’2 cosɵ₁ +v₁^’2 cosɵ₂ )² +(v₁^’2 cosɵ₁ +v₁^’2 cosɵ₂ )²

v₁² = v₁^’2 +v₂^’2 + 2v₁^’ v₂^’(cosɵ₁ cosɵ₂+ sinɵ₁ sinɵ₂)

v₁² = v₁^’2 +v₂^’2 + 2v₁^’ v₂^’cos(ɵ₁ -ɵ₂)

Multiplying both sides by \(\dfrac{1}{2}m\),we can write, 1/2mv₁² = 1/2mv^’₁² + 1/2mv^’₂² + mv^’₁v^’₂cos (ɵ₁ - ɵ₂).

Therefore the elastic conservation of energy is: 1/2mv₁² = 1/2mv^’₁² + 1/2mv^’₂² + mv^’₁v^’₂cos (ɵ₁ - ɵ₂).