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College Physics (Urone)
Found in: Page 291

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Short Answer

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 = 1/2mv12 + 1/2mv22 + mv1v2cos (ɵ1 - ɵ2) as discussed in the text.

The elastic conservation of energy gives 1/2mv12 = 1/2mv12 + 1/2mv22 + mv1v2cos (ɵ1 - ɵ2).

See the step by step solution

Step by Step Solution

Step 1: Definition of conservation of Momentum

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)

Step 2: Calculation of energy

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 = 1/2mv12 + 1/2mv22 + mv1v2cos (ɵ1 - ɵ2).

Therefore the elastic conservation of energy is: 1/2mv12 = 1/2mv12 + 1/2mv22 + mv1v2cos (ɵ1 - ɵ2).

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