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Chapter 8, Problem 51

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QUESTION:

Starting with equations \(m_{1} v_{1}=m_{1} v_{1}^{\prime} \cos \theta_{1}+m_{2} v_{2}^{\prime} \cos \theta_{2}\) and \(0=m_{1} v_{1} \sin \theta_{1}+m_{2} v_{2} \sin \theta_{2}\) 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, \(\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m v_{1}^{\prime}+\frac{1}{2} m v_{2}^{\prime} 2+m v_{1}^{\prime} v_{2}^{\prime} \cos \left(\theta_{1}-\theta_{2}\right)\) as discussed in the text.

Equation Transcription:

Text Transcription:

m_1 v_1 = m_1 v_1 cos theta_1 + m_2 v_2 cos theta_2

0 = m_1 v_1 prime sin theta_1 + m_2 v_2 prime sin theta_2

x

y

1/2 mv_1^2 = 1/2 mv_1 prime + 1/2 mv_2 prime 2 + mv_1 prime v_2 prime cos (theta_1 - theta_2)

Questions & Answers

QUESTION:

Starting with equations \(m_{1} v_{1}=m_{1} v_{1}^{\prime} \cos \theta_{1}+m_{2} v_{2}^{\prime} \cos \theta_{2}\) and \(0=m_{1} v_{1} \sin \theta_{1}+m_{2} v_{2} \sin \theta_{2}\) 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, \(\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m v_{1}^{\prime}+\frac{1}{2} m v_{2}^{\prime} 2+m v_{1}^{\prime} v_{2}^{\prime} \cos \left(\theta_{1}-\theta_{2}\right)\) as discussed in the text.

Equation Transcription:

Text Transcription:

m_1 v_1 = m_1 v_1 cos theta_1 + m_2 v_2 cos theta_2

0 = m_1 v_1 prime sin theta_1 + m_2 v_2 prime sin theta_2

x

y

1/2 mv_1^2 = 1/2 mv_1 prime + 1/2 mv_2 prime 2 + mv_1 prime v_2 prime cos (theta_1 - theta_2)

ANSWER:

Step 1 of 4

The desired result can be showed by manipulating algebraically the equations of  conservation of momentum in x and y directions

. . . . (1)

. . . . (2)

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