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Solved: Starting with equations for conservation of
Chapter 8, Problem 51(choose chapter or problem)
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)