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    • https://phys.libretexts.org/Courses/Berea_College/Introductory_Physics%3A_Berea_College/10%3A_Linear_Momentum_and_the_Center_of_Mass/10.02%3A_Collisions
      From the \(y\) component of momentum conservation, we can find an expression for the speed of the nucleus: \[\begin{aligned} m_p v'_p\sin\theta &= m_N v'_N\sin\phi\\ \therefore v'_N &= \frac{m_p}{m_N}...From the \(y\) component of momentum conservation, we can find an expression for the speed of the nucleus: \[\begin{aligned} m_p v'_p\sin\theta &= m_N v'_N\sin\phi\\ \therefore v'_N &= \frac{m_p}{m_N}v'_p\sin\theta \frac{1}{\sin\phi}\end{aligned}\] which we can substitute into the \(x\) equation for momentum conservation to solve for the angle \(\phi\): \[\begin{aligned} m_p v_p &= m_p v'_p\cos\theta + m_N v'_N\cos\phi\\ m_p v_p &= m_p v'_p\cos\theta + m_N\frac{m_p}{m_N}v'_p\sin\theta \frac{\co…

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