Since developing a love for physics, I have always tried to solve any problem on my own. In this pursuit, I have solved several problems, including those from my textbooks and from the preparation stages for the Physics Olympiad. I have tried to include all the problems that I have solved on my own approach up to this day.
Find out \(v_1\) or \(v_2\) after the collision of two objects.
As we know, if an elastic collision will happen, then the both momentum and kinetic energy will be conserved.
Which means,
\(P_i =P_f...(i)\)
\({E_k}_i = {E_k}_f...(ii)\)
From \((i)\), we get \(P_i =P_f\)
\(\implies m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\)
\(\implies m_1u_1 - m_1v_1 = m_2v_2 - m_2u_2\)
\(\implies m_1(u_1 - v_1) = m_2(v_2 - u_2)\)
\(\implies m_1(u_1 - v_1) = -m_2(u_2 - v_2)\)
\(\implies m_1(u_1 - v_1) = -\frac {m_2(u_2 - v_2)(u_2 + v_2)}{(u_2 + v_2)}\)
\(\implies (u_2 + v_2)(u_1 - v_1) = -\frac {m_2(u_2^2 - v_2^2)}{m_1}\)
From \((ii)\), we get \({E_k}_i = {E_k}_f\)
\(\implies \frac {1}{2} m_1u_1^2 + \frac {1}{2} m_2u_2^2 = \frac {1}{2}m_1v_1^2\)
\(+ \frac {1}{2} m_2v_2^2\)
\(\implies \frac {1}{2}(m_1u_1^2 + m_2u_2^2) = \frac {1}{2}(m_1v_1^2\)
\(+ m_2v_2^2)\)
\(\implies m_1u_1^2 + m_2u_2^2 = m_1v_1^2 + m_2v_2^2\)
\(\implies m_1u_1^2 - m_1v_1^2 = m_2(v_2^2 - u_2^2)\)
\(\implies m_1u_1^2 - m_1v_1^2 = -m_2(u_2^2 - v_2^2)\)
\(\implies m_1v_1^2 = m_1u_1^2 + m_2(u_2^2 - v_2^2)\)
\(\implies v_1^2 = u_1^2 +\frac {m_2(u_2^2 - v_2^2)}{m_1}\)
\(\implies v_1^2 = u_1^2 -(-\frac {m_2(u_2^2 - v_2^2)}{m_1})\)
\(\implies u_2^2 - v_2^2 = (u_2 + v_2)(u_1 - v_1)\)
\(\implies u_1 + v_1 = u_2 + v_2\)
\(\implies v_1 = u_2 - u_1 + v_2...(iii)\)
\(\implies v_2 = u_1 - u_2 + v_1...(iv)\)
From \((i)\), we get \(P_i =P_f\)
\(\implies m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\)
By replacing the value of \((iii)\),
\(\implies m_1u_1 + m_2u_2 = m_1(u_2 - u_1 + v_2)\)
\(+ m_2v_2\)
\(\implies m_1u_1 + m_2u_2 - m_1u_2 + m_1u_1\)
\(= m_1 v_2 + m_2 v_2\)
\(\implies m_1u_1 + m_2u_2 = m_1 u_2 - m_1 u_1\)
\(+ m_1 v_2 + m_2 v_2\)
\(\implies 2m_1 u_1 + m_2 u_2 - m_1 u_2 = (m_1\)
\(+ m_2) v_2\)
\(\implies v_2 = \frac{2m_1 u_1 + (m_2 - m_1)u_2}{m_1 + m_2}\)
By replacing the value of \((iv)\),
\(\implies m_1u_1 + m_2u_2 = m_2(u_1 - u_2 + v_1)\)
\(+ m_1v_1\)
\(\implies m_1v_1 + m_2u_1 - m_2u_2 + m_2v_1\)
\(= m_1u_1 + m_2u_2\)
\(\implies m_1u_1 + m_2u_2 - m_2u_1 + m_2u_2\)
\(= m_1v_1 + m_2v_1\)
\(\implies (m_1 - m_2)u_1 + 2m_2u_2 = (m_1\)
\(+ m_2)v_1\)
\(\implies v_1 = \frac{(m_1 - m_2)u_1 + 2m_2u_2}{m_1 + m_2}\)