Question
Derive an expression for the velocity of the two masses $m_1$ and $m_2$ moving with speeds $u_1$ and $u_2$ undergoing elastic collision in one dimension.
✓
Answer
One dimensional elastic collision is one in which both momentum and K.E. are conserved and the body moves in the same line of motion even after the collision. If $m_1, m_2$ are the masses, $u_1 u_2$ are the initial velocities and $v_1 V_2$ are the final velocities, then $m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2 \ldots$ (i) $\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 v_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2 \cdots$ (ii) (i.e..) $m_1\left(v_1^2-u_1^2\right)=m_2\left(v_2^2-u_2^2\right)$ Form (ii) $m_1\left(v_1-u_1^2\right)=m_2\left(v_2-u_2\right)$ Form (i) Dividing both sides $v_1+u_1=$ $v_2+u_2 v_1=v_2+u_2-u_1$ substituting in (i) we have, $m_1 u_1+m_2 u_2=m_1\left(v_2+u_2-u_1\right)+m_2 v_2 2 m_1 u_1+u_2\left(m_2-m_1\right)$ $=v_2\left(m_1+m_2\right) v^2=\frac{u_2\left(m_2-m_1\right)+2 m_1 u_1}{\left(m_1+m_2\right)}$ Similartly, $v_1=\frac{u_1\left(m_1-m_2\right)+2 m_2 u_2}{\left(m_1+m_2\right)}$
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