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What is elastic collision ? Derive an expression for final velocities of two bodies which undergo elastic in one dimension. |
Answer» Solution :Consider two elastic bodies of masses m, and m, moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure. From the LAW of conservation of linear momentum, Total momentum before collision `(P_i)` = Total momentum after collision `(P_f)` `m_1u_1 + m_2u_2 = m_1v_1 + m_2v_1 `...(1) `m_1(u_1 - v_1) - m_2(v_2 - u_2)`......(2) further  For elastic collision, Total KINETIC energy before collision `KE_i`= Total kinetic energy after collision `KE_f` `1/2 m_1u_1^2 + 1/2 m_2u_2^2 = 1/2 m_1v_1^2+ 1/2 m_2v_2^2`....(3) After simplifying and rearranging the terms, `m_1 (u_1^2 -v_1^2) = m_2 (v_2^2 - u_2^2)` Using the formula ` a^2 -b^2 = (a + b) (a- b) ` ,we can rewrite the above equation as `m_1 (u_1 + v_1) (u_1- v_1) = m_2 (v_1 + u_2) (v_2 - u_2)`....(4) Dividing equation (4) by (2) gives, ` (m_1(u_1 + v_1)(u_1 - v_1) )/(m_1 (u_1 - v_1)) = (m_2(v_2 + u_2)(v_2-u_2) )/(m_2(v_2 - u_2) )` `u_1 + v_1 = v_2+u_1` Rearranging `u_1 - u_2 =v_2 - v_1`.....(5) Equation (5) can be rewritten as `u_1 - u_2 = - (v_1 - v_2)` This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass. Rewriting the above equation for `v_1` and `v_2` `v_1 = v_2 + u_2 -u_1`....(6) or `v_2 = u_1 + v_1 - u_1`.....(7) To find the final velocities `v_1`and `v_2` : Substituting equation (5) in equation (2) gives the velocity of `m_1`as `m_1 (u_1 - v_1)= m_2 (u_1 + v_1 - u_2 - u_2)` `m_1 (u_1 - v_1) = m_2 (u_1 + v_1 - 2u_2)` `m_1u_1 - m_1v_1 = m_2u_1 + m_2v_1 + 2m_2u_2` `m_1u_1 - m_2u_1 + 2m_2u_2 = m_1 v_1 + m_2v_1` (`m_1 - m_2) u_1 + 2m_2u_2 = (m_1 + m_2) v_1` or` v_1= ( (m_1 - m_2)/(m_1 + m_2) ) u_1 + ( (2m_2)/(m_1 + m_2) ) u_2` SIMILARLY, by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final velocity of `m_2`as ` v_2= ( (2m_1)/(m_1 + m_2)) u_1 + ((m_2 - m_1)/(m_1 + m_2) ) u_2`.....(9) |
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