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Two particles of masses m_1 and m_2 moving with velocities u_1 and u_2 , respectively and making an angle theta between them, collide with each other. After collision, the 1st particle travels in the initial direction of motion of the 2nd, and vice-verses. Find the velocities of the two particles aftercollision. Under what condition, would this collision be elastic ? |
Answer» Solution :SUPPOSE `v_1, v_2` are the VELOCITIES of the two particles, respectively, after collision . The particles before and after collision move as shown in Fig.1.44 .It also shows the chosen directions of the x and the y-axis.  For momentum conservation ALONG thex-axis,we get, `m_1u_1 cos""(theta)/(2)+m_2u_2 cos ""(theta)/(2) =m_1v_1 cos ""(theta)/(2) +m_2 v_2 cos ""(theta)/(2) or, m_1u_1+m_2u_2=m_1v_1+m_2v_2` .......(1) Similarly along the y-axis, we get, `-m_1u_1sin""(theta)/2+m_2u_2sin ""theta/2=m_1v_1sin ""theta/2-m_2v_2 sin""theta/2 or, -m_1u_1+m_2u_2=m_1v_1-m_2v_2`.....(2) Adding equations (1) and (2) , `2m_2u_2=2m_1v_1` `or, v_1=(m_2)/(m_1)u_1`.......(3) SUBTRACTING equation (2) from (1) , `2m_1u_1=2m_2v_2` `or, v_2=(m_1)/(m_2)u_1`.......(4) The kinetic energy before collision is, `K_1=1/2m_1u_1^2+1/2u_2^2` and that after collision is `K_2=1/2m_1v_1^2+1/2m_2v_2^2=1/2m_1((m_2)/(m_1)u_2)^2+1/2 m_2((m_1)/(m_2)u_1)^2` =`1/2 (m_2)/(m_1) m_2u_2^2+1/2(m_1)/(m_2) m_1u_1^2` Here `K_1ne K_2` , so the collision is inelastic, in GENERAL. Asa special case , it would be anelastic collision if `K_1=K_2` . It ispossible only when `m_1=m_2` , i.e., the two particles are of equal masses. |
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