Three balls A,B,Cof masses m_1 ,m_2 and m_3 respectively are kept at rest along a straightline. Now A moving in that straight line with velocity u_1strikes C. As a rsult velocity of C becomes u_3. If the collisions areelastic , show that u_3 = 4u_1when m_1gt gt m_2 and m_2gt gt m_3. In case A hits C directly will the velocity of C be higher or lower ?

Three balls A,B,Cof masses m_1 ,m_2 and m_3 respectively are kept at r

1.	Three balls A,B,Cof masses m_1 ,m_2 and m_3 respectively are kept at rest along a straightline. Now A moving in that straight line with velocity u_1strikes C. As a rsult velocity of C becomes u_3. If the collisions areelastic , show that u_3 = 4u_1when m_1gt gt m_2 and m_2gt gt m_3. In case A hits C directly will the velocity of C be higher or lower ?
Answer» Solution :Suppose A acquires a velocity `v_1` after collision with B. From the law of conservation of linear momentum , for collision between A and B . `m_1u_1=m_1v_1+m_2u_2` `or, m_1(u_1-v_1)=m_2u_2` From the law of conservation of kinetic energy of elastic collision, `1/2 m_1u_1^2=1/2m_1v_1^2+1/2m_2u_2^2` `or, m_1(u_1^2-v_1^2)=m_2u_2^2` DIVIDING equation (2) by (1) , `u_1+v_1=u_2 or, v_1=u_2-u_1` Inserting the value of `v_1` in equation (1) , `m_1(2u_1-u_2)=m_2u_2 or, 2m_1u_1=(m_1+m_2)u_2` Hence, `u_2=(2m_1)/(m_1+m_2)u_1=2/(1+(m_2)/m_1)u_1` As `m_1 gt gt m_2 , (m_2)/(m_1)lt lt 1,hence, 1+(m_2)/(m_1)~~1`. THUS , `u_2= 2u_1`. Similary for the collisionof B iwith C , `u_3 ~~2u_2`. thus `u_3= 4u_1`. But , for the direct collision of A with C ,the velocity GAINED by C, `u_3^.~~2u_1`. `therefore u_3^.` isless than `u_3`. Hence, in case of direct collision between A and Cthe speed of C will be less.