Question

Example 8.28 Two particle of mass m and 2m moving in opposite directions collide elastically with velocities v and 2v Find their velocities after collision.
SOL :- Here,$v_1=-v,v_2=2v,m_1=mand{m}_2=2m$

Answer 1

Given $m_1=mm=2m{u}_1=v{u}_2=2v$

total momentum conserved, In elastic collision is given as

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_1{v}_2$

$m_{1}v+2m2v=m{v}_1+2m{v}_2$

$v_1+2v_2=5v$.............(1)

Hence kinetic energy is also conserved. So,

$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}=m_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

$m{v}^2+2m(2v{)}^2=m{v}_1^2+2m{v}_2^2$

$9v^2=v_1^2+2v_2^2$

$9v^2=(v_1+2v_2)(v_1-2v_2)$

Substitute value from eq (1) we get

$9v_2=5v(v_1-2v_2)$

$(v_1-2v_2)=\frac{9}{5}v$..........(2)

solving (1) and (2) we get

$v_1=\frac{17}{5}v$

$v_2=\frac{4}{5}v$