Two bodies of masses 5 kg and 10 kg move towards each other with velocities `10 m*s^(-1) and 14 m*s^(-1)` respectively. If the coefficient of restitution is 0.8, find their velocities after the collision.

Explanation

Let `u_1 and u_2` respectively be the velocities of the first and the second body before the collision. After the collision, their velocities will be `v_1 and v_2` respectively.
From the law of conservation of linear momentum
`m_1u_1+m_2u_2=m_1v_1+m_2v_2`
Here, `m_1=5kg, m_2=10 kg , u_1=10m*s^(-1) and u_2=-14 m*s^(-1)`
[ let us assume `u_1` to be taken along positive x -axis and `u_2` along negative x-axis]
`therefore 5xx10+10xx(-14)=5v_1+10v_2`
`or, v_1+2v_2 =-18`..... (1)
Also, `e=(v_2-v_1)/(u_1-u_2) or, 0.8 =(v_2-v_1)/(10-(-14))`
or, `v_2-v_1 =0.8xx24=19.2 ......(2)`
Hence, from (1) and (2) ,
`3v_2=1.2 or, v_2=(1.2)/(3) =0.4 m*s^(-1)`
and `v_1=-18.8m*s^(-1)` which means after the collision mass goes along positive x-axis .