A lighter particle moving with a speed of `10 ms^(-1)` collides with an object of double its mass moving in the same direction with half its speed. Assume that the collision is a one dimensional clastic collision. What will be the speed of both particles after the collision?

Explanation

Let the mass of the first body be m which moves with an initial velocity, `U_(1)=10ms^(-1)`
Therefore, the mass of second body is 2m and its initial velocity is `u_(2)=1/2u_(1)=1/2(10ms_(-1))`
`v_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)+((2m_(2))/(m_(1)-m_(2)))u_(2)`
`v_(1)=((m-2m)/(m+2m))10+((2xx2m)/(m+2m))5`
`v_(1)=-(1/3)10+(4/3)5=(-10+20)/3=10/3`
`v_(1)=3.33ms_(-1)`
`v_(2)((2m(1))/(m_(1)+m_(2)))u_(1) + ((m_(2)-m_(2))/(m_(1)+m_(2)))u_(2)`
`v_(2)=((2m)/(m+2m))10+((2m-m)/(m+2m))5`
`v_(2)=(2/3)10+(1/3)5=(20+5)/3=25/3`
`v_(2)=8.33ms(-1)`
as the two speeds `v_(1) and v_(2)` are posiitive , they move in the same direction with the veloxities `3.33ms^(-1) and 8.33ms^(-1)`, respectively.