An object A of mass 2 kg is moving on a frictionless horizontal track has perfectly inelastic collision with another object B of mass 3 kg made of the same material and moving in front of A in same direction their common speed after the collision is `4 m//s` Due to the collision the temperature of the two objects which was initially the same, is increased though only by `0.006^(@)` C What was the initial speed `(in m//s)` of the colliding object A before the collision?

Answer

`7`

Explanation

`2(V_(1))+3(V_(2))=(2+3)V`
`Rightarrow2V_(1)+3V_(2)=20` .....(1)
loss in KE = heat energy
`1/2 2(V_(1))^(2)+1/2 3(V_(2))^(2)-1/2 (2+3)V^(2)=(2+3)SDeltaT`
`V_(1)^(2)+3/2V_(2)^(2)-40=15 or V_(1)^(2)+(3V_(2)^(2))/2=55` .....(2)
Solving equation (1) and (2) we get
`V_(1)=1m//s or 7 m//s and V_(2) = 6m//s or 2m//s for collision V_(1) gt V_(2) So V_(1) = 7 m//s and v_(2) = 2m//s`