A ball moving with a velocity v, colliders head on with a stationary second ball of same mass.
After the collision, the velocity of the first ball is reduced to 0.15 v. The kinetic energy of the system is decreased nearly by

Options

A. 0.2
B. 0.25
C. 0.3
D. 0.4

Answer

B

Explanation

Given,

`rArr` Momentum is conserved,
`m_(1)u_(1) + m_(2)u_(2)= m_(1)v_(1)+m_(1)v_(2)`
Given `m_(1)= m_(2) =m,u_(1)= v and u_(2)=0,`
also v = 0.15 v
`mv = m(0,15v)+mv_(2)`
`v-0.15v=v_(2)`
`rArr 0.85v = v_(2)` m / s ...(i)
Here, initial kinetic energy is
`(1)/(2)mu_(1)^(2)=(1)/(2)mv^(2)` ...(ii)
Final kinetic energy is
`(1)/(2)mv_(1)^(2)+(1)/(2)mv_(2)^(2)=(1)/(2),(0.15v)^(2)+(1)/(2)m(0.85v)^(2)` ...(iii)
The decrement in kinetic energy is
`Delta KE=(KE)_(i)-(KE)_(f)`
` rArr DeltaKE=(1)/(2)mv^(2)-[(1)/(2)m(0.15v)^(2)+(1)/(2)m(0. 85)]`
`DeltaKE=(1)/(2)mv^(2)-[(1)/(2)m(0.15v)^(2)+(1)/(2)m(0. 85)]` ...(iv)
% cahnge in kinetic energy `=(DeltaKE)/((KE)_i)xx 100`
`= (1/2mv^2[1-(15)^2-(85)^2])/(1/2mv^2)`
% decrease in KE = 25.5% = 25%