A ball weighing 1 kg is moving horizontally at `12 ms^(-1)`. It collides head on with another of double the mass moving in oppssite direction with double the speed. If the coefficient of restitution is 2/3, the energy lost in the collision is given by :

Options

A. 60 J
B. 120 J
C. 240 J
D. 480 J

Answer

c

Explanation

According to conservation of momnetum,
`1xx12+2xx(-24)=1xxv_(1)+2xxv_(2)`.
` implies v_(1)+2v_(2)=36` …..(i)
Also `e=(v_(2)-v_(1))/(u_(1)-u_(2))implies 2/3=(v_(2)-v_(1))/(12-(-24))`
`implies v_(2)-v_(1)=24`
Adding (i)&(ii),
`3v_(2)=60 implies v_(1)=20 m//s`.
Loss of energy=`1/2xx1xx12^(2)+1/2xx2(-34)`
`-(1/2xx1xx(-4)^(2)+1/2xx2xx(20)^(2))`
=72+576-(8+400)
=240 J.