A particle of mass m moving in the x - direction with speed 2c is hit by another particle of mass 2m moving in the y - direction with with speed v . If the collision is perfectly inelastic , the percentage loss in the enrgy during the collision is close to .

Options

A. 0.44
B. 0.5
C. 0.56
D. 0.62

Answer

C

Explanation

`vec(p_(1))=m_(1)vec(v_(1)) =2mv hat(i)`
`vec(p_(2)) =m_(2)vec(v_(2)) =m(2v)hat(i) =2mv hat(j)`
` :. " Resultant momentum " | vec(p) | = sqrt(p_(1)^(2)+p_(2)^(2))`
`(2m+m) v. = sqrt((2mv)^(2)+(2mv)^(2))`
` : 3v. , = sqrt(4m^(2)v^(2)+4m^(2)v^(2))`
` :. 3v. m = 2sqrt(2) mv `
` :. v = = (2sqrt(2))/3 v `
Now , decrease in K.E = intial K.E - K.E after collision
`1/2 m(2v)^(2) +1/2 (2m) v^(2) - 1/2 [2m+m] (v)^(2)`
` = 2,v^(2) +mv^(2)-1/2 xx3m xx (8v^(2))/9`
` = 3mv ^(2) - (4v^(2)m)/3 `
`= (9mv^(2)-4v^(2)m)/3 = (5mv^(2))/3`
` :. % " loss in K.E " =((5mv^(2))/3)/(2mv^(2)+mv^(2)) xx100 % `
`=(5mv^(2))/(3xx3mv^(2)) xx100 % `
` = 500/9 % `
` = 55.55 % `
` = 56 % `