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

Options

A. 44%
B. 50%
C. 56%
D. 62%

Answer

C

Explanation

Due to conservation of momentum along x-axis,
`m*2v=(m+2m)v_1 or, v_1=(2v)/3`
Now along y-axis
`2m*v =(m+2m)vv_2 or, v_2=(2v)/3`
So, velocity of the particle of mass (m+2m=3m) after collision,
`v^.=sqrt(v_1^2+v_2^2)=(2sqrt(2))/(3)v`
Thus , energy loss
`=E_1-E_2`
`={1/2m(2v)^2+1/2(2m)v^2}-1/2*3m((2sqrt(2))/(3))^2`
`=3mv^2-4/3mv^2=5/3mv^2`
Hence, percentage loss in energy
`=(E_1-E_2)/(E_1)xx100=(5/3mv^2)/(3mv^2)xx100=(500)/(9)=55.6~~56`