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. 0.44
B. 0.5
C. 0.56
D. 0.62

Answer

C

Explanation

Applying conservation of linear momentum just before and just after collision,
`m 2 vhati + 2m v hatj = (m + 2m) vec(v') " "vec(v') = (i+j)`
Hence `|vec(v')| = (2sqrt(2))/(3) v`
Initial kinetic energy of the masses just before collision.
`K_("initial") = 1/2 m (2v)^(2) + 1/2 2 mv^2 = 3mv^2`
Final kinetic energy of the system just after collision.
`K_("final") = 1/2 (3m)((2sqrt(2))/3v)^(2) = 4/3 mv^2`
Loss in kinetic energy, `" " DeltaK = K_("initial") - K_("final") = 3mv^2 - 4/3 mv^2 = 5/3 mv^2`
Hence percentage loss in kinetic energy `" "(DeltaK)/(K)% = (5/3 mv^2)/(3 mv^2) xx 100 = 56%` .