Two identical buggies of each of mass `150 kg` move one after the other friction with same velocity `4m/s`. A man of mass `m` rides the rear buggy. At a certain moment, the man jumps into the front buggy with a velocity v relative to his buggy. As a result of this process, real boggy stops. If the sum of kinetic energies of the man and the front buggy just after collision with the from buggy differs from that just before collision by `2700 J` then
the mass `m` of the man is

Options

A. `60 kg`
B. `75 kg`
C. `50 kg`
D. `90 kg`

Answer

C

Explanation

Since rear buggy stops and velocity of man relative to this buggy is `v`, therefore, absolute velocity of man at the jump is `v`. Applying law of conservation of momentum on the system of rear buggy and man,
`mv=(150+m)4`……….i
Kinetic energy of man and front buggy (just before collision) is
`E_(1) =1/2mv^(2)+1/2xx150xx4^(2)`
Let velocity of the front buggy, after collision be `v_(c)`.
Applying law of conservation of momentum on the system of front buggy and man, we get
`mv+150xx4=(m+150)v_(c)`…………ii
Kinetic energy of the system after collision is
`E_(2)=1/2(m+150)v_(c)^(2)`………iii.
But `E_(1)-E_(2)=2700J`
From above three equations `v_(c)=7 m//s`
`m=50 kg `
`v=16 m//s`