A smooth wedge of mass M is pushed with an acceleration `a=gtantheta` and a block of mass m is projected down the slant with a velocity v relative to the wedge.
`The horizontal force applied on the wedge is:

Options

A. `16 m//s`
B. `8 m//s`
C. `10 m//s`
D. `15 m//s`

Answer

A

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`