A boy of mass `60kg` is standing over a platform of mass `40kg` placed over a smooth horizontal surface. He throws a stone of mass `1kg` with velocity `v=10m//s` at an angle of `45^@` with respect to the ground. Find the displacement of the platform (with boy) on the horizontal surface when the stone lands on the ground. Take `g=10m//s^2`.

Answer

A, C

Explanation

Let v is the horizontal velocity of platform in opposite direction. Then from momentum conservation in opposite direction we have,
`(60+40)v=(1)(10cos45^@)`
`:. v=10/100cos 45^@m//s`
`=10/sqrt2cm//s`
`:.` Displacement of platform `=vt`
`=((v)(2usin45^@))/(g)`
`=(10/sqrt2)=(2xx10xx1/sqrt2)/(10)`
`=10cm`