A mass A is released from the top of a frictionless inclined plane 18 m long and reaches the bottom 3 sec later. At the instant when A is released, a second mass B is projected upwards along the plane from the bottom with a certain initial velocity. The mass B travels a distance up the plane, stops and returns to the bottom so that it arrives simultaneously with A. The two masses do not collide. Find the acc. and initial velocity of B. (How far up the inclined plane does B travel?)

Explanation

As the inclined plane is frictionless, the only force along the plane is the component of weight, i.e., `mg sin theta` So acceleration down the plane `a=g sin theta` = constant and so equations of motion can be applied here. So, applying 2nd eqn. of motion, keeping in mind that mass A starts from rest,
`18=0xx3+1/2 axx3^(2)`

Now if B is projected up the plane with a velocity u, it will go up the plane till its velocity becomes zero (i.e, `u=0`) as for it acc. will be `4m//s^(2)` down the plane So from Ist equation of motion, time taken to go up the plane,
`0=u-at_(1) rArrt_(1)=(u//a)` ....(ii)
and from 3 rd equation of motion distance moves up the plane
`0=u^(2)-2as, rArr s=(u^(2)//2a)` ....(iii)
Now the body B will slide down the plane and so time taken by it to slide a distance s down the plane starting from rest at C, from 2 nd equation of motion,
`s=0+1/2at_(2)^(2) rArrt_(2)=sqrt((2s)/a)`
So, `t_(2)=sqrt(2/axxu^(2)/(2a))=u/a=t_(1)`....(iv)
`["as from Eqn."(iii)s=u^(2)/(2a)]`
But it given that body B reaches the ground with A,
i.e, `t_(!)+t_(2)=3 or (2u)/a=3`
`["as from Eqn."(ii) and(iv), t_(1)=t_(2)=u//a]`
or `u=3/2a=3/2xx4=6m//s["as from Eqn." (i) a=4]`
So, the acceleration of B is `4m//s^(2)` down the plane while its velocity is `6m//s` up the plane. [The distance travelled by B up the plane from Eqn. (iii), i.e, `s=(u^(2)//2a)=4.5m]`